Engineering Mechanics Statics

🧭 Overview

🧠 One-sentence thesis

This section introduces the foundational mathematical and physical concepts—scalars versus vectors, Newton's three laws, units and conversions, mass versus weight, and basic trigonometry—that are essential for solving engineering statics problems.

📌 Key points (3–5)

• Scalar vs vector: scalars have only magnitude (e.g. distance), while vectors have both magnitude and direction (e.g. displacement).
• Newton's laws: the first law describes inertia (no change without force), the second law relates force to mass and acceleration (F = ma), and the third law states that forces come in equal and opposite pairs.
• Units and conversions: SI (metric) and English (imperial) systems both exist; staying in one system and converting carefully prevents costly errors.
• Mass vs weight confusion: mass (kg or slugs) is intrinsic and constant, while weight (N or lb) is the force of gravity and changes with location (Earth vs moon).
• Trigonometry and geometry: the Pythagorean theorem and SOH-CAH-TOA are tools for finding lengths and angles in right triangles, which appear throughout statics problems.

📐 Scalar and vector quantities

📐 What is a scalar?

Scalar quantity: a physical quantity that can be specified completely by a single number and the appropriate unit.

• Scalars are synonyms of "number."
• Examples: time, mass, distance, length, volume, temperature, energy.
• You can add, subtract, multiply, or divide scalars using ordinary algebra.
• Example: a class period of 50 min minus 10 min equals 40 min; 60 cal plus 200 cal equals 260 cal.

🧭 What is a vector?

Vector quantity: a physical quantity that requires both a number (magnitude) and a direction to be specified completely.

• Examples: displacement, velocity, position, force, torque.
• Vectors are represented by mathematical objects called vectors.
• You can add or subtract vectors, and multiply a vector by a scalar or by another vector, but you cannot divide by a vector.
• Example: "I walked 4 km" (scalar) tells you distance; "I walked 2 km north, then 2 km east" (vector) tells you direction and final position.

🔍 How to distinguish scalar from vector

• Scalar: only "how much" (magnitude).
• Vector: "how much" and "which way" (magnitude + direction).
• Don't confuse: saying "I went for a 4 km walk" does not tell you whether you ended up 4 km away or back where you started—that requires vector information.

⚖️ Newton's three laws

⚖️ First law (law of inertia)

Newton's first law: a body at rest remains at rest, and a body in motion continues at constant velocity, unless acted on by an unbalanced (net) force.

• "Velocity" includes both speed and direction.
• If the net force is zero, velocity does not change.
• Example: a rock at rest stays at rest until a net force (push minus friction) moves it; a space capsule in orbit maintains its velocity until a thruster fires.
• Rotational version: a body maintains its rotational velocity until a net moment (torque) changes it (e.g. a spinning top slows due to friction).

⚖️ Second law (force equals mass times acceleration)

Newton's second law: the net force on a body equals the mass of the body times its acceleration: F = m a (both force and acceleration are vectors).

• The direction of the net force equals the direction of the acceleration.
• The magnitude of the net force equals mass times the magnitude of acceleration.
• Rotational version: net moment equals mass moment of inertia times angular acceleration: M = I α (moment and angular acceleration are vectors).
• Example: the heavier a rock and the more forces act on it, the more it accelerates (or decelerates).

⚖️ Third law (action and reaction)

Newton's third law: for every force one body exerts on another, the second body exerts an equal-magnitude, opposite-direction force back on the first.

• All forces come in pairs; each force in the pair acts on a different body.
• Don't confuse: the two forces in a third-law pair act on separate objects, not on the same object.
• Example: a volleyball resting on the ground has two third-law pairs: (1) gravitational force on the ball and gravitational force on the Earth, (2) normal force on the ball and normal force on the ground.
• Example: a rock pushes on the ground with the same force as the ground pushes on the rock, but in opposite directions.

📏 Units and measurement systems

📏 Why units matter

Physical quantity: defined either by how it is measured or by how it is calculated from other measurements.

Units: standardized values used to express measurements.

• Without standardized units, scientists and engineers cannot compare values meaningfully.
• Two major systems: SI (metric) and English (imperial/customary).
• SI is also called the meter–kilogram–second system; English is also called the foot–pound–second (fps) system.

📏 SI base and derived units

• The International System of Quantities (ISQ) defines seven base quantities and their SI base units:

Base Quantity	SI Base Unit
Length	meter (m)
Mass	kilogram (kg)
Time	second (s)
Electrical current	ampere (A)
Thermodynamic temperature	kelvin (K)
Amount of substance	mole (mol)
Luminous intensity	candela (cd)

• Derived units are algebraic combinations of base units.
• Example: area = length × length = m²; speed = length / time = m/s; density = mass / volume = kg/m³.

📏 Common units in statics

Quantity	SI Unit	English Unit
Length	m, km, mm	ft, mi, in
Mass	kg	slug
Force	N (Newton)	lb (pound)
Pressure	Pa (Pascal) = N/m²	psi = lb/in²

• Stay in one unit system throughout a problem to avoid errors.
• Example: in 1999, a conversion error between N and lb caused NASA's $125 million Mars Orbiter to be lost.

🔄 Unit conversions

🔄 How to convert units

Conversion factor: a ratio expressing how many of one unit equal another unit.

• Write the units you have and the units you want.
• Multiply by a conversion factor so unwanted units cancel.
• Example: convert 80 m to km. There are 1000 m in 1 km, so:
  80 m × (1 km / 1000 m) = 0.080 km
  (the meter units cancel, leaving km).

🔄 Key conversions for statics

Quantity	Conversion
Length	1 m = 3.28 ft; 1 mi = 5,280 ft; 1 ft = 12 in; 2.2 km ≈ 1 mi
Mass	1 slug = 14.6 kg
Force	1 lb = 4.448 N
Pressure	1 psi = 6895 Pa

• Don't confuse: 1 lb = 2.2 kg is a shortcut that only works on Earth (it mixes force and mass).
• Always check which unit system your problem uses and convert at the start.

⚖️ Mass versus weight

⚖️ What is mass?

• Mass is an intrinsic property of an object; it does not change with location.
• Units: kg (SI) or slug (English).
• Mass measures "how much matter" an object contains.

⚖️ What is weight?

Weight: the force exerted by gravity on an object.

• Weight is given by F_g = m g, where g is the gravitational field (a vector pointing toward the center of the Earth).
• Near Earth's surface, g ≈ 9.81 m/s² (SI) or g ≈ 32.2 ft/s² (English).
• Units: N (SI) or lb (English).
• Weight changes with location because g changes (e.g. on the moon, g ≈ 1.62 m/s², so weight is six times less, but mass stays the same).

⚖️ How to distinguish mass from weight

• Mass: intrinsic, constant everywhere (kg or slug).
• Weight: force of gravity, depends on location (N or lb).
• Don't confuse: in everyday English, "I weigh 50 kg" is imprecise; in statics, say "my mass is 50 kg" or "I weigh 490 N."
• Example: an astronaut's mass is the same on Earth and the moon, but their weight on the moon is one-sixth their weight on Earth.

📐 Geometry and trigonometry tools

📐 Pythagorean theorem

Right triangle: a triangle containing a 90° angle.

Pythagorean theorem: in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: c² = a² + b².

• c is the longest side (hypotenuse); a and b are the legs (interchangeable).
• Example: a 6 ft ladder leaning against a wall with its base 2 ft from the wall reaches a height of √(6² − 2²) = √(36 − 4) = √32 ≈ 5.66 ft.
• 3-4-5 triangle (Pythagorean triple): if the legs are 3 and 4, the hypotenuse is 5 (3² + 4² = 9 + 16 = 25 = 5²). Many homework problems use this shortcut.

📐 Trigonometric functions (SOH-CAH-TOA)

Trigonometric functions: relate the angles of a right triangle to the ratios of its sides.

• SOH: sin θ = Opposite / Hypotenuse
• CAH: cos θ = Adjacent / Hypotenuse
• TOA: tan θ = Opposite / Adjacent
• Mnemonic: "cos is close"—the side close to the angle is the adjacent side (cosine).
• Example: a 6 ft ladder at a 60° angle reaches a height of 6 ft × sin(60°) ≈ 6 × 0.866 ≈ 5.20 ft.

📐 When to use these tools

• Use the Pythagorean theorem to find unknown side lengths in right triangles.
• Use SOH-CAH-TOA to find angles or to break a vector into components (e.g. resolving a force into horizontal and vertical parts).
• Don't confuse: the Pythagorean theorem finds lengths; trigonometry finds angles or relates angles to lengths.

🗺️ Cartesian coordinate frames

🗺️ Why coordinate frames matter

• A Cartesian coordinate frame provides a standard language for describing the location of a point or the components of a vector.
• In 2D, a point is described by a pair (x, y); a vector is described by its x-component and y-component.
• The x-component and y-component are the orthogonal projections of the vector onto the x-axis and y-axis.

🗺️ Describing vectors in coordinates

• Any vector A in a plane can be written as the sum of its vector components: A = A_x + A_y.
• Example: instead of saying "go 50 km in the direction 37° north of east," you say "go 40 km east and 30 km north" (using x and y components).
• This section introduces 2D frames; 3D frames (x, y, z) will be covered in later sections.

🧭 Overview

🧠 One-sentence thesis

The excerpt does not contain substantive content about XYZ coordinate frames; it consists of unrelated worked examples (dog speed, cross product torque) and an introductory chapter on particles versus rigid bodies.

📌 Key points (3–5)

• The excerpt does not address the title topic "1.2 XYZ Coordinate Frame."
• It includes two worked examples: converting dog speed to SI units and calculating torque using the cross product.
• The main substantive content is from Chapter 2, which distinguishes particles (mass concentrated at a point) from rigid bodies (mass distributed over a volume).
• Common confusion: particles vs. rigid bodies—particles ignore rotation and size; rigid bodies require analysis of moments, torque, and shape.
• When to use particle analysis: when rotational motion is negligible, forces are concurrent, or the object's size is small compared to the scale of motion.

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⚠️ Content mismatch

⚠️ Missing coordinate frame material

• The title "1.2 XYZ Coordinate Frame" suggests the excerpt should explain Cartesian coordinate systems, axes, unit vectors, or coordinate transformations.
• None of that material appears in the provided text.
• The excerpt jumps from a dog-speed calculation to a torque problem, then to Chapter 2 on particles and rigid bodies.

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🐕 Example 1: Dog speed conversion

🐕 What the example shows

• The worked problem converts an average dog's top speed from miles per hour to meters per second.
• Given: 19 miles per hour (average dog top speed).
• Conversion chain: miles → kilometers → meters, and hours → seconds.
• Result: 8.49 m/s.

🐕 Why it matters

• The problem then compares a calculated speed of 2.36 m/s to this benchmark.
• 2.36 m/s is about one-quarter of the dog's top speed, so it is reasonable for a dog not sprinting or a slower breed.
• This is a reasonableness check using unit conversion and real-world data.

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🔧 Example 2: Torque via cross product

🔧 Problem setup

• Find the torque (moment) M on point A due to force F.
• Force vector: F = (5 i – 2 j + k) N.
• Position vector r from A to the point of application: r = (–1 i + 0.25 j + 5 k) m.
• Platform dimensions: width 1 m, height 0.25 m, length 5 m.

🔧 Approach and result

• Use the cross-product formula: M = r × F.
• The excerpt shows the component-by-component calculation (determinant expansion).
• Result: M = (10.25 i + 26 j + 0.75 k) N·m.
• The answer is verified with a cross-product calculator and deemed reasonable.

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🧱 Particles vs. rigid bodies

🧱 What is a particle?

Particle: a body where all the mass is concentrated at a single point in space.

• Particle analysis considers only forces and translational motion.
• Rotation is not considered.
• Example: a comet's trajectory through space—rotation and moments are unimportant, so treat it as a particle.

🧱 What is a rigid body?

Rigid body: a body with mass distributed throughout a finite volume.

• Rigid-body analysis must account for moments (torque) and rotational motion.
• Example: a crowbar prying something—rotation and moments are key, so it must be analyzed as an extended body.

🧱 When to approximate as a particle

• When rotational motions are negligible compared to translational motions.
• When there is no moment exerted on the body (e.g., a concurrent force system where all forces act through a single point).
• Example: a skycam suspended by cables—all tension and gravity forces act through one point, so it can be treated as a particle.

🧱 Scale and context matter

• Particle scale: the object is small compared to the distance or speed involved.
  • Example: a baseball flying through the air (focus on distance traveled, not spin).
  • Example: a skydiver falling (focus on trajectory, not body rotation).
• Rigid-body scale: the object's size, shape, or rotation is important.
  • Example: a baseball bat swinging (length affects how far the ball travels).
  • Example: a bolt being turned by a wrench (torque depends on wrench length).

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🔑 Key distinctions

🔑 Particle vs. rigid body summary

Aspect	Particle	Rigid body
Mass	Concentrated at a point	Distributed over a volume
Shape/size	Negligible	Must be considered
Analysis includes	Forces, translational motion	Forces, moments, rotational motion
Deformation	Assumed none (even though shape is already negligible)	Assumed none
When to use	Rotation negligible; concurrent forces; small scale	Rotation/torque important; size/length matters

🔑 Don't confuse

• Particle analysis ≠ ignoring all physics of the object—it means the object's size and rotation are unimportant for the question at hand.
• Rigid body ≠ "the object is large"—it means you must account for how forces act at different points on the body, creating moments.
• Example: the same baseball can be a particle (when calculating flight distance) or a rigid body (when calculating spin rate).

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🎯 Practical application

🎯 Examples from the excerpt

• Airplane flying at high speed: particle analysis (size is small compared to flight distance and speed).
• Wrench turning a bolt: rigid-body analysis (torque depends on wrench length; rotation is the focus).
• Football in flight: particle analysis (trajectory matters more than spin, unless you focus on the spin itself).

🎯 Takeaway

• Choose particle analysis when the object's dimensions and rotation are negligible relative to the motion you care about.
• Choose rigid-body analysis when size, shape, or rotation are central to the problem.
• The excerpt notes that "you would have done particle analyses in your high school physics classes"; rigid-body analysis (starting in Chapter 3) adds complexity by including shape and size.

🧭 Overview

🧠 One-sentence thesis

This section demonstrates how equilibrium equations, couples, distributed loads, and lever arms solve real-world statics problems involving everyday objects like couches, steering wheels, bridges, and seesaws.

📌 Key points (3–5)

• What the examples cover: reaction forces on supports, couple moments from hand forces, distributed loads from objects on beams/bridges, and lever-arm ratios for balanced moments.
• Core method: all examples use equilibrium equations (sum of forces = 0, sum of moments = 0) or couple/distributed-load formulas to find unknowns.
• Common confusion: couple vs single-force moment—a couple uses two equal-opposite forces at a distance; a single force creates a moment about a pivot point.
• Why distance matters: in moment calculations, both force magnitude and distance from the pivot affect the result; farther distance can compensate for smaller force.
• Review step: every example checks whether the answer makes physical sense (e.g., positive direction, larger force on the side with more weight).

🛋️ Reaction forces and equilibrium

🛋️ Couch with family members (Example 3.6.1)

Problem: A 5 m couch (weight 120 N) supports a child (30 kg at 1 m from one end), mother (60 kg at 1.5 m), and father (70 kg at 4.5 m). Find the reaction forces at the two roller supports (A and B).

How it works:

1. Convert masses to forces using gravity (9.81 m/s²).
2. Sum moments about point A to solve for reaction at B: moment = force × distance.
3. Sum vertical forces (ΣFy = 0) to solve for reaction at A.

Result:

• N_B = 913 N (larger)
• N_A = 776 N (smaller)

Why N_B is larger: Even though the mother and child together (80 kg) outweigh the father (70 kg), the father sits much closer to support B (only 0.5 m away), while the child is 1 m from A. Distance amplifies the moment effect.

Don't confuse: The magnitude of the force alone does not determine the reaction; the distance from the support is equally important in the moment equation.

🔍 Sign check

• Both reactions are positive (upward), which makes sense—supports push up to balance downward weights.
• Switching the moment center (summing about B instead of A) yields the same answers, confirming correctness.

🔄 Couples and steering wheels

🔄 What is a couple?

Couple: A pair of equal-magnitude, opposite-direction forces separated by a distance, producing a pure rotational moment with no net translational force.

Formula: M = F × d (where d is the perpendicular distance between the two forces).

Key insight: The moment from a couple is the same no matter where you measure it on the object—it does not depend on choosing a pivot point.

🚗 Water valve (Example 3.6.2)

• Diameter = 10 in (0.42 ft radius), force = 7.5 lb on each side.
• Using cross product (vector method): M_A = 3.15 ft·lb, M_B = 3.15 ft·lb → total M = 6.3 ft·lb.
• Shortcut: M = F × d = 7.5 lb × (10 in) = 6.25 ft·lb (slightly more accurate because no rounding).

🚗 Steering wheel with two hands (Example 3.6.5)

• Diameter = 40 cm, force = 115 N per hand.
• Couple moment: M = 115 N × 0.4 m = 46 Nm.
• One-hand comparison: To produce the same 46 Nm with one hand at the edge (radius = 0.2 m, perpendicular), you need F = 46 Nm / 0.2 m = 230 N—exactly double the two-hand force.

Don't confuse: A couple (two hands) vs a single-force moment (one hand). The couple formula is simpler (M = F × d); the single-force moment depends on the angle and distance from the pivot (M = F × r × sin θ).

🚗 Unequal hand forces (Example 3.6.6)

• Right hand = 200 N, left hand = 150 N, diameter = 15 in (0.381 m).
• Right turn: M₁ = –66.675 N·m (negative k-direction).
• Left turn (forces reversed): M₂ = +66.675 N·m (positive k-direction).
• Same magnitude, opposite direction—makes sense because the radius and force magnitudes are the same, only directions change.

📦 Distributed loads

📦 What is a distributed load?

Distributed load: A force spread continuously over a length (measured in force per unit length, e.g., N/m), rather than concentrated at a single point.

Why it matters: Real objects (shelves, bridges, beams) experience weight spread along their length. To analyze them, we replace the distributed load with a single resultant force acting at a specific location.

📐 Key formulas

• Resultant force: F_r = ∫ w(x) dx (area under the load curve).
• Location of resultant: x_r = [∫ x · w(x) dx] / [∫ w(x) dx] (weighted average position).

🛒 Shelf with curved load (Example 3.6.3)

• Load function: w = 4x⁴ + 2 N/m over 1 m length.
• F_r = ∫₀¹ (4x⁴ + 2) dx = 2.8 N.
• x_r = 0.59 m from the wall.
• Why 0.59 m? The load increases toward the right end (x⁴ term), so the resultant is closer to the right than the left—not at the midpoint (0.5 m).

🚗 Cars on a bridge (Example 3.6.8)

• 5 identical cars (each 2250 N) arranged asymmetrically: 3 in one lane, 2 in the other.
• Each car is modeled as a distributed load (rectangle + triangle).
• 3-car side: F_R3 = 6750 N at x̄ = 6.17 m.
• 2-car side: F_R2 = 4500 N at x̄ = 6.33 m (from left).
• Total: F_R5 = 11,250 N at (6.234 m, 3.6 m).
• Why x̄ = 6.234 m? Closer to the 3-car side (6.17 m) because more weight is there; the weighted average shifts toward the heavier side.
• Why ȳ = 3.6 m? Less than halfway across the 8 m width because the 3-car lane (at y = 2 m) has more weight than the 2-car lane (at y = 6 m).

Don't confuse: The resultant location is not the geometric center; it is the weighted center based on force distribution.

⚖️ Lever arms and balanced moments

⚖️ Seesaw problem (Example 3.6.7)

Problem: Two children (35 kg and 50 kg) want to balance a seesaw by applying equal force. What ratio of arm lengths is needed?

How it works:

• For balance, moments must be equal: m₁ · R₁ = m₂ · R₂.
• Gravity cancels out (appears on both sides).
• Solve for ratio: R₁ / R₂ = 50 kg / 35 kg = 10 / 7.

Why it makes sense: The lighter child (35 kg) must sit farther away (R₁ = 10 units) to produce the same moment as the heavier child (50 kg) at R₂ = 7 units. The ratio matches the inverse of their weight ratio.

Key principle: To balance unequal forces, the lighter force must act at a greater distance from the pivot.

🔧 Problem-solving structure

🔧 Six-step method (used in all examples)

1. Problem: State the scenario and what to find.
2. Draw: Sketch + free-body diagram (FBD) showing forces, distances, and coordinate system.
3. Knowns and Unknowns: List given values and what you need to solve for.
4. Approach: Choose the method (equilibrium equations, couple formula, distributed-load integrals).
5. Analysis: Perform calculations step-by-step.
6. Review: Check if the answer makes physical sense (sign, magnitude, direction).

🔧 Common review checks

• Sign check: Are reaction forces positive (upward)? Are moments in the expected direction?
• Magnitude check: Does the larger force/moment appear on the side with more weight or closer distance?
• Symmetry check: If the problem is symmetric, do the results reflect that?
• Alternative method: Can you verify the answer using a different approach (e.g., summing moments about a different point)?

Example: In the water-valve problem, the vector cross-product method gave 6.3 ft·lb, while the shortcut M = F × d gave 6.25 ft·lb—the small difference is due to rounding in unit conversion.

🧮 Calculation techniques

🧮 Converting units

• Always convert to consistent units before calculating (e.g., inches → feet, cm → meters).
• Rounding too early can introduce error; keep extra digits during intermediate steps.

🧮 Vector cross product for moments

• Express position vector r and force F in component form.
• Compute M = r × F using the determinant method (i, j, k components).
• The k-component gives the moment magnitude; sign indicates direction (clockwise or counterclockwise).

Example (water valve):

• r_A = [0.42, 0, 0] ft, F_A = [0, 7.5, 0] lb → M_A = 3.15 k̂ ft·lb.
• r_B = [–0.42, 0, 0] ft, F_B = [0, –7.5, 0] lb → M_B = 3.15 k̂ ft·lb.
• Total M = 6.3 k̂ ft·lb.

🧮 Distributed-load integration

• For a uniform load (constant w), F_r = w × length, located at the midpoint.
• For a triangular load, F_r = ½ × w_max × length, located at ⅓ from the base.
• For a general curve w(x), integrate to find F_r and x_r.

Example (shelf): w = 4x⁴ + 2 N/m is not uniform, so you must integrate; the x⁴ term shifts the resultant toward x = 1 m.

🎯 Key takeaways

🎯 Distance amplifies moment

• A smaller force farther from the pivot can produce the same moment as a larger force closer in.
• This explains why the couch reaction at B (closer to the father) is larger, and why the lighter child needs a longer seesaw arm.

🎯 Couples simplify rotation

• When two equal-opposite forces act, you don't need to pick a pivot point—the couple moment is the same everywhere.
• Steering wheels, valve handles, and jar lids are everyday examples.

🎯 Distributed loads need integration

• Real loads (people on a couch, cars on a bridge, objects on a shelf) are not point forces.
• Replace the distribution with a single resultant force at the weighted-average location.

🎯 Always review your answer

• Check signs (directions), magnitudes (does the heavier side have more force?), and units.
• If possible, solve the same problem a different way to confirm.

🧭 Overview

🧠 One-sentence thesis

Free-body diagrams (FBDs) for rigid bodies require precise placement of force arrows at their actual points of application, and equilibrium equations (including moments) allow engineers to solve for unknown forces and moments in static systems.

📌 Key points (3–5)

• Rigid body vs particle FBDs: Rigid bodies require showing exactly where forces are applied, not just their direction and magnitude.
• System vs part FBDs: A system FBD shows only external forces; part FBDs replace removed objects with internal forces (equal and opposite between parts).
• Equilibrium equations: For 2D rigid bodies, three equations are available (ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0); for 3D, six equations (three force, three moment).
• Common confusion: Internal forces appear on part FBDs but not on system FBDs; they must be equal in magnitude and opposite in direction between connected parts.
• Friction and impending motion: Static friction adjusts up to a maximum (μₛ × N); objects can slip or tip depending on which requires less force.

🎨 Drawing Free-Body Diagrams

🎨 Core steps for a part FBD

A free-body diagram isolates a body from all surroundings and shows all forces and moments acting directly on it.

Four-step process:

1. Draw the shape of the body (floating in space, separated from surfaces).
2. Add a coordinate frame (define positive x and y directions).
3. Replace all surfaces and connections with force arrows (gravity, normal, friction, applied forces).
4. Label each force uniquely and show known magnitudes/directions.

📍 Where forces act on rigid bodies

• Gravity: Acts at the center of mass (sum of all gravitational forces on particles).
• Normal force: Acts perpendicular to the contact surface.
• Friction: Acts parallel to the contact surface, opposing motion.
• Applied forces: Act at the exact point of application (e.g., where a hand pushes).
• Tension in cables: Acts along the cable direction, always pulling.

Example: A book pushed across a table has normal force on the bottom surface, friction along the bottom, gravity at the center of mass, and applied force where the hand touches.

Don't confuse: Particle FBDs (forces can be drawn anywhere on the body) vs rigid body FBDs (arrow head must point to the actual location).

🔗 System FBDs with multiple parts

When analyzing multiple connected objects:

System-level FBD:

• Draw all parts together as one unit.
• Include only external forces (gravity, applied loads, reactions from supports).
• Do NOT include forces between the parts (those are internal to the system).
• Use consistent labels (e.g., subscript 1 for part 1, subscript 2 for part 2).

Part-level FBDs:

• Draw each part separately, floating in space.
• Copy external forces from the system FBD with identical labels.
• Add internal forces where parts were connected (replace the removed part with force arrows).
• Internal forces must have the same label on both part FBDs but opposite directions (Newton's third law).

Example: Two stacked books require three FBDs: (1) system with both books showing only floor normal force, friction, and total gravity; (2) bottom book with top book replaced by downward force; (3) top book with bottom book replaced by upward force (same magnitude, opposite direction).

🔢 Equilibrium Equations

🔢 Available equations for rigid bodies

For a rigid body in static equilibrium, both the sum of forces and the sum of moments must equal zero.

Two dimensions (2D):

• ΣFₓ = 0 (sum of horizontal force components)
• ΣFᵧ = 0 (sum of vertical force components)
• ΣM = 0 (sum of moments about any point)

Three dimensions (3D):

• ΣFₓ = 0, ΣFᵧ = 0, ΣFᵧ = 0
• ΣMₓ = 0, ΣMᵧ = 0, ΣMᵧ = 0

These equations equal zero because the body is static (not accelerating or rotating).

🎯 Solving strategy

1. Draw the FBD with all known and unknown forces.
2. Choose coordinate axes (aligning with some forces simplifies math).
3. Break all forces into x, y (and z) components.
4. Write force equilibrium equations (sum components in each direction = 0).
5. Choose a moment point (picking a point where unknown forces pass through reduces unknowns in that equation).
6. Write moment equation(s).
7. Solve the system of equations for unknowns.

Tip: The number of unknowns you can solve equals the number of independent equations available.

Example: A beam supported at two points with people standing on it—vertical force equilibrium gives one equation relating the reaction forces; taking moments about one support eliminates that reaction from the equation, allowing you to solve for the other reaction.

🧱 External Forces on Rigid Bodies

🧱 Types of external forces

External forces include gravitational, normal, frictional, spring, and applied forces (including tension and reactions).

Force type	Formula	Direction	Notes
Gravity	Fₘ = mg	Downward	Acts at center of mass
Normal	N = mg cos θ (on incline)	Perpendicular to surface	Balances perpendicular component of weight
Friction	Fₓ = μN	Parallel to surface	Opposes motion or tendency to move
Spring	Fₛ = -kx	Along spring axis	Opposes displacement from relaxed position
Tension	T (in cable)	Along cable	Always pulls, never pushes

⚖️ Normal force details

• On a horizontal surface: N = mg (equal and opposite to weight).
• On an incline at angle θ: N = mg cos θ (balances the perpendicular component of weight).
• The surface deforms slightly (like a trampoline) until the restoring force equals the load.

Don't confuse: Normal force magnitude with weight—on an incline, normal force is less than the object's weight.

🪢 Tension in flexible connectors

• Tension is uniform throughout a massless rope/cable (if no friction at bends).
• Direction changes around corners but magnitude stays constant.
• To create large tension with small perpendicular force: T = F⊥ / (2 sin θ), where θ is the angle from horizontal—as θ approaches zero, tension becomes very large.

Example: Pulling a car from mud with a chain—push perpendicular to a taut chain to create large tension.

🔄 Friction fundamentals

• Static friction: Exists before slipping; adjusts to match applied force up to maximum Fₛ,max = μₛN.
• Kinetic friction: Exists during sliding; Fₖ = μₖN (usually μₖ < μₛ).
• Always opposes relative motion between surfaces.

⚠️ Friction and Impending Motion

⚠️ Static vs kinetic friction

Dry friction opposes surfaces sliding; static friction adjusts up to a maximum, then kinetic friction takes over once sliding begins.

Before slipping (static):

• Friction force equals the applied force (up to a limit).
• Maximum static friction: Fₛ,max = μₛ × N.
• The point just before slipping is called impending motion.

During slipping (kinetic):

• Friction force is constant: Fₖ = μₖ × N.
• Kinetic coefficient μₖ is almost always less than static coefficient μₛ.

Example: Pushing a box—friction matches your push until you push hard enough to overcome μₛN, then the box suddenly slides more easily because μₖN < μₛN.

🔀 Slipping vs tipping

When a force is applied to an object, it will either slip or tip, whichever requires less force.

To determine which occurs first:

1. Calculate the force required to cause slipping: Fₛₗᵢₚ = μₛN.
2. Calculate the force required to cause tipping (take moments about the tipping edge; normal force shifts to that edge at impending tip).
3. Compare: the smaller force happens first.

Slipping: Occurs when applied force exceeds maximum static friction.

Tipping: Occurs when the moment from applied force and friction exceeds the moment from gravity and normal force; at impending tip, the normal force acts at the edge of the base.

Example: Pushing a tall box—if the box is tall and narrow, tipping may occur before slipping; if short and wide, slipping occurs first.

Don't confuse: The normal force is distributed across the bottom surface but can be modeled as a single equivalent point load that shifts position to resist rotation.

🔗 Connections and Reactions

🔗 Types of supports and their reactions

Different connection types prevent different motions and thus generate different reaction forces/moments:

Support type	Motions prevented	Reactions generated
Roller	Vertical motion	Normal force (perpendicular to surface)
Pin joint	Motion in all directions	Normal forces in x and y directions
Fixed connection	Motion and rotation	Normal forces in x and y, plus reaction moment

Example: A roller allows rotation and sliding along the surface but prevents motion perpendicular to the surface (one reaction force). A pin prevents all translation but allows rotation (two reaction forces). A fixed support prevents everything (two forces and one moment).

🎯 Choosing a moment point strategically

• Any point can be chosen for the moment equation.
• Forces passing through the chosen point contribute zero moment about that point.
• Strategy: Choose a point where multiple unknown forces meet to reduce the number of unknowns in the moment equation.

Example: For a beam with unknown reactions at both ends, taking moments about one end eliminates that reaction from the equation, leaving only one unknown reaction in the moment equation.

🧭 Overview

🧠 One-sentence thesis

External forces—including gravity, normal force, friction, spring force, and applied forces—must be quantified and balanced to keep objects stationary or predict their behavior in statics problems.

📌 Key points (3–5)

• What external forces include: gravitational force (weight), normal force, frictional force, spring force, and applied forces (reaction forces, tension, moments).
• Normal force adjusts to context: on horizontal surfaces it equals weight; on inclines it equals the perpendicular component of weight.
• Tension is a pulling force: flexible connectors (ropes, cables) can only pull along their length, never push; tension is uniform along a massless rope.
• Common confusion: normal force vs weight—normal force is not always equal to the full weight (e.g., on inclines or when other vertical forces act).
• Why it matters: calculating each force correctly allows us to solve for unknown forces and ensure equilibrium (net force = zero).

🌍 Types of external forces

⚖️ Gravitational force (weight)

Gravitational force (weight): the pervasive force pulling objects downward, calculated as F_g = mg (mass times gravitational acceleration).

• Acts at all times and must be counteracted to prevent falling.
• In statics, weight is the "load" that other forces must balance.
• Example: a bag of dog food on a table—its weight pulls down; the table must push up with equal magnitude to keep it stationary.

🔼 Normal force

Normal force: the perpendicular (normal-to-surface) force exerted by a surface to support a load resting on it.

• The word "normal" means perpendicular to the contact surface.
• On a horizontal surface: the normal force equals the weight in magnitude and is opposite in direction: N = mg.
• On an incline: the normal force equals only the perpendicular component of weight: N = mg cos θ (where θ is the angle of the incline to the horizontal).
• The surface deforms slightly (like a trampoline or table sagging) until the restoring force equals the load's weight.
• Don't confuse: normal force is not always equal to the full weight—it depends on the geometry and other forces.

🪢 Tension

Tension: a pulling force transmitted along the length of a flexible connector (rope, cable, chain, tendon).

• Flexible connectors can only pull, never push ("you can't push a rope").
• Tension acts parallel to the connector's length and pulls outward at both ends.
• For a stationary mass: tension equals the weight of the supported mass: T = mg.
• Uniform along the rope: if the rope is massless and there is no friction, tension is the same at all points along the rope.
• Example: holding a 5.00-kg mass on a rope—tension T = (5.00 kg)(9.80 m/s²) = 49.0 N.

🔀 Tension around corners and perpendicular forces

• If there is no friction, tension magnitude is unchanged when a flexible connector bends around a corner; only its direction changes.
• Creating large tension: applying a perpendicular force to a taut connector creates tension T = F_perp / (2 sin θ), where θ is the angle between the horizontal and the bent connector.
• As θ approaches zero (connector becomes more horizontal), tension becomes very large—even a small weight causes sagging because horizontal alignment would require infinite tension.
• Example: pulling a car out of mud with a chain—pushing perpendicular to a nearly straight chain creates large tension.

🧲 Other external forces (brief mention)

• Friction: a resistive force opposing motion (details not fully covered in this excerpt).
• Spring force: F_S = -kx (force proportional to displacement, with negative sign indicating restoring direction).
• Applied forces: measured or calculated forces such as reaction forces, moments from motors, etc.

🧮 Calculating forces on inclines

📐 Weight components on an incline

• When an object rests on an incline at angle θ to the horizontal, gravity splits into two components:
  • Parallel to the plane: w_x = mg sin θ (causes acceleration down the incline if unopposed).
  • Perpendicular to the plane: w_y = mg cos θ (balanced by the normal force).
• The angle θ of the incline is the same as the angle between the full weight vector and the perpendicular component.
• Use trigonometry: cos θ = w_y / w and sin θ = w_x / w.

🔍 Normal force on an incline

• The normal force equals the perpendicular component of weight: N = mg cos θ.
• Don't memorize blindly—draw the right-angle triangle formed by the weight vectors and use trigonometry to reason through the components.
• Example: on a steeper incline (larger θ), cos θ is smaller, so the normal force is less than the full weight.

🔗 Key principles for statics

⚖️ Equilibrium condition

• For a stationary object, net external force = zero.
• Example: a mass on a rope—tension T upward and weight w downward must balance: T - w = 0, so T = w = mg.

🧩 Fundamental forces background

• Normal, friction, spring, and applied forces are all types of electromagnetic forces (charged/neutral particles attracting or repelling).
• Gravitational force is one of the four fundamental forces of nature (gravitational, electromagnetic, weak nuclear, strong nuclear).
• This background is beyond the scope of statics, but it explains why objects don't fall through surfaces (electron repulsion) and why they are pulled down (gravity).

📋 Summary of force equations

Force type	Equation	Notes
Gravity	F_g = mg	Always acts downward
Normal (horizontal)	N = mg	Perpendicular to surface
Normal (incline)	N = mg cos θ	Perpendicular component only
Friction	F_f = μN	Resistive, opposes motion
Spring	F_S = -kx	Restoring force
Tension	T = mg (for stationary mass)	Pulling force along connector
Tension (perpendicular force)	T = F_perp / (2 sin θ)	Large when θ is small

🧭 Overview

🧠 One-sentence thesis

Tension in flexible connectors (ropes, cables, chains) transmits force uniformly along their length and can be amplified by applying perpendicular forces at small angles, while free-body diagrams isolate objects from their surroundings to model all external forces acting on them.

📌 Key points (3–5)

• Tension uniformity: Tension is equal at all points along a frictionless flexible connector; determining tension at one location gives you tension everywhere.
• Perpendicular force amplification: Applying a force perpendicular to a taut connector creates large tension, especially when the angle θ approaches zero.
• Free-body diagram purpose: FBDs "free" a body from all surroundings and replace surfaces/contacts with force arrows to simplify analysis.
• Common confusion: In rigid-body FBDs, force location matters—arrows must point to the exact application point, not just anywhere on the object.
• System vs part FBDs: A system FBD shows only external forces on combined objects; part FBDs replace removed objects with interaction forces.

🔗 Tension in flexible connectors

🔗 Uniform tension property

Tension anywhere in a frictionless rope between two points is equal throughout.

• Once you determine tension at one location, you know it at all locations along the rope.
• This applies to ropes, cables, wires, and chains when friction is absent.
• Example: If a rope connects a hand to a mass, the tension measured near the hand equals the tension near the mass.

🔄 Tension around corners

• Flexible connectors transmit forces around corners (hospital traction, tendons, bicycle brake cables).
• If there is no friction: tension magnitude is undiminished; only direction changes.
• The tension is always parallel to the flexible connector at every point.

⚡ Creating large tension with perpendicular force

The excerpt gives the relationship:

T = F⊥ / (2 sin θ)

where:

• T is tension in the connector
• F⊥ is the perpendicular force applied at the middle
• θ is the angle between the horizontal and the bent connector

Why tension becomes large:

• As θ approaches zero (connector becomes more horizontal), sin θ becomes very small.
• Dividing by a very small number produces a very large tension.
• An infinite tension would result if the connector were perfectly horizontal (θ = 0, sin θ = 0).
• Even the small weight of the connector itself causes it to sag, because keeping it perfectly horizontal would require infinite tension.

Example: Pulling a car out of mud with a chain—each time the car moves forward, the chain is tightened to keep it as straight as possible; since θ is small, T becomes large.

🔁 Don't confuse: tension direction vs magnitude

• Tension magnitude stays the same along a frictionless connector.
• Tension direction changes when the connector bends around a corner.
• The force is always parallel to the connector at each point.

🎨 Free-body diagrams for rigid bodies

🎨 What is a free-body diagram

A free-body diagram (FBD) is a tool that "frees" the body from all other objects and surfaces so it can be studied in isolation, with all forces and moments acting on it drawn in.

• The body is separated from surroundings.
• All external forces and moments are drawn as arrows.
• Key dimensions and angles are labeled.
• This simplified diagram allows easier writing of equilibrium equations or equations of motion.

🆚 Particle FBD vs rigid-body FBD

Aspect	Particle FBD	Rigid-body FBD
Force location	Not critical (particle is a point)	Must be precise—arrow head points to exact application point
Example	Force acts "on" the particle	Force from your hand acts on my hand, not at my center of mass

Common confusion: In rigid-body FBDs, where the force is applied matters; you cannot just draw forces anywhere on the object.

📐 Rules for constructing a part FBD

The excerpt lists these steps:

1. Draw the shape of the body being analyzed, separated from all surroundings.
2. Add a coordinate frame (which way is positive x and positive y?).
3. Replace surfaces with force arrows (floor, hand, objects touching it become arrows).
4. Point forces in the correct direction (arrow head points to where force acts).
5. Use unique names for each force (different labels for each).
6. Label key dimensions and angles on the diagram.

🧲 Common external forces in FBDs

🌍 Gravitational force

• Acts on every particle in an object; summed at the center of mass to avoid drawing millions of arrows.
• Always points downward toward the center of the Earth.
• Magnitude: weight in pounds (English system) or 9.81 × mass in kg (metric system, resulting in Newtons).
• We concentrate gravity at the center of mass because we know the total mass and its location (often the geometric center).

🔲 Normal forces (reaction forces)

Normal forces: exerted by objects in direct contact with the body, preventing the two objects from occupying the same space.

• Always act perpendicular to the surfaces in contact (hence "normal").
• Only objects in direct contact can exert normal forces.
• Example: A book on a table has a normal force on the bottom of the book pointing upward.

Joints and connections:

• Joints cause reaction forces or moments for each type of motion or rotation they prevent.
• Roller: allows rotation and movement along surface; normal force in y direction prevents vertical motion.
• Pin joint: allows rotation; normal forces in x and y directions prevent motion in all directions.
• Fixed connection: normal forces prevent motion in all directions; reaction moment prevents rotation.

🛑 Friction forces

• Exerted by objects in direct contact, resisting sliding between surfaces.
• Always act parallel to the plane between the two surfaces (a shear force).
• Always oppose motion (very important in dynamics).
• The excerpt distinguishes:
  • Smooth surfaces: assume no friction force.
  • Rough surfaces: assume bodies will not slide relative to one another; friction force is just large enough to prevent sliding.

🪢 Tension in cables

Tension forces: exerted by cables, wires, or ropes attached to the body, always in the direction of the cable.

• These forces always pull on the body.
• Cables, ropes, and flexible tethers cannot be used for pushing.
• The tension force acts along the direction of the cable.

🌀 Spring force

Spring force: a restoring force proportional to and in the opposite direction of the displacement from the relaxed position.

Hooke's law (in words):

• Force = –k × displacement
• k is the spring stiffness constant.
• The force is parallel to the spring axis.
• The sense of the force is opposite to the displacement direction.
• Displacement must be measured from the relaxed position (x = 0 when spring is relaxed).

Why negative sign:

• The force acts in the opposite direction of the motion traveled.
• In application, you set the direction to match whether the spring is pushing or pulling.

Three cases:

1. Spring relaxed: exerts no force.
2. Spring compressed by displacement: exerts restoring force in the opposite direction.
3. Spring stretched by displacement: exerts restoring force in the opposite direction.

🎯 Other forces

• Pressure from fluids
• Magnetic forces
• Applied forces and moments (distributed loads, motors, pushing/pulling on an object)

📦 System free-body diagrams

📦 What is a system FBD

• Composed of multiple parts, so you can consider multiple "levels":
  • System level: all objects on the same FBD, showing only external forces.
  • Part level: individual FBD for each part, replacing removed parts with interaction forces.

Why use system FBDs:

• Helpful when you have more unknowns than equations using equilibrium equations.
• Splitting the system into individual parts provides more information.

🔀 External vs internal forces

• System FBD: look at parts combined together; add only external forces (gravity, applied, normal, frictional, spring).
• Part FBD: look at each part separately; include interaction between objects by replacing a part with forces (generally 2 forces: vertical and horizontal).

Example: Two books stacked on top of each other require 3 FBDs:

1. System-level FBD with both books.
2. Part FBD for the bottom book, with the top book replaced by force arrows.
3. Part FBD for the top book, with the bottom book replaced by force arrows.

🏷️ Labeling in system FBDs

• Use unique, consistent labels (a letter or number per part).
• The system should be drawn floating in space, separated from surroundings.

⚠️ Don't confuse: external vs internal forces

• External forces: act on the system from outside (gravity, applied forces, normal forces from surfaces not part of the system).
• Internal forces: forces between parts within the system; these are not included in the system-level FBD but are included in part FBDs as interaction forces.

🛠️ Practical tips for FBDs

📝 Book on a table example

To model a book being pushed across a table, apply these forces:

• Normal force: on the bottom of the book (perpendicular to table surface).
• Friction force: along the bottom surface between book and table (parallel to surface, opposing motion).
• Gravitational force: at the center of mass (downward).
• Applied force: at the point of application, such as your hand pushing on the book.

If the book is pulled by a string instead:

• The image is the same, but the applied force and friction force change direction (because friction always opposes motion).

🪜 Ladder example

The excerpt shows a ladder supporting a person:

• The ladder is separated from all other objects.
• All forces acting on the ladder are drawn in with key dimensions and angles.
• Assuming friction only at the base, the FBD shows normal and friction forces at the base, and other relevant forces.

🧭 Overview

🧠 One-sentence thesis

Free-body diagrams (FBDs) simplify complex mechanical problems by representing forces and moments as arrows, and equilibrium equations allow us to solve for unknown forces by requiring that all forces and moments sum to zero for static bodies.

📌 Key points (3–5)

• What FBDs do: represent forces acting on a body as arrows, isolating the object from its surroundings to analyze forces systematically.
• System vs part FBDs: system-level diagrams show only external forces on combined objects; part-level diagrams add internal forces between objects (equal and opposite by Newton's laws).
• Equilibrium equations: for static rigid bodies, the sum of forces in each direction and the sum of moments about any axis must equal zero.
• Common confusion: external vs internal forces—internal forces (interactions between parts) appear only on part FBDs, not on system FBDs; they must have matching labels but opposite directions.
• Why it matters: FBDs and equilibrium equations together allow calculation of unknown forces and moments in structures and mechanical systems.

🎨 What free-body diagrams are

🎨 Purpose and representation

A free-body diagram isolates a body and shows all forces acting on it as arrows.

• The body is drawn "floating in space" with no surfaces or other objects touching it.
• Every force is replaced by an arrow showing direction and point of application.
• Labels indicate known magnitudes or variable names for unknowns.
• A coordinate frame must be included to define directions.

🔧 Common force types

The excerpt lists typical forces to include:

Force type	Direction	Notes
Tension (cables, ropes)	Along the cable	Always pulling, never pushing
Gravity	Downward	Acts at center of mass
Normal	Perpendicular to surface	Contact force
Friction	Parallel to surface	Opposes sliding
Applied forces	Varies	External pushes or pulls
Other	Varies	Pressure, spring, magnetic forces may exist

• Don't confuse: tension forces cannot push; flexible tethers only pull.

📐 Final labeling step

After identifying and drawing forces, label key dimensions and angles on the diagram (distances, heights, separations between points).

🧩 System vs part FBDs

🧩 When to use system FBDs

• System FBD: treats multiple objects as one combined unit.
• Shows only external forces (gravity, applied, normal, friction, spring).
• Do not include internal forces (interactions between the parts).
• Useful when you have enough equations to solve for unknowns at the system level.

Example: Two books stacked together—draw one FBD showing the combined weight and the normal force from the floor, ignoring the contact force between the books.

🔀 When to use part FBDs

• Part FBD: draws each object separately.
• Now you must include internal forces—replace the other object with force arrows (typically horizontal and vertical components).
• Copy external forces from the system FBD with identical labels and directions.
• Internal forces between two parts must have the same label (same magnitude) but opposite directions (Newton's third law: equal and opposite).

Example: For the two-book stack, draw three FBDs:

1. System-level (both books together, external forces only).
2. Bottom book alone (external forces + forces from top book pressing down).
3. Top book alone (external forces + forces from bottom book pushing up—same magnitude as in FBD 2, opposite direction).

🏷️ Labeling rules

• Use unique labels (letters or numbers) for each part (e.g., A and B, or 1 and 2).
• External force labels must be identical between system and part FBDs.
• Internal force labels must be identical between the two part FBDs, but arrows point in opposite directions.
• Don't confuse: changing labels between system and part FBDs makes equations unsolvable.

📍 Center of mass tip

• You may combine gravitational forces into one system-level force at the system's center of mass, or
• Model gravity as separate forces at each part's center of mass (better if you need to separate objects for calculations).

⚖️ Equilibrium equations for rigid bodies

⚖️ What equilibrium means

For a rigid body in static equilibrium (non-deformable, forces not concurrent), the sum of forces and the sum of moments acting on the body must equal zero.

• "Static" means the body is not accelerating (not moving or moving at constant velocity).
• Rigid bodies can have forces applied at different points, so moments (rotational effects) must be considered.

📊 Equations in two dimensions

For a 2D problem, three equilibrium equations:

• Sum of force components in x direction equals zero.
• Sum of force components in y direction equals zero.
• Sum of moments about the z-axis (clockwise or counterclockwise rotation) equals zero.

In symbols (rewritten in words):

• Sum of F_x = 0
• Sum of F_y = 0
• Sum of M_z = 0

📦 Equations in three dimensions

For a 3D problem, six equilibrium equations:

• Sum of force components in x, y, and z directions each equals zero (three equations).
• Sum of moment components about x, y, and z axes each equals zero (three more equations).

In symbols (rewritten in words):

• Sum of F_x = 0, Sum of F_y = 0, Sum of F_z = 0
• Sum of M_x = 0, Sum of M_y = 0, Sum of M_z = 0

🔄 Why moments add equations

• Particles (point masses) have only force equations.
• Rigid bodies have shape and size, so forces at different points create rotation (moments).
• Adding moment equations allows solving for more unknowns compared to particle problems.

🛠️ How to find equilibrium equations

🛠️ Step-by-step process

1. Draw the free-body diagram: show all force vectors, known values, and variable names for unknowns.
2. Choose coordinate axes: x, y, (and z for 3D)—axes must be perpendicular but need not be horizontal/vertical; aligning axes with some force vectors simplifies analysis.
3. Break forces into components: resolve each force vector into components along x, y, and z directions.
4. Write force equations: sum of components in each direction equals zero.
5. Choose a moment point: pick a point to take moments about—any point works, but choosing a point through which unknown forces pass eliminates those unknowns from the moment equation (forces through a point exert no moment about that point).
6. Write moment equations: sum the moments exerted by each force about the chosen point and axis, set equal to zero.
7. Solve the system: use the equilibrium equations to solve for unknowns—number of solvable unknowns equals number of independent equations.

🎯 Choosing the moment point strategically

• Forces passing through the chosen point contribute zero moment about that point.
• Example: if two unknown reaction forces act at point A, taking moments about point A eliminates both from the moment equation, reducing the number of unknowns.

📐 Example scenario

The excerpt describes a beam with 200 lbs pushing down, supported by two reaction forces pushing up:

• In the vertical (y) direction: the 200 lbs downward must be balanced by the sum of the two upward reaction forces.
• If the load is centered, the two reactions are equal.
• If the load is off-center, use the sum of moments equation and the distances to determine the size of each reaction force.

Example: A car with mass 1500 lbs, center of mass 4 ft behind the front wheels—use equilibrium equations to find normal forces on front and back wheels.

🔍 Common confusions and tips

🔍 External vs internal forces

• External forces: act on the system from outside (gravity, applied loads, contact with surfaces).
• Internal forces: interactions between parts within the system (contact forces between stacked objects).
• On a system FBD, show only external forces.
• On part FBDs, show both external (copied from system FBD) and internal forces (equal and opposite between the two parts).

🔍 Label consistency

• Use the same label for the same force across system and part FBDs.
• Use the same label for internal force pairs, but reverse the arrow direction.
• Don't confuse: inventing new labels for the same force creates unsolvable equations.

🔍 Coordinate frame consistency

• Include a coordinate frame on every FBD (system and each part).
• If you rotate an object, rotate the coordinate frame consistently to avoid sign errors.

🔍 Moment direction

• In 2D, moments are about the z-axis (into or out of the page).
• In 3D, moments can be about x, y, or z axes.
• Sign convention: choose clockwise or counterclockwise as positive and stick to it.

🧪 Practical application

🧪 Why FBDs and equilibrium matter

• FBDs are used in "step 2" of nearly every mechanics problem.
• Equilibrium equations allow calculation of unknown reaction forces, internal forces, and moments in structures.
• Example applications: determining support reactions in beams, analyzing forces in trusses, calculating loads on wheels of a vehicle.

🧪 Looking ahead

• The excerpt notes that FBDs and equilibrium equations will be used extensively in later chapters (Chapters 5 and 6).
• In dynamics (moving bodies), the right side of equilibrium equations will equal mass times acceleration (for translation) and moment of inertia times angular acceleration (for rotation), rather than zero.

🧭 Overview

🧠 One-sentence thesis

This section demonstrates how to apply equilibrium equations and friction principles to solve real-world statics problems involving forces, moments, and the choice between slipping and tipping.

📌 Key points (3–5)

• Purpose of examples: Apply equilibrium equations (sum of forces = 0, sum of moments = 0) to determine unknown forces and locations in multi-body systems.
• Friction calculations: Use static and kinetic friction coefficients with normal forces to find required forces for motion or equilibrium.
• Slipping vs tipping analysis: Compare the force needed to overcome friction (slipping) against the force needed to rotate the body (tipping) to predict which occurs first.
• Common confusion: Normal force is not always equal to weight—it can change based on applied forces in the y-direction and can redistribute to resist rotation.
• Real-world scenarios: Bridge loading, inclined planes, and objects against walls all require systematic free-body diagrams and equilibrium analysis.

🌉 Multi-body equilibrium problems

🌉 Bridge loading scenario (Example 4.5.1)

Setup: Three people stand on an 11-foot bridge supported by rollers at both ends; the left support can withstand a maximum of 225 lbs.

Method:

• Draw a free-body diagram showing all downward forces (people's weights) and upward reaction forces at supports.
• Use sum of forces in y-direction to find the reaction force at the right support.
• Use sum of moments about one support to relate positions and forces.
• Solve for the unknown position where the third person can stand without exceeding the left support's limit.

Key insight: The moment equation incorporates distances, so the location of forces matters, not just their magnitude.

Don't confuse: The reaction forces are not simply half the total weight—they depend on where the loads are positioned along the bridge.

📦 Free-body diagram construction (Example 4.5.2)

Scenario: A box on a 15° inclined plane is pushed downward with 20 N while in static equilibrium.

Forces to include:

• Weight/gravitational force (downward)
• Normal force (perpendicular to the surface)
• Friction force (up the plane, opposing the push)
• Applied force (20 N down the plane)

Why all four: Static equilibrium means the net force is zero, so friction must balance the applied force and the component of weight along the plane.

🔧 Friction force calculations

🔧 Finding normal force with angled applied force (Example 4.5.3a)

Setup: A 12 kg box is pushed along level ground with 150 N at 30° above horizontal.

Key steps:

1. Calculate gravitational force: F_g = mass × gravity = 12 kg × 9.81 m/s² = 117.72 N
2. Resolve the applied force into components: vertical component = 150 N × sin(30°) = 75 N
3. Apply equilibrium in y-direction: F_N = F_g + vertical component of applied force
4. Result: F_N = 117.72 N + 75 N = 192.7 N

Why it's larger than weight: The downward component of the angled push increases the normal force beyond just the weight.

🔧 Coefficient of friction from equilibrium (Example 4.5.3b)

Method:

• In x-direction: friction force = horizontal component of applied force = 150 N × cos(30°)
• Friction equation: F_f = μ × F_N
• Solve for μ: μ = (150 N × cos(30°)) / 192.7 N = 0.67

Reasonableness check: The excerpt notes that 0.67 is reasonable for wood-on-wood contact (typical range 0.5–0.7).

🧱 Vertical surface friction (Example 4.5.4)

Scenario: A person pushes a 5 kg brick against a vertical wall to prevent it from sliding down; coefficient of static friction = 0.49.

Logic chain:

1. Friction must support the weight: F_f = F_g = 49.05 N (upward)
2. Friction depends on normal force: F_f = μ × F_N
3. Normal force equals the applied horizontal push: F_N = F_A
4. Solve: F_A = F_f / μ = 49.05 N / 0.49 = 100.1 N

Distributed load: If the hand is 16 cm long, intensity w = 100.1 N / 0.16 m = 625.64 N/m

Key insight: On a vertical surface, friction acts upward to oppose gravity, and the normal force comes entirely from the horizontal push.

📐 Slipping versus tipping analysis

📐 Comparing failure modes

The question: When a pushing force increases on a box, will it slide (slip) or rotate (tip) first?

Two calculations needed:

Failure mode	What to calculate	Condition
Slipping	Force to overcome static friction	Pushing force > μ_s × F_N
Tipping	Force to create sufficient moment	Moment from push + friction > moment from weight + normal force

Decision rule: Whichever requires the smaller pushing force will occur first.

📐 Tipping mechanism

How normal force redistributes:

• At rest: normal force is uniformly distributed across the bottom surface.
• As pushing force increases: the distributed normal force shifts toward the edge, creating a counter-moment.
• At impending tip: the equivalent point load of the normal force reaches the very edge of the box.
• Beyond this point: the normal force cannot shift further, so the box rotates.

Example conclusion (Example 5): If the calculated pushing force to tip (e.g., 50 N) is less than the force to slip (e.g., 70 N), the box will tip first.

Don't confuse: The normal force magnitude can stay the same while its location shifts—it's the moment arm that changes, not necessarily the force size.

✅ Problem-solving workflow

✅ Standard approach across all examples

1. Draw: Create a sketch of the real scenario, then a free-body diagram showing all forces.
2. List knowns and unknowns: Identify given values (masses, angles, coefficients) and what you need to find.
3. Choose equations: Equilibrium equations (sum F = 0, sum M = 0), friction equations (F_f = μ F_N), trigonometry for components.
4. Analyze: Solve the equations systematically, often finding one unknown to use in the next equation.
5. Review: Check if the answer makes physical sense (e.g., is the coefficient between 0 and 1? Is the normal force reasonable?).

✅ Common techniques

• Moments about a point: Choose a pivot point (often a support) to eliminate unknown forces from the moment equation.
• Component resolution: Break angled forces into x and y components using sine and cosine.
• Sequential solving: Use sum of forces in one direction to find a reaction, then use that result in the moment equation.

Example: In the bridge problem, finding the right support reaction first (from sum of forces) allows you to then use sum of moments to find the unknown position.

🧭 Overview

🧠 One-sentence thesis

The distinction between particles and rigid bodies determines whether an analysis must account for rotation and moments (rigid bodies) or can focus solely on forces and translational motion (particles).

📌 Key points (3–5)

• Particles: all mass concentrated at a single point; analysis considers only forces and translational motion, not rotation.
• Rigid bodies: mass distributed throughout a finite volume; analysis must include moments and rotational motions.
• When to approximate as a particle: when rotational motions are negligible compared to translational motions, or when forces act through a single point (concurrent force system).
• Common confusion: particles vs rigid bodies—particles have mass only; rigid bodies have mass and shape/size; neither deforms.
• Scale matters: particle analysis suits large-scale motion where object size is small relative to distance or speed; rigid body analysis is needed when length, size, or rotation must be considered.

🔍 Core definitions

🔍 What is a particle?

Particle: a body where all the mass is concentrated at a single point in space.

• Particle analysis accounts for:
  • Forces acting on the body
  • Translational motion
• Particle analysis does not consider rotation.
• Example: A comet's trajectory through space—rotation and moments are unimportant, so it is treated as a particle.

🔍 What is a rigid body?

Rigid body: a body with mass distributed throughout a finite volume.

• Rigid body analysis must account for:
  • Moments (torques)
  • Rotational motions
• More complex than particle analysis.
• Example: A crowbar—rotation and moments are key to understanding how it works, so it must be analyzed as an extended body.

🔍 Neither deforms

• Both particles and rigid bodies are assumed to be non-deforming (they do not change shape).
• Particles: already assume shape is negligible, so "no deformation" is implicit.
• Rigid bodies: shape and size matter, but the body does not deform during analysis.

🎯 When to use each approach

🎯 Approximating real bodies as particles

In reality, no bodies are truly particles, but some can be approximated as particles to simplify analysis.

When to treat a body as a particle:

• Rotational motions are negligible compared to translational motions.
• The system involves no moment exerted on the body.
• Forces act through a single point (concurrent force system).

Example: A skycam—gravitational and tension forces all act through a single point, making it a concurrent force system that can be analyzed as a particle.

🎯 When rigid body analysis is required

Rigid body analysis is necessary when:

• The length or size of the object must be considered.
• Rotation is involved.
• Torque calculations are needed.

Example: Turning a bolt with a wrench—the length of the wrench and the rotation of the bolt require rigid body analysis.

🏀 Scale and context

🏀 Particle analysis: large-scale motion

• Suitable when the object is small in comparison to the distance traveled or the speed involved.
• Typical examples from the excerpt:
  • A sky diver falling through the sky
  • A football flying through the air
  • An airplane flying at high speed through the air
  • A baseball traveling through the air (focus on distance, not spin)

Why it works: The object's size is negligible relative to the motion being studied, so rotation and shape can be ignored.

🏀 Rigid body analysis: size and rotation matter

• Required when the object's dimensions or rotation are central to the problem.
• Typical examples from the excerpt:
  • A bat swinging to hit a ball (length of the bat affects how far the ball travels)
  • Calculating the spin on a ball as it flies (focus on rotation)
  • A bolt being turned (rotation is the key motion)

Why it's needed: The object's shape, size, or rotational behavior directly affects the outcome.

🧩 Conceptual summary

🧩 Mass vs mass + shape

Aspect	Particle	Rigid Body
What it has	Mass	Mass + shape + size
Motion considered	Translational only	Translational + rotational
Forces/moments	Forces only	Forces + moments (torques)
Deformation	Assumed non-deforming	Assumed non-deforming

One way to think of it: Particles have mass; rigid bodies have mass and shape.

🧩 Baseball example breakdown

The excerpt uses baseball to illustrate both approaches:

• Particle analysis: How far the ball travels—speed is much greater than the size of the ball, so treat it as a particle.
• Rigid body analysis (bat): How the bat swings to hit the ball—the length of the bat changes how far the ball travels.
• Rigid body analysis (ball spin): Calculating the spin on the ball as it flies—focus on how it is rotating.

Don't confuse: The same object (the baseball) can be analyzed as a particle or a rigid body depending on what aspect of the motion you care about.

🧩 Background note

• Particle analyses are typically covered in high school physics.
• Starting in Chapter 3 (per the excerpt), the course expands to include rigid bodies, bringing shape and size into the problem.

🧭 Overview

🧠 One-sentence thesis

Trusses are rigid structures made entirely of two-force members connected at joints, and their internal forces can be systematically determined using either the method of joints (particle analysis at each connection) or the method of sections (rigid-body analysis of cut portions).

📌 Key points (3–5)

• What trusses are: engineering structures composed entirely of two-force members (bodies with forces applied at exactly two locations) that connect at joints and carry loads in either tension or compression.
• Two analysis methods available: method of joints (analyze each pin connection as a particle using force equilibrium) vs. method of sections (cut through the truss and analyze sections as rigid bodies using force and moment equilibrium).
• When to use each method: method of joints is fastest for finding all member forces; method of sections is better for targeting just one or two specific members without solving the entire structure.
• Common confusion—tension vs. compression signs: the standard convention assumes all members are in tension (positive); negative results indicate compression, but the physical direction of the force on a joint follows Newton's third law (pulling on a member pulls on the joint; pushing on a member pushes on the joint).
• Zero-force members: some members carry no load but keep the truss rigid; recognizing them by inspection simplifies calculations.

🏗️ What is a truss and why it matters

🏗️ Definition and structure

Truss: An engineering structure made entirely of two-force members that is independently rigid (no part can move relative to the rest when separated from supports).

• Trusses are commonly found in roof frames and bridge sides.
• They are statically determinate, meaning all forces can be found using equilibrium equations alone.
• The goal of truss analysis is to identify both external forces (reactions and applied loads) and internal forces in each member (magnitude and whether tension or compression).

🔩 Two-force members

Two-force member: A body that has forces (and only forces, no moments) acting on it in exactly two locations.

• For static equilibrium, the forces at the two locations must be equal in magnitude, opposite in direction, and collinear (acting along the line connecting the two points).
• Why collinear? If the forces were not collinear, they would form a couple (a pair of equal and opposite forces separated by a distance), which would create a moment with nothing to counteract it. To keep the sum of moments zero, the perpendicular distance between forces must be zero—hence the forces must lie along the same line.
• This results in each member being in either tension (pulling apart) or compression (pushing together).

Example: Imagine a beam with forces only at its two ends. To have zero net force, the forces must be equal and opposite. To have zero net moment, they must act along the same line connecting the ends; otherwise, they would twist the beam.

🧩 Parts of a truss

Component	Description
Joints	Connection points where members meet; often labelled with letters (A, B, C, etc.); external forces are applied here
Members	The beams (metal or wood) connecting joints; labelled by the joints they connect (e.g., member AB connects joints A and B)
External forces	Reaction forces (from supports) and applied forces (loads from traffic, roof weight, etc.)

• Applied loads travel from the deck → stringers → beams → joints of the truss.
• Stringers: run parallel to the direction of travel.
• Beams: run perpendicular to stringers, transferring load to the truss joints.

⚖️ Tension and compression sign convention

• Standard convention: assume all members are in tension (positive, +).
• If the calculated force is negative (−), the member is actually in compression.
• Physical interpretation at joints:
  • Tension in a member → the member pulls on the joint (force arrows point away from the joint on the free-body diagram).
  • Compression in a member → the member pushes on the joint (force arrows point toward the joint).
• By Newton's third law, the force magnitude in member AB is the same at joint A and joint B, but the directions are opposite (equal and opposite reactions).

Don't confuse: The sign (+ or −) tells you tension or compression; the arrow direction on a joint diagram shows how the member acts on that joint (pulling or pushing).

🔍 Method of Joints

🔍 What it is

Method of joints: A particle-analysis technique that solves for unknown forces by analyzing each joint (pin connection) individually using only force equilibrium equations.

• Treat each joint as a particle (no moment equations, only force balance).
• Available equations at each joint: sum of forces in x = 0, sum of forces in y = 0.
• Because you have only two equations per joint, you can solve a joint only when it has two or fewer unknowns.

🔍 Step-by-step process

1. Label all members and joints (e.g., members AB, BC; joints A, B, C).
2. Solve for external reactions: Treat the entire truss as a single rigid body; draw a free-body diagram; write equilibrium equations (sum F_x = 0, sum F_y = 0, sum M = 0); solve for reaction forces at supports.
3. Draw free-body diagrams for each joint: Include external forces (reactions, applied loads) and internal forces from each connected member. Assume all member forces are tensile (pulling on the joint, arrows point away from the joint).
4. Write equilibrium equations for each joint: sum F_x = 0 and sum F_y = 0.
5. Solve systematically: Start at a joint with ≤2 unknowns; solve for those forces; move to the next joint that now has ≤2 unknowns; repeat until all member forces are found.
6. Interpret signs: Positive result = tension; negative result = compression.

Example: At joint A with reaction R_ay upward and two members AB and AG connected, assume F_AB and F_AG are tensile. Write sum F_y = 0: R_ay + F_AB sin(60°) = 0, so F_AB = −R_ay / sin(60°). If F_AB comes out negative, member AB is in compression.

🔍 When to use

• Best for: Finding forces in all members of the truss.
• Advantage: Usually the fastest and easiest method when you need a complete solution.
• Limitation: Requires solving many joints sequentially; not efficient if you only need one or two member forces.

✂️ Method of Sections

✂️ What it is

Method of sections: A rigid-body analysis technique that solves for specific member forces by imagining a cut through the truss, splitting it into two sections, and analyzing one section using force and moment equilibrium.

• Treat each section as a rigid body (not a particle), so you have three equilibrium equations: sum F_x = 0, sum F_y = 0, sum M = 0.
• The cut turns internal member forces into external forces on the section's free-body diagram.

✂️ Step-by-step process

1. Label all members.
2. Solve for external reactions (same as method of joints): analyze the whole truss as a rigid body.
3. Make a strategic cut: Imagine cutting through the truss along a line (not necessarily straight) that passes through the member(s) you want to solve for. Cut through as few members as possible (ideally three or fewer, since you have three equations).
4. Draw a free-body diagram for one (or both) of the resulting sections: Include all external forces on that section (reactions, applied loads) and the internal forces in the cut members. Assume all cut member forces are tensile (pulling on the section).
5. Write equilibrium equations for the section: sum F_x = 0, sum F_y = 0, sum M = 0 (choose a convenient moment center to eliminate unknowns).
6. Solve for the unknown member forces.
7. Interpret signs: Positive = tension; negative = compression.

Example: To find force in member E, make a cut through members including E. Draw the left section with reaction forces and the three cut member forces (assumed tensile). Take moments about a joint where two of the cut members intersect (their forces create zero moment there), leaving only F_E in the moment equation. Solve for F_E directly.

✂️ When to use

• Best for: Finding forces in one or a few specific members without solving the entire truss.
• Advantage: Can target a member directly; sometimes you don't even need to solve for reactions first (if the cut section doesn't include the supports).
• Limitation: Not efficient if you need forces in all members.

✂️ Comparison with method of joints

Aspect	Method of Joints	Method of Sections
Analysis type	Particle (each joint)	Rigid body (each section)
Equations per unit	2 (F_x, F_y)	3 (F_x, F_y, M)
Best for	All member forces	One or two specific members
Speed	Fastest for complete solution	Fastest for targeted solution
Can combine?	Yes, use both methods in the same problem as needed	Yes

🚫 Zero-force members

🚫 What they are

Zero-force members: Members in a truss that carry no load (force = 0) but are necessary to keep the structure rigid and maintain its shape.

• They can be identified by inspection using equilibrium at a joint.
• Recognizing them simplifies calculations (one fewer unknown at that joint).

🚫 How to identify

Look at a joint and check the equilibrium equations:

• Case 1: If only one member connects to a joint in a particular direction (x or y) and there is no external force in that direction, then that member must be a zero-force member.
  • Example: At joint E, only member EH has a y-component, and no external load acts in y. Then sum F_y = 0 implies F_EH = 0.
• Case 2: If two members meet at a joint with no external load, and they are collinear (lie along the same line), then both are zero-force members (or both carry the same force, but if no load, both are zero).

Don't confuse: If an external load (applied force or reaction) acts at the joint, the member may not be zero-force even if it looks isolated in one direction.

🚫 Practical note

• Zero-force members are more common in theoretical analysis than in real loaded structures (where deck loads are distributed to many joints).
• They are still structurally important: removing them would allow the truss to deform or collapse.

Example: In a truss with joints E, H, M, K, C, P, if joint E has only member EH in the vertical direction and no vertical load, then F_EH = 0. Similarly, if joints M and C have only one vertical member each (MK and CP) with no load, those are also zero-force members.

🔄 Putting it all together

🔄 General workflow for truss analysis

1. Identify the structure: Label joints and members; recognize it as a truss (all two-force members).
2. Choose your method:
   • Need all member forces? → Method of joints.
   • Need just one or two? → Method of sections.
   • Can combine both if helpful.
3. Solve for reactions (usually required first): Treat whole truss as rigid body; use sum F_x = 0, sum F_y = 0, sum M = 0.
4. Apply the chosen method:
   • Joints: Start at a joint with ≤2 unknowns; solve; move to next joint; repeat.
   • Sections: Make a strategic cut; draw free-body diagram of one section; solve using 3 equilibrium equations.
5. Check for zero-force members (optional but helpful): Inspect joints to spot members with zero load; simplifies subsequent steps.
6. Interpret results: Positive = tension (member pulls on joints); negative = compression (member pushes on joints).

🔄 Common truss types

• Bridge trusses: Pratt, Howe, Warren, K-truss, etc. (load applied at bottom or top chord joints).
• Roof trusses: King post, Queen post, Fink, Howe, etc. (load applied at top chord from roof weight).

🔄 Load path in bridges

Applied load (vehicles) → deck → stringers (parallel to traffic) → floor beams (perpendicular to traffic) → truss joints → truss members (in tension or compression) → supports.

🔄 Constraints and reactions

• Truss supports are modelled the same way as in earlier rigid-body analysis (pin support: two reaction components; roller support: one reaction component perpendicular to the surface).
• Solve for these reactions first by treating the entire truss as one rigid body.

Don't confuse: The reaction forces are external to the truss; the member forces are internal (but become external when you cut the truss in the method of sections).

🧭 Overview

🧠 One-sentence thesis

When a beam is subjected to transverse loads, three internal forces—normal (axial) force, shear force, and bending moment—develop at any cross-section, and their magnitudes can be calculated by cutting the beam and applying equilibrium equations, then visualized through shear and moment diagrams.

📌 Key points (3–5)

• Three types of internal forces: Normal force (N) acts along the beam axis; shear force (V) acts perpendicular to the axis; bending moment (M) causes rotation.
• Sign convention matters: Positive shear on the right side acts upward; positive moment causes the beam to sag (concave upward); tension is positive normal force.
• How to calculate: Find external reactions first, make a cut at the point of interest, add internal forces using the positive sign convention, then solve with equilibrium equations.
• Shear and moment diagrams: Graphical plots showing how V and M vary along the entire beam; the slope of the moment diagram equals the shear force, and the slope of the shear diagram equals the negative of the distributed load.
• Common confusion: Don't mix up "left side" vs "right side" when applying sign conventions—the same internal force appears equal and opposite on each side of a cut for equilibrium.

🔍 Types of internal forces and moments

🔍 Normal force (N)

Normal force: the algebraic sum of all axial forces acting on either side of a section.

• Also called "axial force" because it acts along the axis of the beam.
• For a horizontal beam, normal force is horizontal; for a vertical column, it is vertical.
• Units: N or lb.
• Sign convention: Positive if it tends to pull the member apart (tension); negative if it tends to crush the member (compression).
• Example: In a truss member analyzed in the previous chapter, the normal force was the only internal force because loads were purely axial.

🔍 Shear force (V)

Shear force: the algebraic sum of all transverse forces acting on either side of a section.

• Acts perpendicular to the beam axis.
• For a horizontal beam, shear force is vertical; for a vertical column, it is horizontal (sometimes called "transverse").
• Units: N or lb.
• Sign convention: Positive shear on the right side of a cut acts upward (or equivalently, on the left side acts downward); negative shear is the reverse.
• The phrase "on either side" means you can sum forces on the left or the right of the cut and get the same magnitude (but opposite direction).
• Example: A simply supported beam with a downward point load will have positive shear to the left of the load and negative shear to the right.

🔍 Bending moment (M)

Bending moment: the algebraic sum of all moments of forces acting on either side of a section.

• Causes the beam to bend or rotate.
• Units: Nm or ft-lb.
• Sign convention: Positive moment causes the beam to sag (concave upward, like a smile); negative moment causes the beam to hog (concave downward, like a frown).
• Example: A cantilever beam with a downward load at the free end will have a negative bending moment at the fixed support because it causes upward concavity.

🔍 2D vs 3D analysis

• The excerpt focuses on 2D analysis: one normal force, one shear force, one bending moment.
• In 3D, there are: 1 normal force, 2 shear forces (in two perpendicular directions), and 3 bending moments (two bending moments plus torsion).
• For this chapter, assume negligible loading in the third dimension.

🧮 Calculating internal forces at a point

🧮 Step-by-step procedure

1. Find external and reaction forces: Use equilibrium equations (sum of forces = 0, sum of moments = 0) on the entire beam to solve for reactions at supports.
2. Make a cut: Slice the beam at the point of interest where you want to know the internal forces.
3. Add internal forces using positive sign convention: On the free-body diagram of one side of the cut, draw N, V, and M according to the positive convention (on the right side: normal out, shear up, moment causing sag).
4. Apply equilibrium equations: Use sum of forces in x, sum of forces in y, and sum of moments to solve for N, V, and M.

🧮 Positive sign convention details

• On the right side of a cut:
  • Normal force: positive points to the right (tension).
  • Shear force: positive points upward.
  • Moment: positive causes upward concavity (sagging).
• On the left side of a cut: The directions are reversed for equilibrium.
• Why it matters: If your calculated value is negative, it means the force or moment acts in the opposite direction from what you drew.
• Don't confuse: The sign convention for the cut is different from the overall beam behavior—positive shear on the whole beam is "right side down," but on a cut section, positive shear on the right is "up" to balance the overall motion.

🧮 Example walkthrough

The excerpt provides an example of a cantilever beam with a distributed load:

• Step 1: Calculate reactions at the fixed support (A_x = 0, A_y = 87.5 lb upward, C = 612.5 lb upward at the roller).
• Step 2: Make a cut at midpoint B (2 ft from the left end).
• Step 3: Draw the left portion with internal forces N, V, M using positive convention.
• Step 4: Sum forces in y: 87.5 lb - 200 lb (from distributed load on the 2 ft segment) - V = 0 → V = -112.5 lb (negative means it acts downward, not upward).
• Sum moments about the left end: -(200 lb)(1 ft) - V(2 ft) + M = 0 → M = -25 ft·lb (negative means it causes downward concavity, not upward).
• Result: N = 0, V = -112.5 lb, M = -25 ft·lb.

📊 Shear and moment diagrams

📊 What they are

Shear diagram: a graphical representation of the variation of shear force along the entire length of the beam.

Moment diagram: a graphical representation of the variation of bending moment along the entire length of the beam.

• The x-axis represents position along the beam (lined up with the beam's length).
• The y-axis represents the magnitude of V (for shear diagram) or M (for moment diagram).
• Positive values are plotted above the x-axis; negative values below.
• These diagrams help identify maximum shear and moment, which are critical for design.

📊 Relationship between load, shear, and moment

The excerpt provides key calculus relationships:

• The derivative of moment with respect to x equals shear: dM/dx = V(x).
• The derivative of shear with respect to x equals the negative of the distributed load: dV/dx = -w(x).
• The second derivative of moment equals the negative of the distributed load: d²M/dx² = -w(x).
• In integral form:
  • Change in moment = integral of shear: ΔM = ∫V(x)dx.
  • Change in shear = integral of load: ΔV = ∫w(x)dx.
• What this means:
  • If the distributed load is constant, the shear diagram will be linear (slope = -w), and the moment diagram will be parabolic (quadratic).
  • If there is no distributed load (w = 0), the shear is constant, and the moment is linear.
• Example: A uniformly distributed load of 100 lb/ft over 5 m will cause the shear to decrease linearly from 0 to -100×5 = -500 lb, and the moment to be a downward-opening parabola.

📊 How to produce diagrams

Method 1 (using equations and integration): Find the equation for V and M in each segment, then integrate using the relationships above.

Method 2 (point-by-point): Calculate internal forces at key points (where loads are applied, at supports, at ends of distributed loads), plot these points, then connect them with the appropriate shape (linear, parabolic, etc.).

Method 3 (equilibrium equations for each segment):

1. Draw a free-body diagram of the entire beam.
2. Calculate reactions.
3. Make cuts in each segment (between load changes).
4. For each segment, write expressions for V(x) and M(x) using equilibrium equations.
5. Plot these expressions on the diagrams.

The excerpt uses Method 3 in its examples.

📊 Example: Cantilever with point load

• Beam: 3 ft cantilever with 5 lb downward load at the free end.
• Reactions at fixed support: B_x = 0, B_y = 5 lb upward, M_B = 15 ft·lb counterclockwise.
• Shear: Cut at distance x from the free end. Sum forces in y: -5 lb - V = 0 → V = -5 lb (constant along the entire beam).
• Moment: Sum moments about the free end: -V·x - M = 0 → M = -5x (linear, starting at 0 at the free end and reaching -15 ft·lb at the fixed end).
• Diagrams: Shear is a horizontal line at -5 lb. Moment is a straight line from 0 to -15 ft·lb.
• The negative signs indicate the shear acts downward and the moment causes upward concavity (hogging).

📊 Example: Cantilever with distributed load

• Beam: 5 m cantilever with uniformly distributed load of 20 kN/m over the entire length.
• Reactions: B_y = 100 kN upward, M_B = 250 kN·m.
• Shear: V(x) = -20x (linear, starting at 0 at the free end and reaching -100 kN at the fixed end).
• Moment: M(x) = -10x² (parabolic, starting at 0 at the free end and reaching -250 kN·m at the fixed end).
• Diagrams: Shear is a straight line with negative slope. Moment is a downward-opening parabola.

🛠️ Tips for drawing diagrams

🛠️ General rules

• Positive V means increasing M: If shear is positive, the moment diagram has a positive slope.
• Negative V means decreasing M: If shear is negative, the moment diagram has a negative slope.
• When V = 0, M is at a maximum or minimum: The moment diagram has a horizontal tangent (inflection point) where shear crosses zero.

🛠️ Starting and ending values

• Cantilever beam:
  • At the free end: V = 0 and M = 0 (if no load or moment applied there).
  • At the fixed support: V and M are nonzero (equal to the reactions).
• Simply supported beam:
  • At both supports: M = 0 (pinned or roller supports cannot resist moment).
  • V starts and ends at the reaction forces.

🛠️ Jumps and inflection points

• In the shear diagram: Shear "jumps" up or down at point loads (matching the direction of the load) and at reaction forces.
• In the moment diagram: Moment "jumps" up or down at applied moments (matching the direction of the moment).
• Distributed loads: Cause the shear to have a sloped line (positive load → negative slope in V) and the moment to be curved (parabolic for uniform load).

🛠️ Shape relationships

The excerpt provides a helpful figure showing the derivative/integral relationships:

• Going down (derivative): Cubic → Quadratic → Linear → Constant → Zero.
• Going up (integral): Zero → Constant → Linear → Quadratic → Cubic.
• If the moment diagram is quadratic (parabolic), the shear diagram is linear, and the load is constant.
• If the shear diagram is constant, the moment diagram is linear, and the load is zero.

🛠️ Checking your work

• The shear diagram should return to zero at the end if you account for all reactions.
• The moment diagram should return to zero at simply supported ends.
• The area under the shear diagram between two points equals the change in moment between those points.
• The area under the load diagram between two points equals the change in shear between those points.

🔑 Key Takeaways

Basically: Internal forces (normal, shear, bending moment) develop inside a beam when it is loaded; you find them by cutting the beam and using equilibrium equations. Shear and moment diagrams plot these forces along the entire beam and help identify critical stress points.

Application: Use this method to analyze bridges, diving boards, shelves, or any beam structure to ensure it can safely carry the applied loads. The diagrams show where the beam experiences maximum stress.

Looking Ahead: These concepts are foundational for structures courses, where you will design beams and select materials based on internal force magnitudes.

🧭 Overview

🧠 One-sentence thesis

Zero-force members are theoretical structural elements that carry no load but maintain truss shape, and the method of sections provides a faster way to analyze specific members compared to solving every joint sequentially.

📌 Key points (3–5)

• What zero-force members are: two-force members that do not carry any load but help keep the structure in a certain shape.
• When they appear: typically in trusses when there are no loads applied at certain joints; real-world structures may have loads that activate these members.
• Method comparison: the method of sections is faster for finding the load in one specific member, while the method of joints is better when calculating loads in all members.
• Common confusion: zero-force members are "more theoretical than actual" because real structures often have loads on the deck or at joints that activate these members.
• Sign conventions: negative results in equilibrium equations indicate the assumed direction of the force is opposite to the actual direction (e.g., compression instead of tension).

🏗️ Zero-force member concept

🏗️ What they are

Zero-force members: two-force members that do not carry any load but help keep the structure into a certain shape.

• They exist in theory when analyzing trusses with no loads at certain joints.
• Their purpose is structural stability and shape maintenance, not load-bearing.
• Example: in a truss bridge with no loads on the joints, several members may be zero-force members.

🌉 Real-world vs theoretical

• The excerpt states zero-force members are "more theoretical than actual."
• In practice, loads on the deck or at joints mean these members often end up carrying loads.
• Example: a bridge side might theoretically have 7 zero-force members, but deck loads likely activate all of them.

Condition	Zero-force status
No loads at joints	Members remain zero-force
Loads on deck/joints	Members likely carry loads

🔧 Method of sections application

🔧 When to use sections vs joints

The excerpt demonstrates both methods through a flower cart truss problem:

• Method of sections: make a cut through the truss to isolate the member of interest, then solve using equilibrium equations.
• Method of joints: solve joints sequentially, starting from supports.

Speed comparison:

• For finding one specific member: method of sections is faster because you can cut directly to that member.
• For finding all member loads: method of joints is faster because you solve systematically without multiple cuts.

✂️ How the method of sections works

The flower cart example shows the process:

1. Draw the full free-body diagram with all external forces.
2. Make a cut through the truss that passes through the member you want to find.
3. Choose one piece (top or bottom) from the cut—pick the piece with fewer external forces for simpler calculations.
4. Apply equilibrium equations to the isolated piece to solve for the internal force in the cut member.

Example: To find force in member CG, a cut is made through it, the top half is chosen (fewer external forces), and equilibrium equations yield the force magnitude and type (tension or compression).

📐 Sign convention and interpretation

• Forces are initially drawn with an assumed direction (e.g., tension pulling away from the joint).
• Negative result: the actual direction is opposite to what was drawn.
• Example: if FCG is drawn as tension but the calculation gives -515 N, the member is actually in compression (515 N).

Don't confuse: A negative sign does not mean the force is "wrong"—it only means the direction assumption was opposite to reality. The magnitude is still correct.

🧮 Equilibrium equation strategy

🧮 Solving reaction forces

The flower cart problem demonstrates systematic use of equilibrium:

• Sum of forces in x-direction = 0: solves for horizontal reaction RAx (equal and opposite to applied force P).
• Sum of moments about point A = 0: solves for vertical reaction RBy using moment arms.
• Sum of forces in y-direction = 0: solves for remaining vertical reaction RAy.

Why moments are useful: Taking moments about a point eliminates unknown forces at that point, simplifying the equation.

🔍 Verification checks

The excerpt includes review steps to confirm correctness:

• RAx should equal and opposite to the applied force P in the x-direction.
• The moment of RBy about point A should equal and opposite to the moment of P about A.
• For the cut member, its x-component should balance the applied force if it's the only cut member in that direction.

Example: The x-component of FCG (4/√17 × 515 N = 500 N) equals the applied force P, confirming the solution.

📝 Problem-solving structure

📝 Systematic approach steps

The flower cart example follows a clear structure:

1. Problem statement: define the scenario, geometry, and what to find.
2. Draw: create sketch and free-body diagram.
3. Knowns and unknowns: list given values and what needs to be solved.
4. Approach: state which method(s) will be used and why.
5. Analysis: perform calculations with equilibrium equations.
6. Review: verify results using alternative checks or physical reasoning.

🎯 Strategic method selection

The excerpt emphasizes choosing the right method based on the question:

• Single member load: method of sections avoids solving unnecessary joints.
• All member loads: method of joints is more systematic and avoids multiple cuts.
• Strategy changes with scope: what is fastest for one member may not be fastest for the entire structure.

🧭 Overview

🧠 One-sentence thesis

When analyzing a single beam, three types of internal forces and moments—normal force, shear force, and bending moment—describe what happens at any cut point along the beam, with shear force and bending moment typically changing throughout the beam while normal force usually remains constant.

📌 Key points (3–5)

• What internal forces means: the term "internal forces" actually refers to both internal forces and internal moments acting inside a beam.
• Three types at any cut: normal (axial) force, shear force, and bending moment describe the internal state at any point along a beam.
• How they differ along the beam: shear force and bending moment change throughout the beam when transverse forces are applied; normal force usually stays constant because axial forces along the beam are uncommon.
• Common confusion: this chapter focuses on a single beam, not trusses—trusses were analyzed in the previous chapter by looking at joints or sections.
• Analogy to reactions: making a cut in a beam is similar to analyzing a fixed reaction—you describe the point using one horizontal force, one vertical force, and a moment.

🔍 What this chapter covers

🔍 Shift from trusses to single beams

• The previous chapter examined normal (axial) force in beams joined into trusses, using method of joints or method of sections.
• This chapter shifts focus to what happens along a single beam.
• The excerpt emphasizes: "we look at what happens along a single beam."

📐 The three internal quantities

When you make a cut in an object, similar to a fixed reaction, we describe what is happening at that point using one horizontal force (called normal force), one vertical force (called shear force), and a bending moment.

• Normal force: the horizontal (axial) force at the cut.
• Shear force: the vertical force at the cut.
• Bending moment: the moment acting at the cut.
• These three quantities fully describe the internal state at any point in the beam.

🔄 How internal forces vary along the beam

🔄 Shear force and bending moment change

• Shear force and bending moment change throughout the beam because additional transverse (perpendicular) forces are applied along its length.
• The excerpt states: "the shear force and bending moment change throughout the beam because additional transverse forces are applied."
• Example: if a beam supports multiple loads at different points, the shear force and bending moment will vary from point to point.

➡️ Normal force usually stays constant

• Normal force typically stays the same along the beam.
• Why: "it's uncommon to have applied axial forces along the beam."
• Don't confuse: normal force can vary in trusses (previous chapter), but in a single beam with typical loading, it remains constant.

📊 Chapter structure and tools

📊 What the chapter includes

The excerpt lists three sections:

1. 6.1 Types of Internal Forces: defines shear force, normal force, and bending moment.
2. 6.2 Shear/Moment Diagrams: graphing the shear force and bending moment along the beam.
3. 6.3 Examples: peer-submitted examples.

📊 Important equations

• The excerpt mentions "important equations for this chapter" but does not provide them in the text shown.
• The chapter will use equilibrium methods to calculate these internal quantities at any cut point.

🧭 Overview

🧠 One-sentence thesis

Internal forces (normal force, shear force, and bending moment) vary along a beam under transverse loading, and analyzing these forces through cuts and diagrams helps engineers identify critical stress points for safe structural design.

📌 Key points (3–5)

• Three internal force types: Normal (axial) force N, shear force V, and bending moment M describe what happens inside a beam at any point.
• Sign convention matters: Positive shear is "right side down" on the whole beam (or "up on the right" when cut); positive moment causes upward concavity (sagging).
• How to find internal forces: Solve external reactions, make a cut at the point of interest, add internal forces using the positive sign convention, then apply equilibrium equations.
• Shear/moment diagrams: Graphical representations show how V and M vary along the entire beam; the slope of M equals V, and the slope of V equals negative distributed load intensity.
• Common confusion: Sign convention can be tricky—remember that when you cut the beam, the internal forces must balance the external loads on that side; negative results mean the arrow points the opposite direction.

🔧 Three types of internal forces

🔧 Normal force (N)

Normal force: the algebraic sum of the axial forces acting on either side of a section.

• Also called "axial force" because it acts along the axis of the beam.
• For a horizontal beam, normal force is horizontal; for a vertical column, it would be vertical.
• Units: N or lb.
• Positive convention: Tension (pulling apart); negative means compression (crushing together).
• In many beam problems with only transverse loads, N remains zero or constant throughout.

✂️ Shear force (V)

Shearing force: the algebraic sum of all the transverse forces acting on either side of the section.

• Acts perpendicular to the beam axis (vertical for horizontal beams).
• Units: N or lb.
• Positive convention: On the right side of a cut, positive shear points upward; equivalently, "right side down" when viewing the whole beam.
• Changes at points where concentrated forces are applied and varies linearly under distributed loads.
• Example: A beam with a downward point load will show a sudden drop (jump) in the shear diagram at that location.

🔄 Bending moment (M)

Bending moment: the algebraic sum of all the forces' moments acting on either side of the section.

• Causes the beam to bend or bow.
• Units: Nm or ft-lb.
• Positive convention: Causes upward concavity (sagging, like a smile shape); negative causes downward concavity (hogging, like a frown).
• The moment diagram's slope at any point equals the shear force at that point.
• Example: A simply supported beam with a center load will have maximum positive moment at the center where shear crosses zero.

🎯 Sign convention rules

🎯 Why sign convention is essential

• Engineers need a standard so everyone interprets diagrams the same way.
• The excerpt emphasizes: "So that there is a standard within the industry, a sign convention is necessary."

📐 Positive sign convention summary

Force/Moment	On the right side of cut	Visual cue
Normal (N)	Points right (tension)	Pulling apart
Shear (V)	Points up	Right side down on whole beam
Moment (M)	Counterclockwise	Causes upward concavity (sagging)

• Don't confuse: When you look at the whole beam, positive shear appears as "right side down," but when you cut and isolate one side for equilibrium, positive shear on the right cut face points upward to balance the external loads.
• The excerpt shows identical sign conventions drawn two ways to clarify this.

🔪 Calculating internal forces at a point

🔪 Four-step method

1. Find external and reaction forces: Use equilibrium equations (ΣFx = 0, ΣFy = 0, ΣM = 0) on the whole beam.
2. Make a cut at the point of interest.
3. Add internal forces (N, V, M) to the free-body diagram of one side using the positive sign convention.
4. Apply equilibrium equations to solve for N, V, and M.

🧮 Worked example from excerpt

• A cantilever beam 7 ft long with distributed load 100 lb/ft.
• Reaction forces: Ax = 0, Ay = 87.5 lb (+j), C = 612.5 lb (+j).
• Cut at midpoint B (2 ft from left):
  • Force from distributed load on left segment: Fw = 100 lb/ft × 2 ft = 200 lb, acting 1 ft from left.
  • ΣFy = 87.5 lb - 200 lb - V = 0 → V = -112.5 lb (negative means it acts upward, not downward).
  • ΣM = -200 ft·lb - V(2 ft) + M = 0 → M = -25 ft·lb (negative means clockwise, not counterclockwise).
• Key insight: Negative signs indicate the actual direction is opposite to the assumed positive convention; always check and redraw arrows if needed.

⚠️ Common pitfall

• If your calculated value is negative, it means the force or moment acts in the opposite direction to what you drew using the positive convention—don't forget to note this in your final answer.

📊 Shear and moment diagrams

📊 What are they?

Shearing force diagram: a graphical representation of the variation of the shearing force along the beam.

Bending moment diagram: a graphical representation of the variation of the bending moment along the beam.

• These diagrams plot V(x) and M(x) versus position x along the beam.
• Convention: Positive shear can be drawn above or below the axis (must label); positive moment is drawn above the x-axis, negative below.

🔗 Relationship between load, shear, and moment

The excerpt provides three key equations:

• dM/dx = V(x): The slope of the moment diagram equals the shear force.
• dV/dx = -w(x): The slope of the shear diagram equals the negative of the distributed load intensity.
• d²M/dx² = -w(x): The second derivative of moment equals the negative load intensity.

In integral form:

• ΔM = ∫V(x)dx: Change in moment equals the area under the shear diagram.
• ΔV = ∫w(x)dx: Change in shear equals the area under the load diagram.

Practical meaning:

• Constant distributed load → linear shear diagram → parabolic moment diagram.
• Zero distributed load → constant shear → linear moment.
• Example: A uniformly loaded cantilever will show a straight-line shear diagram and a curved (parabolic) moment diagram.

🖊️ How to draw shear/moment diagrams

🖊️ Method (using equilibrium equations)

1. Draw the beam's free-body diagram with all loads and reactions.
2. Calculate reaction forces.
3. Make cuts at key points (where loads change, at reactions, at distributed load boundaries).
4. For each segment, write expressions for V(x) and M(x) using equilibrium on one side of the cut.
5. Plot V(x) and M(x) on aligned axes below the beam diagram.
6. Check: Both diagrams should return to zero at free ends (for cantilevers) or at supports (for simply supported beams).

📈 Diagram construction tips from excerpt

• For shear diagram: Start at one end (often zero); jump up/down by the magnitude of point forces; slope linearly under distributed loads (slope = load intensity).
• For moment diagram: Integrate the shear diagram; jump up/down at applied moments; the diagram should return to zero at the far end if the beam is in equilibrium.
• Visual check: Where V = 0, M has a local maximum or minimum; positive V means M is increasing; negative V means M is decreasing.

🧪 Example 1 from excerpt: Cantilever with end load

• Cantilever beam, 3 ft long, 5 lb load at free end.
• Reactions at fixed end B: Bx = 0, By = 5 lb, MB = 15 ft·lb.
• Cut anywhere along beam (0 < x < 3 ft from free end):
  • V = -5 lb (constant, negative means downward on left side).
  • M = -(5 lb)x (linear, negative moment).
• Shear diagram: Horizontal line at -5 lb, jumps to 0 at the reaction.
• Moment diagram: Straight line from 0 at free end to -15 ft·lb at fixed end.
• Insight: Constant shear → linear moment; the negative sign matches the sign convention (downward concavity).

🧪 Example 2 from excerpt: Cantilever with distributed load

• Cantilever 5 m long, uniformly distributed load throughout.
• Shear varies linearly from 0 at free end to -100 kN at fixed end (V = -wx, where w is load intensity).
• Moment varies parabolically (M = -wx²/2), from 0 at free end to maximum negative at fixed end.
• Insight: Linear shear → parabolic moment, consistent with dM/dx = V.

🎨 Diagram shape rules

🎨 General patterns

• +V → M increasing; -V → M decreasing; V = 0 → M at max or min.
• Jumps in V occur at point loads (magnitude = load); jumps in M occur at applied moments.
• Inflection points in M (where curvature changes) correspond to V = 0.

🏗️ Boundary conditions by support type

Beam type	Shear diagram	Moment diagram
Cantilever	Nonzero V and M at fixed end; zero at free end	Nonzero at fixed end; zero at free end
Simply supported	Starts and ends with reaction forces	Starts and ends at zero

🔄 Load-to-diagram progression

The excerpt provides a helpful visual:

• 0 (no load) → constant shear → linear moment.
• Constant load → linear shear → parabolic moment.
• Linear load → parabolic shear → cubic moment.
• Moving down the chart: differentiate (load → shear → moment).
• Moving up the chart: integrate (moment ← shear ← load).

Don't confuse: The derivative relationship goes from moment to shear to load; the integral relationship goes the opposite direction.

✅ Key takeaways from excerpt

✅ Practical application

• Bridge design: Different loads (cars, trucks, lampposts) create varying internal forces; use this method to find critical stress points.
• Solar panel support (from student example): Calculate internal forces at bolt locations to ensure safe attachment.
• Diving board (from student example): Determine where the board experiences maximum bending to prevent failure.

✅ Looking ahead

• The excerpt notes: "You will use this more in your structures class."
• These diagrams help identify where to reinforce a beam or where failure is most likely.
• Online tools (SkyCiv, ClearCalcs, BeamGuru) can check your work, but the excerpt warns they are "not acceptable to use on the exam."

🧭 Overview

🧠 One-sentence thesis

When a beam is subjected to transverse loads, three internal forces and moments develop—normal (axial) force, shear force, and bending moment—which must be calculated at any point along the beam to predict structural behavior and design safe members.

📌 Key points (3–5)

• Three internal forces/moments in 2D beams: normal force (N, horizontal/axial), shear force (V, vertical/transverse), and bending moment (M, rotational).
• Sign convention is standardized: on the right side of a cut, positive shear is upward, positive normal is tension (pulling apart), and positive moment causes upward concavity (sagging).
• How to calculate internal forces: solve external reactions first, make a cut at the point of interest, draw a free-body diagram with internal forces using the positive sign convention, then apply equilibrium equations.
• Common confusion—left vs right side: the phrase "on either side" means you can sum forces on the left or right of the cut; the sign convention flips depending on which side you analyze.
• Distributed loads relate to diagrams: the slope of the shear diagram equals the negative of the distributed load, and the slope of the moment diagram equals the shear force.

🔍 What are internal forces and moments?

🔍 The three types in 2D beams

Normal force (N): the algebraic sum of the axial forces acting on either side of the section.

Shearing force (V): the algebraic sum of all the transverse forces acting on either side of the section of a beam or a frame.

Bending moment (M): the algebraic sum of all the forces' moments acting on either side of the section of a beam or a frame.

• When you make a cut in a beam, you describe what happens at that point using one horizontal force (normal), one vertical force (shear), and a bending moment—similar to a fixed reaction.
• Normal and shear have units of N or lb; bending moment has units of Nm or ft-lb.
• For a horizontal beam: normal is horizontal, shear is vertical, moment is rotational.
• For a vertical column: normal (axial) is vertical, shear (transverse) is horizontal.
• This analysis assumes 2D loading with negligible forces in the third dimension.

🌐 Extension to 3D

In three dimensions, a beam has:

• 1 normal force (N)
• 2 shear forces (V₁ and V₂)
• 3 bending moments (M₁, M₂, and T for torsion)

📐 Why internal forces matter

• These forces show the maximum bending moments and shearing forces needed for sizing structural members during design.
• Predicting the magnitudes of these forces is essential to understand how structures behave under load.
• Example: A bridge with different loads (cars, trucks, lampposts) requires calculating internal loads at particular points of interest to ensure safety.

✅ Sign convention for internal forces

✅ Why a standard sign convention is necessary

A sign convention ensures that engineers agree on what is positive and what is negative, creating consistency across the industry.

✅ Positive sign rules

Force/Moment	Positive direction	Description
Normal force (N)	Tension (pulling apart)	Tends to tear the member at the section; compressive (crushing) is negative
Shear force (V)	Right side up (or left side down)	Tends to move the left side upward or the right side downward; opposite is negative
Bending moment (M)	Upward concavity (sagging)	Causes the beam to sag upward; downward concavity (hogging) is negative

• On the right side of a cut: shear-up is positive.
• Looking at the whole beam: positive shear is right side down; when you cut into the beam, for static equilibrium, positive shear must be up on the right (equal and opposite to the overall motion).
• Don't confuse: the sign convention appears to flip depending on whether you're looking at the whole beam or one side of a cut—this is because internal forces must balance external forces for equilibrium.

✅ Axial force details

• Positive (tensile): tends to tear the member; the member is in axial tension.
• Negative (compressive): tends to crush the member; the member is in axial compression.

✅ Shear force details

• Positive: moves the left side of the section upward or the right side downward.
• Negative: moves the left side downward or the right side upward.

✅ Bending moment details

• Positive: causes concavity upward (sagging).
• Negative: causes concavity downward (hogging).

🧮 How to calculate internal forces at a point

🧮 Four-step method

1. Find the external and reaction forces: solve the entire structure for support reactions using equilibrium equations.
2. Make a cut at the point of interest along the beam.
3. Draw a free-body diagram (FBD) of one side of the cut: add the internal forces (N, V) and moment (M) using the positive sign convention.
4. Use equilibrium equations to solve for the unknown internal forces and moments.

🧮 Worked example: distributed load on a beam

Given: A beam with a distributed load of 100 lb/ft over 7 ft, supported at A (left) and C (right, 4 ft from A).

Step 1: Solve external forces

• Sum of horizontal forces: Aₓ = 0
• Sum of vertical forces: Aᵧ + C − (100 lb/ft)(7 ft) = 0
• Sum of moments about A: −(100 lb/ft × 7 ft)(7 ft / 2) + (4 ft)C = 0

Solving:

• C = (100 lb/ft × 49 ft²) / (2 × 4 ft) = 612.5 lb (upward)
• Aᵧ = (100 lb/ft × 7 ft) − 612.5 lb = 87.5 lb (upward)
• Aₓ = 0

Step 2: Make a cut at midpoint B (2 ft from A).

Step 3: Draw FBD of the left side with internal forces N, V, M using positive sign convention.

Step 4: Apply equilibrium

• For the 2 ft segment, the distributed load force is Fᵥ = (100 lb/ft)(2 ft) = 200 lb, acting 1 ft from the left.
• Sum of vertical forces: 87.5 lb − 200 lb − V = 0 → V = −112.5 lb (negative means upward, not downward)
• Sum of moments about A: −(100 lb/ft × 2 ft)(1 ft) − V(2 ft) + M = 0 → M = 200 ft·lb − 225 ft·lb = −25 ft·lb (negative means reverse direction, hogging)

Result: N = 0, V = −112.5 lb, M = −25 ft·lb (clockwise).

🧮 Key insight: "on either side"

• The phrase "on either side" means you can sum forces on the left or right of the cut.
• Both approaches give the same answer if you apply the sign convention correctly for each side.
• Don't confuse: the internal forces on the left and right faces of the cut are equal and opposite (Newton's third law), but the sign convention accounts for this.

📊 Shear and moment diagrams

📊 What are shear/moment diagrams?

Shearing force diagram: a graphical representation of the variation of the shearing force on a portion or the entire length of a beam or frame.

Bending moment diagram: a graphical representation of the variation of the bending moment on a segment or the entire length of a beam or frame.

• These diagrams show how internal shear force and bending moment change along the whole beam.
• Convention for shear diagrams: can be drawn above or below the x-centroidal axis, but must indicate if positive or negative.
• Convention for moment diagrams: positive bending moments are drawn above the x-centroidal axis; negative moments are drawn below.

📊 Relationship between distributed loads and diagrams

The mathematical relationships are:

• The derivative of moment with respect to position equals shear: dM/dx = V(x)
• The derivative of shear with respect to position equals the negative of the distributed load: dV/dx = −w(x)
• The second derivative of moment equals the negative of the distributed load: d²M/dx² = −w(x)

Or in integral form:

• Change in moment = integral of shear: ΔM = ∫V(x)dx
• Change in shear = integral of distributed load: ΔV = ∫w(x)dx

Practical meaning:

• If there is a constant distributed load, then the slope of the shear diagram will be linear (straight line).
• The slope of the moment diagram will be parabolic (curved).

📊 Why diagrams are useful

• Shear and moment diagrams aid immeasurably during design.
• They show the maximum bending moments and shearing forces needed for sizing structural members.
• Engineers can quickly identify critical points where internal forces are largest.

🔑 Summary and application

🔑 Core takeaway

The internal forces (and moments) for a 2D beam are shear, normal, and bending moment. There is a positive sign convention to use when making a cut along a beam to determine the forces inside: on the left, shear down, normal out, moment up; on the right, shear up, normal out, moment up.

🔑 Real-world application

Example: A bridge that has different loads applied (from cars, trucks, lampposts, etc.). Use this method to calculate the internal loads at a particular point of interest to ensure the structure can safely carry the loads.

🔑 Looking ahead

In the next section, the excerpt mentions calculating the internal force across the whole beam and displaying the results graphically—this refers to constructing complete shear and moment diagrams for the entire beam length.

🧭 Overview

🧠 One-sentence thesis

Shear and moment diagrams graphically display how internal shear forces and bending moments vary along a beam, enabling engineers to identify critical stress points for safer design.

📌 Key points (3–5)

• What they are: graphical representations of internal shear force (V) and bending moment (M) along the entire length of a beam.
• Mathematical relationships: the derivative of moment equals shear (dM/dx = V), and the derivative of shear equals negative distributed load (dV/dx = -w); conversely, the integral of shear gives change in moment, and the integral of load gives change in shear.
• How to produce them: calculate reactions, make cuts at key points, write equilibrium equations for V and M as functions of distance x, then plot these functions.
• Common confusion: sign conventions—positive shear and positive moment have specific directional meanings; positive moments are drawn above the neutral axis, negative moments below.
• Shape rules: constant distributed load → linear shear slope and parabolic moment slope; zero load → constant shear and linear moment; where V = 0, M reaches a maximum or minimum.

📊 What are shear and moment diagrams?

📊 Shear force diagram

A graphical representation of the variation of the shearing force on a portion or the entire length of a beam or frame.

• The x-axis represents position along the beam (lined up with the beam's length).
• The y-axis represents the magnitude of the internal shear force at each location.
• Sign convention: can be drawn above or below the x-centroidal axis, but must indicate whether positive or negative.
• Positive shear: upward internal force to the right of the cross section, downward force to the left.
• Negative shear: downward internal force to the right, upward force to the left.

📊 Bending moment diagram

A graphical representation of the variation of the bending moment on a segment or the entire length of a beam or frame.

• The x-axis represents position along the beam.
• The y-axis represents the magnitude of the internal bending moment.
• Sign convention: positive bending moments are drawn above the x-centroidal axis; negative bending moments are drawn below.
• Positive moment: causes the beam to bow downwards (smile shape).
• Negative moment: causes the beam to bow upwards (frown shape).

🔗 Mathematical relationships between load, shear, and moment

🔗 Derivative relationships

The excerpt derives three key equations from equilibrium on an infinitesimal beam segment of length dx:

Equation	Meaning in words
dM/dx = V(x)	The slope of the moment diagram at any point equals the shear force at that point
dV/dx = -w(x)	The slope of the shear diagram equals the negative of the distributed load intensity
d²M/dx² = -w(x)	The second derivative of moment equals the negative of the distributed load intensity

• Why this matters: these relationships let you predict the shape of one diagram from another.
• Example: if distributed load w is constant, then dV/dx is constant (linear shear), and d²M/dx² is constant (parabolic moment).

🔗 Integral relationships

The same relationships can be expressed as integrals:

Equation	Meaning in words
ΔM = ∫V(x)dx	The change in moment between two points equals the area under the shear diagram between those points
ΔV = ∫w(x)dx	The change in shear equals the area under the load diagram

• These are useful for calculating values at specific points without writing full equations.
• Example: if shear is constant at -5 lb over 3 ft, then ΔM = (-5 lb)(3 ft) = -15 ft·lb.

🔗 Shape prediction from load type

Load type	Shear diagram shape	Moment diagram shape
No distributed load (w = 0)	Constant (horizontal line)	Linear (straight sloped line)
Constant distributed load	Linear (sloped line)	Parabolic (curved)
Linear distributed load	Parabolic	Cubic

• Don't confuse: the shape of the moment diagram is the integral of the shear diagram shape, not the same shape.

🛠️ How to produce shear and moment diagrams

🛠️ General procedure (equilibrium method)

The excerpt describes a step-by-step method using equilibrium equations:

1. Draw a free-body diagram of the entire structure.
2. Calculate reactions using equilibrium equations (ΣFx = 0, ΣFy = 0, ΣM = 0). For cantilever beams, you may skip this if starting from the free end.
3. Make cuts at key locations and add internal forces N, V, and M using positive sign convention. You need multiple cuts if there are multiple loads along the beam.
4. For shear: write an equilibrium equation (ΣFy = 0) to find V as a function of distance x from a reference point (often the reaction or free end).
5. For moment: write an equilibrium equation (ΣM = 0) to find M as a function of x.
6. Plot the V(x) and M(x) equations on aligned graphs, with the beam position on the x-axis.

• Key point: you must use the positive sign convention for internal forces when making cuts.
• Example: in Example 1, a cantilever beam with a 5 lb downward load at the free end yields V = -5 lb (constant) and M = -5x (linear), where x is measured from the free end.

🛠️ Alternative graphical method

The excerpt also describes a method that builds the diagrams step-by-step without writing equations:

For shear diagram:

• Start at zero on one end.
• Move along the beam; keep the line steady except:
  • Jump up by the magnitude of any upward point force.
  • Jump down by the magnitude of any downward point force.
  • Slope linearly through any uniformly distributed load (slope = magnitude of w; positive slope for upward load, negative for downward).
  • For non-uniform distributed loads, the shear plot shape is the integral of the load function.
• Ignore moments and horizontal forces.
• You should return to zero at the other end (a check on your work).

For moment diagram:

• Start at zero on one end.
• The moment diagram is primarily the integral of the shear diagram, except:
  • Jump up by the magnitude of any negative (clockwise) applied moment.
  • Jump down by the magnitude of any positive (counter-clockwise) applied moment.
• Ignore forces in the free body diagram.
• You should return to zero at the other end.

🛠️ Key locations for cuts

You need to calculate internal forces at:

• Points where loads are applied.
• Start and end of distributed loads.
• Reaction points.
• Any point of interest for design.

📐 Sign conventions and diagram features

📐 Sign conventions recap

Quantity	Positive convention	Negative convention
Shear force	Upward force to the right of cut, downward to the left	Downward force to the right, upward to the left
Bending moment	Drawn above neutral axis; beam bows downward (smile)	Drawn below neutral axis; beam bows upward (frown)

• Common confusion: the sign of V or M in your equation tells you the direction relative to the assumed positive convention, not just magnitude.
• Example: V = -5 lb means the shear acts opposite to the assumed positive direction.

📐 Starting and ending values

Beam type	Shear diagram	Moment diagram
Cantilever (fixed end)	Starts with nonzero V and M at fixed end; ends at zero at free end	Same
Simply supported	Starts and ends with reaction forces (jumps at supports)	Starts and ends at zero

• Why: at a free end, no external force or moment is applied, so internal V and M must be zero. At a simple support (pin or roller), no moment is resisted, so M = 0.

📐 Jumps and inflection points

• In shear diagram: vertical jumps occur where point forces are applied (including reactions); jump direction matches force direction.
• In moment diagram: vertical jumps occur where applied moments are present; jump direction matches moment direction.
• Inflection points in M: occur where the slope changes sign; these correspond to V = 0 in the shear diagram.
• Max/min of M: occur where V = 0.

🔍 Interpreting the diagrams

🔍 Reading relationships between plots

The excerpt provides several rules for checking consistency:

• Positive V → M is increasing (positive slope).
• Negative V → M is decreasing (negative slope).
• V = 0 → M is at a maximum or minimum (horizontal tangent).
• Positive load intensity → V is increasing.
• Negative load intensity → V is decreasing.
• Inflection point in M (where concavity changes) → zero crossing in V.

🔍 Derivative/integral progression

The excerpt includes a visual showing the relationship:

• Going down (taking derivatives): moment → shear → load intensity.

  • Quadratic → linear → constant → zero.

• Going up (taking integrals): load intensity → shear → moment.

  • Zero → constant → linear → quadratic → cubic.

• Example: if the moment diagram is parabolic (quadratic), the shear diagram will be linear, and the load will be constant.

🔍 Practical use

• Why it matters: shear and moment diagrams identify where internal forces are largest, which are the critical points for stress analysis and design.
• The excerpt states: "This can help you identify the major stress points to provide a safer design."
• Engineers use these diagrams to size beams, choose materials, and place reinforcements.

📝 Worked examples summary

📝 Example 1: Cantilever with point load

• Setup: cantilever beam, 3 ft long, 5 lb downward load at free end.
• Reactions: at fixed end B, Bₓ = 0, Bᵧ = 5 lb, M_B = 15 ft·lb.
• Shear: V = -5 lb (constant along entire beam).
• Moment: M = -5x (linear, where x is measured from free end); at x = 0, M = 0; at x = 3 ft, M = -15 ft·lb.
• Diagram shapes: shear is a horizontal line at -5 lb; moment is a straight line from 0 to -15 ft·lb.
• Check: reaction force at fixed end brings shear back to zero (jump from -5 to 0).

📝 Example 2: Cantilever with uniform distributed load

• Setup: cantilever beam, 5 m long, uniformly distributed load along entire length.
• Reactions: at fixed end B, vertical reaction Bᵧ = 100 kN, moment M_B.
• Shear function: V varies linearly from 0 at free end to -100 kN at fixed end.
• Moment function: M is parabolic (quadratic in x); varies from 0 at free end to a maximum negative value at fixed end.
• Diagram shapes: shear is a sloped straight line; moment is a parabolic curve.
• Why parabolic: because the distributed load is constant, dV/dx is constant (linear V), and dM/dx = V (so M is the integral of a linear function, which is quadratic).

📝 Additional examples (3–6)

The excerpt references Examples 3–6 with diagrams but does not provide detailed solution steps. These show various combinations of point loads, distributed loads, and support types, illustrating the range of shear and moment diagram shapes.

🧩 Tips and common patterns

🧩 General rules (with exceptions noted)

• +V means increasing M: when shear is positive, the moment diagram slopes upward.
• -V means decreasing M: when shear is negative, the moment diagram slopes downward.
• V = 0 indicates max or min M: where the shear crosses zero, the moment reaches a local extremum.

🧩 Checking your work

• End values: for simply supported beams, both V and M should return to zero at the ends (after accounting for reaction jumps). For cantilevers, V and M should be zero at the free end.
• Area under shear = change in moment: you can verify moment values by integrating (or summing areas) under the shear diagram.
• Consistency: the slope of M at any point should match the value of V at that point.

🧩 Online tools (not for exams)

The excerpt mentions online beam calculators (SkyCiv, ClearCalcs, BeamGuru) as learning aids to confirm diagram shapes, but notes these are not acceptable for exams or homework.

🧭 Overview

🧠 One-sentence thesis

This excerpt provides worked examples of internal force analysis, shear and moment diagrams, and beam cutting techniques, demonstrating how to apply equilibrium equations to real-world structural problems.

📌 Key points (3–5)

• What the examples cover: calculating reaction forces, internal forces (N, V, M), and drawing shear/moment diagrams for beams under various loads (point loads, distributed loads, triangular loads).
• Core method: use equilibrium equations (sum of forces = 0, sum of moments = 0) to find reactions, then "cut" the beam at points of interest to solve for internal forces.
• Relationship between shear and moment: the moment diagram is the integral of the shear diagram; constant shear → linear moment, linear shear → quadratic moment.
• Common confusion: distributed loads must be resolved into equivalent point loads (magnitude = load intensity × length) acting at the centroid of the load distribution before applying equilibrium equations.
• Sign conventions: negative results for reaction or internal forces indicate the assumed direction was wrong; the force acts in the opposite direction.

🔧 Core method: equilibrium and beam cuts

🔧 Finding reaction forces

• Step 1: Draw a free-body diagram showing all applied loads and unknown reactions.
• Step 2: Apply equilibrium equations:
  • Sum of forces in x = 0 (horizontal equilibrium)
  • Sum of forces in y = 0 (vertical equilibrium)
  • Sum of moments about any point = 0 (rotational equilibrium)
• Step 3: Solve the system of equations for the unknown reactions.
• Example: In Example 6.3.2, a simply supported beam with two point loads (500 N at 2 m, 300 N at 5.5 m) and total length 8 m yields reactions A_y = 468.75 N and C_y = 331.25 N by taking moments about A.

✂️ Cutting the beam to find internal forces

• What "cutting" means: imagine slicing the beam at a point of interest and analyzing one side as a separate free body.
• Internal forces exposed: normal force N (axial), shear force V (perpendicular to beam), and bending moment M (rotational).
• Apply equilibrium to the cut section: use the same three equilibrium equations on the exposed section.
• Example: In Example 6.3.1, cutting at point C (0.2 m from A) on a beam with distributed load 220 N/m gives V_c = -206.25 N and M_c = 64.35 N·m.
• Don't confuse: the cut can be made anywhere along the beam, but you must include all forces and moments acting on the chosen side up to the cut location.

📊 Distributed loads and equivalent forces

📊 Resolving distributed loads

Distributed load: a load spread over a length, measured in force per unit length (e.g., N/m).

• Convert to equivalent point load: multiply load intensity by the length over which it acts.
  • Formula: F_equivalent = (load intensity) × (length)
• Location of equivalent load: acts at the centroid of the load distribution.
  • Uniform rectangular load: centroid at midpoint (length / 2)
  • Triangular load: centroid at 1/3 of the base length from the zero end
• Example: In Example 6.3.1, a 220 N/m load over 1.9 m becomes F_R = 418 N acting at 0.95 m from one end.
• Example: In Example 6.3.5, a triangular load from 0 to 4 N/m over 5.5 m becomes 11 N acting at 1.83 m from the start of the load (1/3 × 5.5 m).

📊 Handling partial distributed loads in cuts

• When a cut passes through a distributed load region, only the portion of the load up to the cut is included.
• The equivalent load magnitude and location must be recalculated for the partial length.
• Example: In Example 6.3.5, when cutting in the triangular load region, the slope of the load is calculated as m = (56 - 4x) / 5.5, and the equivalent force becomes F_eq = (4x² / 11) - (112x / 11) + (784 / 11).

📈 Shear and moment diagrams

📈 Relationship between shear and moment

• Key principle: the moment M is the integral of the shear V; equivalently, dM/dx = V.
• Shape rules:
  • Constant shear (horizontal line) → linear moment (sloped line)
  • Linear shear (sloped line) → quadratic moment (curved parabola)
  • Quadratic shear → cubic moment
• Example: In Example 6.3.4, section A-B has a distributed load creating linear shear; integrating gives a quadratic (curved) moment diagram.
• Don't confuse: the slope of the moment diagram at any point equals the shear value at that point; if shear is negative, moment slope is negative (decreasing).

📈 Constructing the diagrams

• Shear diagram (V-x):
  • Plot shear force as a function of position x along the beam.
  • Point loads cause sudden jumps (discontinuities) in shear.
  • Distributed loads cause linear or curved changes in shear.
• Moment diagram (M-x):
  • Plot bending moment as a function of position x.
  • Moment must return to zero at free ends (for static beams).
  • Maximum moment often occurs where shear crosses zero.
• Example: In Example 6.3.3, a symmetric beam with a central load has shear = +15 N from A to B, then -15 N from B to C; moment is linear in each section, peaking at the center (0.6 N·m).

📈 Checking your work

• Static equilibrium checks:
  • Shear and moment must return to zero at free ends.
  • Reaction forces must balance applied loads.
• Sign consistency: if a calculated force is negative, redraw the arrow in the opposite direction in the final diagram.
• Example: In Example 6.3.2, the final shear/moment diagrams return to zero at both ends, confirming correctness.

🧮 Worked example patterns

🧮 Simply supported beams

• Setup: beam supported by a pin at one end and a roller at the other.
• Reactions: pin provides x and y forces; roller provides only y force.
• Example: Example 6.3.6 (diving board) has a pin at 0 ft and roller at 4 ft; a 140 lb load at 10 ft gives R_Ay = -210 lb (downward) and R_By = 350 lb (upward).
• Why the pattern: the roller at 4 ft must counteract the moment from the 140 lb load at 10 ft, creating a large upward force; the pin then balances the net vertical force.

🧮 Cantilever beams (fixed support)

• Setup: beam rigidly fixed at one end, free at the other.
• Reactions: fixed support provides x force, y force, and a moment.
• Example: Example 6.3.7 (wrench) has a fixed support at A; a 200 N downward force at 0.3 m gives A_y = 200 N (upward) and M_A = -60 N·m (clockwise).
• Internal forces: shear remains constant (200 N) along the wrench; moment varies linearly (M(x) = 200x - 60 N·m).

🧮 Symmetric loading

• Pattern: if loads and geometry are symmetric, reactions are equal.
• Example: Example 6.3.3 has a central load on a symmetric beam; reactions at both ends are equal (15 N each).
• Example: Example 6.3.8 (light frame) has symmetric rods; tensions are equal (T₁ = T₂ = 26.98 N).
• Don't confuse: symmetry in geometry does not guarantee symmetric reactions if loads are asymmetric (see Example 6.3.6, where the load is not centered).

🔍 Common pitfalls and checks

🔍 Sign conventions and direction

• Negative results: a negative value means the assumed direction was wrong; flip the arrow.
• Example: In Example 6.3.1, A_y = -52.25 N indicates the reaction at A points downward, not upward as initially drawn.
• Moment sign: positive moment is typically counterclockwise; negative is clockwise (or vice versa, depending on convention—be consistent).

🔍 Distributed load centroids

• Rectangular load: centroid at the midpoint of the length.
• Triangular load: centroid at 1/3 of the base from the zero end.
• Don't confuse: the centroid is measured from the start of the load, not from the beam origin; add any offset to find the position from the beam origin.
• Example: In Example 6.3.5, a triangular load starts at 4 m + 4.5 m = 8.5 m from A; its centroid is at 8.5 m + 1.83 m = 10.3 m from A.

🔍 Reviewing your answer

• Equilibrium check: sum all forces and moments on the entire beam; they should equal zero.
• Boundary conditions: shear and moment at free ends must be zero; moment at a pin or roller is zero (unless a couple is applied).
• Example: In Example 6.3.9, substituting the calculated internal forces back into equilibrium equations for the left side confirms the solution.

🧭 Overview

🧠 One-sentence thesis

This section demonstrates how to apply equilibrium equations, couples, distributed loads, and lever arms to solve real-world statics problems involving forces, moments, and reaction forces.

📌 Key points (3–5)

• Purpose of examples: Real-world problems submitted by students to illustrate Chapter 3 concepts (reaction forces, couples, distributed loads, lever arms).
• Core method: All examples use equilibrium equations (sum of forces = 0, sum of moments = 0) and/or couple/distributed load formulas.
• Common pattern: Each example follows a six-step structure: Problem → Draw → Knowns/Unknowns → Approach → Analysis → Review.
• Key distinction: Couples produce moments without net force (equal and opposite forces); distributed loads require integration to find resultant force and location.
• Review step importance: Every solution checks whether the answer makes physical sense (sign, magnitude, direction).

📐 Problem-solving structure

📐 Six-step framework

Every example follows the same structure:

1. Problem: State the scenario and what to find
2. Draw: Sketch and free-body diagram
3. Knowns and Unknowns: List given values and what to solve for
4. Approach: Identify which equations/methods to use
5. Analysis: Perform calculations
6. Review: Check if the answer makes physical sense

🎯 Why this structure matters

• The "Review" step consistently checks signs, magnitudes, and whether results align with physical intuition.
• Example: In the couch problem, the review notes that reaction force B is larger than A even though the combined weight near A is greater—because distance from the pivot matters, not just force magnitude.

⚖️ Reaction forces and equilibrium

⚖️ Couch example (Example 3.6.1)

Setup: A 5 m couch with two roller supports holds a family (child 30 kg at 1 m, mother 60 kg at 1.5 m, father 70 kg at 4.5 m); couch weighs 120 N.

Method:

• Sum of moments about point A = 0 to find reaction force at B
• Sum of vertical forces = 0 to find reaction force at A

Key insight from review:

• N_B (913 N) > N_A (776 N) even though more weight is closer to A
• Distance matters: father is 0.5 m from B but child is 1 m from A
• Both reactions are positive (upward), as expected

🔍 Don't confuse

Larger nearby weight ≠ larger reaction force; the lever arm (distance from support to force) is equally important in moment calculations.

🔄 Couples and moments

🔄 What is a couple

Couple: Two equal and opposite parallel forces separated by a distance, producing a pure moment (rotation) without net translational force.

Formula: M = F × d (force times perpendicular distance)

🚰 Water valve example (Example 3.6.2)

• Wheel diameter 10 inches, force 7.5 lb on each side
• Method: Calculate moment from each force using cross product, then add
• Result: M = 6.3 ft·lb
• Alternative shortcut: M = F × d = 7.5 lb × (10 in) = 6.25 ft·lb (slightly more accurate without rounding)

🚗 Steering wheel examples (Examples 3.6.4–3.6.6)

Example	Setup	Key finding
3.6.4 (Mason jar)	5.5 Nm moment, 11 cm diameter	Each finger applies 50 N
3.6.5 (Two hands)	115 N each hand, 40 cm diameter	Couple moment = 46 Nm; one hand needs 230 N (double)
3.6.6 (Unequal forces)	Right 200 N, left 150 N, 15 in diameter	Right turn: -66.675 N·m; left turn: +66.675 N·m (same magnitude, opposite direction)

🎯 Key insight

When both hands apply equal force on opposite sides of a wheel, the moment is a couple. Using one hand requires double the force to produce the same moment because the lever arm is halved.

📊 Distributed loads

📊 What is a distributed load

Distributed load: Force spread continuously over a length (e.g., objects on a shelf), described by a function w(x) in force per unit length.

Key formulas:

• Resultant force: F_r = ∫ w(x) dx (area under the load curve)
• Location of resultant: x_r = [∫ x·w(x) dx] / [∫ w(x) dx] (weighted average position)

📚 Shelf example (Example 3.6.3)

• Load function: w = 4x⁴ + 2 N/m over 1 m shelf
• Resultant force: F_r = 2.8 N
• Location: x_r = 0.59 m from the wall
• Review note: The function increases along the shelf, so the resultant is closer to the right end (0.59 m vs 0.5 m midpoint)—makes sense.

🚗 Bridge with cars example (Example 3.6.8)

Setup: 5 identical cars (1000 N/m distributed load) on a bridge; 3 in one lane, 2 in the other.

Method:

• Each car modeled as rectangle + triangle distributed load
• One car: F = 2250 N located at 1.17 m from its front
• Three cars: F_R3 = 6750 N at 6.17 m from left
• Two cars: F_R2 = 4500 N at 6.33 m from left
• All five: F_R5 = 11250 N at (6.234 m, 3.6 m)

Key insight: The x-coordinate is closer to the side with more cars; the y-coordinate is less than halfway because that lane has more vehicles.

🎪 Lever arms and balance

🎪 Seesaw example (Example 3.6.7)

Problem: Two children (35 kg and 50 kg) should apply equal force on a seesaw.

Method:

• For equal moments: m₁ · R₁ = m₂ · R₂
• Gravity cancels from both sides
• Ratio: R₁ : R₂ = 50 : 35 = 10 : 7

Review: The lighter child must sit farther away (10 vs 7) to balance the heavier child—matches the weight ratio and makes physical sense.

🔍 Don't confuse

Equal force ≠ equal moment. A lighter person farther from the pivot can produce the same moment as a heavier person closer to the pivot.

✅ Common review checks

✅ What to verify in every solution

• Sign: Does the direction (positive/negative, clockwise/counterclockwise) make sense?
• Magnitude: Is the number reasonable for the scenario?
• Symmetry: If the problem has symmetry, do results reflect it?
• Limiting cases: If one variable changes, does the answer change as expected?

Example from 3.6.5: When using one hand instead of two on a steering wheel, the force requirement doubles—exactly as expected because the lever arm is halved.

Example from 3.6.6: Left and right turns produce moments of equal magnitude but opposite sign—correct because forces and positions have the same magnitudes but opposite directions.

🧭 Overview

🧠 One-sentence thesis

External forces in statics—gravity, normal force, friction, spring force, and applied forces—are all quantifiable forces that must balance to keep objects stationary, and understanding how to calculate and resolve them is essential for analyzing rigid bodies.

📌 Key points (3–5)

• What external forces include: gravitational force (weight), normal force, frictional force, spring force, and applied forces (including tension and moments).
• Fundamental nature: normal, friction, spring, and applied forces are all types of electromagnetic forces at the atomic level; gravity is a separate fundamental force.
• Normal force adjusts to context: on a horizontal surface N = mg, but on an incline N = mg cos θ because weight splits into perpendicular and parallel components.
• Tension is always a pull: flexible connectors (ropes, cables) can only pull along their length, never push; tension equals the supported weight when the system is stationary.
• Common confusion: weight components on an incline—the angle θ of the incline equals the angle between total weight and the perpendicular component, so use trigonometry (not memorization) to resolve w_x = mg sin θ and w_y = mg cos θ.

🌍 Categories of external forces

🌍 The five main types

The excerpt lists five categories of external forces used in statics:

Force type	Calculation method	Notes
Gravitational (weight)	F_g = mg	Always present, acts downward
Normal	Calculated from equilibrium	Perpendicular to contact surface
Frictional	F_f = μN	Resistive force opposing motion
Spring	F_S = -kx	Restoring force proportional to deformation
Applied	Measured or calculated	Includes reaction forces, tension, and moments from motors

⚛️ Fundamental forces background

• At the atomic level, normal force, friction, spring force, and applied forces are all electromagnetic forces—charged and neutral particles attracting or repelling each other.
• Example: your laptop doesn't fall through the table because electrons in both objects repel each other (electromagnetic) while gravity pulls both down.
• The four fundamental forces (gravitational, electromagnetic, weak nuclear, strong nuclear) are beyond the scope of statics, but knowing this background helps understand how external forces operate.
• Don't confuse: the named forces in statics (normal, friction, etc.) are practical categories for calculation, not separate fundamental forces.

⬆️ Normal force

⬆️ What normal force is

Normal force: a force perpendicular to the surface of contact between a load and its support, equal in magnitude and opposite in direction to the component of weight perpendicular to that surface.

• The word "normal" means perpendicular to a surface.
• Symbol: N (with vector notation N̅; not the newton unit).
• Whatever supports a load—animate or inanimate—must supply an upward force equal to the weight component pressing into the surface.

🛋️ How objects support loads

• When a load (e.g., a bag of dog food) is placed on a table, the table sags slightly under the load.
• The deformation creates a restoring force, much like a compressed spring or trampoline.
• The greater the deformation, the greater the restoring force.
• The table sags until the restoring force (normal force) equals the weight of the load; at that point, net external force is zero and the load is stationary.
• Example: a card table sags noticeably; a sturdy oak table deforms less, but still deforms.

📐 Normal force on a horizontal surface

• When an object rests on a horizontal surface, the normal force in vector form is:
  • N̅ = -mg̅ (opposite direction to weight)
• In scalar form:
  • N = mg
• This assumes the surface is horizontal and the object is stationary.

🔺 Normal force on an incline

• When an object rests on an incline at angle θ to the horizontal, gravity splits into two components:
  • w_y: perpendicular to the plane (into the surface)
  • w_x: parallel to the plane (down the slope)
• The normal force equals the perpendicular component:
  • N = mg cos θ
• The parallel component w_x = mg sin θ causes the object to accelerate down the incline (if not held in place).

🧮 How to resolve weight components without memorizing

• Draw the right triangle formed by the three weight vectors: total weight w, perpendicular component w_y, and parallel component w_x.
• The angle θ of the incline is the same as the angle between w and w_y.
• Use trigonometry:
  • cos θ = w_y / w → w_y = w cos θ = mg cos θ
  • sin θ = w_x / w → w_x = w sin θ = mg sin θ
• Don't confuse: the incline angle θ is not the angle between w_x and the horizontal; it's the angle between the total weight and the perpendicular component.

🪢 Tension

🪢 What tension is

Tension: a pulling force along the length of a flexible connector (rope, cable, chain, wire, tendon) that acts parallel to the connector.

• The word "tension" comes from Latin "to stretch."
• A flexible connector can only pull, never push—"you can't push a rope."
• Tension pulls outward along the two ends of a connector.

⚖️ Tension in a stationary system

• Consider a person holding a mass on a rope. If the mass is stationary (acceleration = 0), then net force = 0.
• The only external forces on the mass are its weight w and the tension T from the rope.
• Force balance: T - w = 0 → T = w = mg
• Example: for a 5.00 kg mass, T = (5.00 kg)(9.80 m/s²) = 49.0 N.
• If you cut the rope and insert a spring, the spring would extend to exert 49.0 N, directly measuring the tension.

🔗 Tension is uniform along a rope

• A perfectly flexible connector (one requiring no force to bend) transmits force parallel to its length.
• By Newton's third law, the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the rope's own weight).
• The tension is equal at all locations along the rope between the hand and the mass.
• Once you determine tension at one location, you know it everywhere along the rope.

🔄 Tension around corners

• Flexible connectors transmit forces around corners (e.g., hospital traction systems, tendons, bicycle brake cables).
• If there is no friction, the tension magnitude is unchanged; only its direction changes.
• The tension remains parallel to the flexible connector at every point.

📏 Creating large tension with perpendicular force

• If a perpendicular force F⊥ is applied at the middle of a taut flexible connector, the tension is:
  • T = F⊥ / (2 sin θ)
• θ is the angle between the horizontal and the bent connector.
• As θ approaches zero (connector becomes more horizontal), T becomes very large.
• Even a small weight causes a flexible connector to sag, because infinite tension would be needed to keep it perfectly horizontal (sin θ = 0).
• Example: pulling a car out of mud with a chain—push perpendicular to a taut chain to create large tension; the smaller θ, the larger T.
• Don't confuse: this is analogous to a tightrope walker, but the tensions shown are those transmitted to the anchors (car and tree), not at the point where F⊥ is applied.

🧲 Friction and spring forces

🧲 Friction

Friction: a resistive force opposing motion.

• Calculation: F_f = μN (where μ is the coefficient of friction and N is the normal force).
• The excerpt notes friction is a resistive force but does not elaborate further in this section.

🔩 Spring force

• Calculation: F_S = -kx (where k is the spring constant and x is the displacement from equilibrium).
• The negative sign indicates the force is a restoring force—it acts opposite to the direction of deformation.
• Spring force is proportional to deformation: greater stretch or compression → greater restoring force.

🧭 Overview

🧠 One-sentence thesis

Rigid body free-body diagrams (FBDs) differ from particle FBDs by requiring precise placement of force arrows at their actual points of application, enabling engineers to isolate and analyze forces acting on objects or systems of objects.

📌 Key points (3–5)

• Key difference from particle FBDs: In rigid body FBDs, the location where each force is applied matters—force arrows must point to the exact application point, not just the center of mass.
• Core purpose: FBDs "free" the body from surrounding objects and surfaces so it can be studied in isolation with all forces and moments drawn in.
• System vs. part FBDs: A system FBD shows external forces on multiple objects together; part FBDs show individual objects with interaction forces replacing the removed parts.
• Common confusion: Internal forces (forces between parts of the same body) are not included in a single-body FBD; only external forces acting directly on the body are shown.
• Why it matters: FBDs are the first step in solving statics, dynamics, and strength-of-materials problems—they simplify complex scenarios into solvable equilibrium or motion equations.

🎯 What makes rigid body FBDs different

🎯 Precision in force location

• Unlike particle FBDs (where all forces can be shown at a single point), rigid body FBDs require you to draw the force arrow's head at the exact location where the force is applied.
• Example: In a high-five scenario, the applied force arrow points to the hand contact point, not to the center of mass of the person.
• This precision is necessary because the location of a force affects rotational effects (moments) on the rigid body.

🎯 Center of mass for gravity

• Gravity acts on every particle in an object, but drawing millions of tiny arrows is impractical.
• Instead, the gravitational force is concentrated at the center of mass (often the geometric center).
• This simplification works because we typically know the total mass and its center location.

🔧 How to construct a part FBD

🔧 The four-step process

1. Draw the shape of the body being analyzed.
2. Add a coordinate frame (define positive x and positive y directions).
3. Replace surfaces and contacts with force arrows (floor becomes normal and friction forces; hands become applied forces).
4. Label each force uniquely (use different names for each force).

📍 Rules for force placement and direction

Force type	Direction rule	Location rule
Gravitational	Always downward toward Earth	At the center of mass
Normal	Perpendicular to the contact surface	At the contact point
Friction	Parallel to the surface; opposes motion	Along the contact surface
Spring	Opposite to displacement direction	Along the spring axis
Applied (push/pull)	Toward or away from the object	At the point of application
Tension (cables)	Along the cable direction; always pulling	At the attachment point

📍 Important force behaviors

• Normal forces: Always act 90 degrees from the surface. If the ground is angled, the normal force is perpendicular to that angled surface.
• Friction: Acts parallel to the plane between two surfaces (a shear force). Always opposes the direction of motion or tendency to move.
• Spring force: Shown as negative because it acts opposite to the displacement. The force is proportional to displacement from the relaxed position (Hooke's law: force equals negative k times displacement, where k is the spring's stiffness).
• Tension in cables: Always pulls; cables cannot push. The force acts along the cable direction.

📖 Example: Book pushed across a table

For a book being pushed by a hand:

• Normal force (green arrow): on the bottom surface of the book, pointing up perpendicular to the table.
• Friction force (yellow arrow): along the bottom surface, opposing the motion (pointing opposite to the push).
• Gravitational force (pink arrow): at the center of mass, pointing downward.
• Applied force (blue arrow): at the point where the hand contacts the book, pointing in the direction of the push.

If the book were pulled by a string instead, the applied force and friction force would reverse direction (friction still opposes motion).

🚫 What not to include

• Do not include internal forces: Forces between parts of the same body are not shown on a single-body FBD.
• Do not include forces that don't act directly on the body: Only forces acting on the body being analyzed are drawn.

🔗 System FBDs and multiple objects

🔗 What is a system FBD

A system free-body diagram treats multiple parts together, showing only the external forces acting on the combined system.

• When objects interact (e.g., two books stacked), you can analyze them at two levels:
  • System level: All objects together, with only external forces (gravity, applied, normal, friction, spring).
  • Part level: Each object separately, with interaction forces (forces between the objects) replacing the removed parts.

🔗 Why use system FBDs

• Helpful when you have more unknowns than equations at the system level.
• By splitting the system into individual part FBDs, you can find additional equations to solve for unknowns.
• Each part FBD includes interaction forces (typically two forces: one vertical, one horizontal) where the other part was removed.

📖 Example: Two stacked books

You need three FBDs:

1. System-level FBD: Both books together, showing external forces (gravity on each book, normal force from the table, friction from the table, any applied force).
2. Part FBD for the bottom book: The top book is replaced by force arrows (the normal force and friction force the top book exerts on the bottom book).
3. Part FBD for the top book: The bottom book is replaced by force arrows (the normal force and friction force the bottom book exerts on the top book).

🔗 Steps to make a system FBD

1. Draw the system FBD with all parts together.
2. Use unique, consistent labels for each part (e.g., a letter or number per part).
3. Show only external forces acting on the system as a whole.
4. The system should appear "floating" in space, separated from all surroundings.

🛠️ Common force details

🛠️ Reaction forces at joints and connections

• Joints or connections between bodies cause reaction forces or moments.
• You have one force or moment for each type of motion or rotation the connection prevents.

Connection type	What it allows	What it prevents	Reaction forces/moments
Roller	Rotation and movement along surface	Motion perpendicular to surface	Normal force in one direction (e.g., y direction)
Pin joint	Rotation	Motion in all directions	Normal forces in x and y directions
Fixed connection	Nothing (fully constrained)	Motion in all directions and rotation	Normal forces in all directions plus a reaction moment

🛠️ Friction assumptions

• Smooth surfaces: Assume no friction force; only normal force is present.
• Rough surfaces: Assume bodies will not slide relative to one another, no matter what. The friction force is always just large enough to prevent sliding.

🛠️ Tension in flexible connectors

• Cables, wires, or ropes exert a tension force in the direction of the cable.
• Tension always pulls on the body; flexible connectors cannot push.
• If there is no friction, tension transmission is undiminished around corners; only the direction changes, always parallel to the connector.

🛠️ Creating large tension

• To create a large tension in a flexible connector, exert a force perpendicular to a taut connector.
• The tension T is related to the perpendicular force by: T equals the perpendicular force divided by (2 times sine of the angle theta), where theta is the angle between the horizontal and the bent connector.
• As theta approaches zero, T becomes very large. Even a small weight causes a connector to sag, because infinite tension would result if the connector were perfectly horizontal (theta equals zero, sine of theta equals zero).
• Example: Pulling a car out of mud with a chain—push perpendicular to the chain to create large tension; since theta is small, T is large.

🛠️ Other forces

• Pressure from fluids, spring forces, and magnetic forces may also act on a body and should be included if present.

📐 Final steps and labeling

📐 Key dimensions and angles

• After identifying and drawing all forces, label any key dimensions and angles on the diagram.
• This information is necessary for writing equilibrium equations or equations of motion.

📐 Purpose of the FBD

• The simplified diagram allows you to more easily write equilibrium equations for statics or strengths-of-materials problems, or equations of motion for dynamics problems.
• Constructing the FBD is the first step in solving most mechanics problems.

🧭 Overview

🧠 One-sentence thesis

Rigid body equilibrium equations extend particle equilibrium by adding moment equations, enabling engineers to solve for more unknowns when forces act at different points on a non-deformable body.

📌 Key points

• What equilibrium equations do: calculate unknown forces and moments using known values and the requirement that sums equal zero for static bodies.
• Rigid body vs particle: rigid bodies add moment equations (rotation) to force equations (translation), increasing the number of solvable unknowns.
• 2D vs 3D: two-dimensional problems have 3 equations (two force directions + one moment axis); three-dimensional problems have 6 equations (three force directions + three moment axes).
• Common confusion: moments are vectors that must be broken into scalar components just like forces; in 2D, only the z-axis moment matters (clockwise/counterclockwise rotation).
• Why it matters: the additional moment equations allow solving for reaction forces, support forces, and other unknowns that particle equilibrium cannot handle.

🧮 Core equilibrium equations

🧮 Particle equilibrium (review)

Particle equilibrium equations: sum of forces in x, y, and z directions equals zero.

• The excerpt reviews particle equilibrium from section 2.3: sum of F_x = 0, sum of F_y = 0, sum of F_z = 0.
• These equations assume forces are concurrent (act at a single point), so no rotation is considered.
• Rigid body equilibrium builds on this foundation by adding moments.

🔄 Rigid body equilibrium in 2D

Rigid body two-dimensional equilibrium: sum of F_x = 0, sum of F_y = 0, sum of M_O = 0.

• Three equations total: two for translation (x and y force components) and one for rotation (moment about the z-axis).
• The moment equation accounts for clockwise or counterclockwise rotation.
• The excerpt emphasizes that "the addition of moments (as opposed to particles, where we only looked at the forces) adds another set of possible equilibrium equations, allowing us to solve for more unknowns."
• Example: A beam with people standing on it—vertical forces must sum to zero, and moments about any point must also sum to zero to find reaction forces at supports.

🌐 Rigid body equilibrium in 3D

Rigid body three-dimensional equilibrium: sum of F_x = 0, sum of F_y = 0, sum of F_z = 0, and sum of M_x = 0, sum of M_y = 0, sum of M_z = 0.

• Six equations total: three for translation (x, y, z force components) and three for rotation (moments about x, y, z axes).
• The body may have moments about each of the three axes, so each axis gets its own moment equation.
• The excerpt states: "If we look at a three-dimensional problem, we will increase the number of possible equilibrium equations to six."

⚖️ Why the right side equals zero

• The excerpt specifies: "Because these are static bodies, the right side of the equations equals 0."
• For a rigid body in static equilibrium, the sum of both forces and moments must be zero.
• Don't confuse: in dynamics (motion problems), the right side equals mass times acceleration for translation and rotation, not zero.

🛠️ How to apply equilibrium equations

🛠️ Step 1: Draw the free-body diagram

• Show all force vectors acting on the body.
• Provide values for known magnitudes, directions, and points of application.
• Assign variable names for any unknowns (magnitudes, directions, or distances).
• The excerpt emphasizes: "As with particles, the first step in finding the equilibrium equations is to draw a free-body diagram."

📐 Step 2: Choose coordinate axes

• Axes must be perpendicular to one another but do not have to be horizontal or vertical.
• The excerpt advises: "If you choose coordinate axes that line up with some of your force vectors, you will simplify later analysis."
• Example: If several forces act along a sloped surface, aligning one axis with the slope reduces the number of components to calculate.

➗ Step 3: Break forces into components

• Decompose all force vectors into components along the x, y, and z directions.
• Write one equation for each direction: sum of x-components = 0, sum of y-components = 0, sum of z-components = 0.
• The excerpt states: "We break our one vector force equation into two scalar component equations" (for 2D) or three (for 3D).

🔁 Step 4: Write moment equations

• Choose a point to take moments about—any point works, but choosing wisely reduces unknowns.
• The excerpt notes: "It is usually advantageous to choose a point that will decrease the number of unknowns in the equation. Remember that any force vector that travels through a given point will exert no moment about that point."
• Sum the moments exerted by each force about the chosen point and axis, then set the sum equal to zero.
• For 2D: one moment equation (about the z-axis).
• For 3D: three moment equations (about x, y, and z axes).

✅ Step 5: Solve the equations

• The number of unknowns you can solve for equals the number of independent equations.
• The excerpt states: "Once you have your equilibrium equations, you can solve these equations for the unknowns."
• Example: In 2D, you have 3 equations, so you can solve for up to 3 unknowns (e.g., two reaction forces and one distance).

🔍 Moments as vectors

🔍 Why moments need scalar components

Moments, like forces, are vectors. This means that our vector equation needs to be broken down into scalar components before we can solve the equilibrium equations.

• Just as force vectors are split into x, y, z components, moment vectors must be split into components about x, y, z axes.
• In 2D, rotation is only about the z-axis (clockwise or counterclockwise), so only one moment component exists.
• In 3D, rotation can occur about all three axes, so three moment components exist.

🔄 2D moment equation

• The excerpt explains: "In a two-dimensional problem, the body can only have clockwise or counterclockwise rotation (corresponding to rotations about the z-axis)."
• The one moment vector equation becomes a single scalar equation: sum of M_z = 0.
• Don't confuse: even though the body is flat (2D), the moment axis (z) is perpendicular to the plane of the body.

🌐 3D moment equations

• The excerpt states: "The second set of three equilibrium equations states that the sum of the moment components about the x, y, and z axes must also be equal to zero."
• Each axis gets its own moment equation: sum of M_x = 0, sum of M_y = 0, sum of M_z = 0.
• Example: A three-dimensional frame may have moments trying to twist it about multiple axes; all must sum to zero for equilibrium.

📊 Worked example: beam with people

📊 Setup

• The excerpt provides a visual example: a beam with 200 lbs pushing down, supported by two reaction forces pushing up.
• The problem: determine the size of the reaction forces at different support locations.

📊 Force equilibrium (vertical direction)

• The excerpt states: "If we only consider the y (vertical) direction, the 200 lbs pushing down on the beam must be balanced by the reaction forces pushing up."
• Sum of F_y = 0: reaction forces must total 200 lbs upward to balance the 200 lbs downward.
• If the forces on top are balanced evenly between the supports, the two reaction forces are equal.

📊 Moment equilibrium (rotation)

• The excerpt explains: "If they are at different locations, we use the sum of the moments equation and the distances of the people to determine the size of the reaction forces."
• By taking moments about one support, the distance of the applied load from that support creates a moment that must be balanced by the other reaction force.
• Example: If the 200 lbs is closer to one support, that support carries more load; the moment equation quantifies this.

🚗 Worked example: car on wheels

🚗 Problem statement

• The excerpt gives Example 1: "The car below has a mass of 1500 lbs, with the center of mass 4 ft behind the front wheels of the car. What are the normal forces on the front and the back wheels of the car?"
• Unknowns: normal forces at front and back wheels.
• Known: total weight (1500 lbs), location of center of mass, distance between wheels.

🚗 Approach

• Draw a free-body diagram showing the weight acting downward at the center of mass and normal forces acting upward at the front and back wheels.
• Use sum of F_y = 0 to relate the two normal forces to the total weight.
• Use sum of M_O = 0 (taking moments about one wheel) to relate the normal forces to the distances.
• Solve the two equations for the two unknowns.

🚗 Why moments matter

• The center of mass is not midway between the wheels, so the normal forces are not equal.
• The moment equation accounts for the lever-arm effect: the weight's distance from each wheel determines how much load each wheel carries.
• Don't confuse: without the moment equation, you could only find the sum of the normal forces, not their individual values.

🪑 Worked example: person in chair

🪑 Problem statement

• The excerpt gives Example 2: "While sitting in a chair, a person exerts the forces in the diagram below. Determine all forces acting on the chair at points A and B. (Assume A is frictionless and B is a rough surface)."
• Unknowns: forces at points A and B (normal and possibly friction).
• Known: applied forces from the person, geometry of the chair.

🪑 Approach

• Draw a free-body diagram of the chair showing the person's forces and the reaction forces at A and B.
• At A (frictionless), only a normal force perpendicular to the surface exists.
• At B (rough surface), both normal and friction forces may exist.
• Use sum of F_x = 0, sum of F_y = 0, and sum of M_O = 0 to solve for the unknowns.

🪑 Choosing the moment point

• The excerpt advises choosing a point that reduces unknowns.
• Taking moments about point A eliminates the unknown force at A from the moment equation, leaving fewer unknowns in that equation.
• Example: If you take moments about A, the force at A exerts no moment (zero lever arm), so the equation involves only the forces at B and the applied forces.

🔑 Key takeaways

🔑 What rigid body equilibrium adds

• The excerpt summarizes: "The equilibrium equations for rigid bodies are a way to determine unknown forces and moments using known forces and moments, separating the motion in 2 (or 3) directions for translation and rotation."
• Rigid bodies consider shape and length, so moments can be calculated.
• More equations mean more unknowns can be solved compared to particle equilibrium.

🔑 Application

• The excerpt gives an application: "Calculate the reaction forces from the combined weight of an object."
• Example: Finding how much load each support of a bridge carries when a vehicle is on it.

🔑 Looking ahead

• The excerpt notes: "This method will be used extensively in Ch 5 and 6."
• Equilibrium equations are foundational for analyzing structures, machines, and other engineering systems.

🧭 Overview

🧠 One-sentence thesis

Friction opposes sliding between surfaces and can be predicted using coefficients of friction, and when forces are applied to objects, they will either slip or tip depending on which requires less force.

📌 Key points (3–5)

• Static vs kinetic friction: static friction resists motion before slipping starts and can equal the applied force up to a maximum; kinetic friction acts during sliding and is usually smaller.
• Impending motion: the critical point just before an object begins to slip, where static friction reaches its maximum value (coefficient of static friction times normal force).
• Slipping vs tipping: when a pushing force increases, an object will either slide (slip) or rotate and fall over (tip)—whichever requires less force happens first.
• Common confusion: friction force is not constant—static friction adjusts to match the applied force until it reaches its maximum, then switches to kinetic friction once sliding begins.
• Coefficients matter: the static coefficient of friction is always higher than the kinetic coefficient, meaning it takes more force to start motion than to keep it going.

🧱 Types of friction

🧱 Dry friction (Coulomb friction)

Dry friction: the force that opposes one solid surface sliding across another solid surface.

• Always opposes relative motion between surfaces.
• Can either resist motion or cause motion depending on the situation.
• The most common model is Coulomb friction, which divides into static and kinetic types.

🛑 Static friction

Static friction: occurs prior to the box slipping and moving; the friction force equals the applied force in magnitude but opposite in direction.

• Acts before the object starts sliding.
• The friction force adjusts to match the pushing force exactly (equal magnitude, opposite direction).
• As the pushing force increases, static friction increases to match it.
• Has a maximum limit: the static coefficient of friction multiplied by the normal force.
• Example: pushing a box with 50 N—static friction will be 50 N in the opposite direction, keeping the box stationary.

⚡ Kinetic friction

Kinetic friction: occurs beyond the point of impending motion when the box is sliding.

• Acts once the object is already moving/sliding.
• The magnitude equals the kinetic coefficient of friction times the normal force.
• Almost always less than the maximum static friction force.
• Does not change with the applied force—it remains constant as long as the object slides.
• Example: once the box starts moving, kinetic friction stays at a fixed value (e.g., 40 N) even if you push harder.

🔄 Impending motion

Impending motion: the point just before the box slips; the maximum static friction force before slipping.

• The transition moment between static and kinetic friction.
• At this instant, static friction reaches its maximum possible value.
• Formula: maximum static friction = static coefficient × normal force.
• Beyond this point, the object begins to slide and kinetic friction takes over.

🔀 Slipping versus tipping

🔀 The two failure modes

When you push on a box sitting on a rough surface, one of two things will happen as force increases:

1. Slipping: the pushing force exceeds maximum static friction, and the box slides across the surface.
2. Tipping: the pushing force and friction create a strong enough couple (moment) that the box rotates and falls on its side.

Which happens first? Calculate the force required for each scenario—whichever requires less force will occur first.

📏 Determining slip force

• A body slips when the pushing force exceeds the maximum static friction force.
• Maximum static friction = static coefficient of friction × normal force.
• If pushing force > (μ_static × normal force), then slipping occurs.
• Example: if normal force is 100 N and μ_static is 0.7, then pushing force must exceed 70 N to cause slipping.

🔄 Determining tip force

• Normal forces are distributed across the bottom of an object.
• These forces redistribute to resist rotation when moments are applied.
• The equivalent point load of the normal force shifts left or right to counter the moment from the pushing force.
• Tipping occurs when: the normal force cannot shift any further (it reaches the edge of the box) and can no longer counter the moment from the pushing and friction forces.

How it works:

• At rest: normal force is uniformly distributed on the bottom.
• As pushing increases: the distributed normal force shifts, moving the equivalent point load toward one edge.
• The gravity force and normal force create a couple that counters the couple from pushing force and friction force.
• If the required shift goes beyond the edge of the box, the box tips over.

Don't confuse: the normal force is a result of physical contact—it cannot act beyond the surfaces in contact (beyond the edge).

🧮 Comparison approach

Failure mode	Condition	Key calculation
Slipping	Pushing force > max static friction	Force needed = μ_static × normal force
Tipping	Moment from push/friction > moment from gravity/normal	Use moment equilibrium; normal force acts at edge at impending tip

Example from the excerpt: A box is pushed—calculate both the force to slip and the force to tip. If slip requires 120 N but tip requires 100 N, the box will tip first.

🔧 Practical applications

🔧 Coefficient properties

• The static coefficient is always higher than the kinetic coefficient.
• Both depend on the materials in contact (can be looked up in tables).
• Example: wood on wood might have μ_static = 0.5–0.7 and μ_kinetic slightly lower.

🔧 Real-world implications

• Starting vs maintaining motion: it takes more force to start an object moving than to keep it moving (because static > kinetic).
• Design considerations: knowing whether an object will slip or tip helps in designing stable structures and predicting failure modes.
• Future relevance: the excerpt notes this becomes important in dynamics (the study of motion).

🔧 Problem-solving strategy

1. Draw a free-body diagram showing all forces (gravity, normal, friction, applied forces).
2. Use equilibrium equations: sum of forces = 0, sum of moments = 0.
3. For friction problems: determine if static or kinetic friction applies.
4. For slip vs tip: calculate the force required for each and compare—the smaller force determines what happens first.

Don't confuse: friction force in static equilibrium is not automatically at its maximum—it only reaches maximum at impending motion.

🧭 Overview

🧠 One-sentence thesis

These worked examples demonstrate how to apply equilibrium equations and friction laws to solve real-world problems involving objects on surfaces, inclines, and distributed loads, showing that friction can prevent both slipping and tipping depending on the applied forces and geometry.

📌 Key points (3–5)

• Core method: all examples use equilibrium equations (sum of forces equals zero, sum of moments equals zero) combined with friction formulas (friction force equals coefficient times normal force).
• Friction direction: the sign of the calculated friction force reveals its actual direction—negative values mean friction acts opposite to the assumed direction in the diagram.
• Slip vs tip: an object may tip over before it slips if the maximum static friction exceeds the horizontal pulling force; checking both conditions is necessary.
• Common confusion: friction does not always oppose motion in the direction you first assume—it opposes relative motion, so it can act "upward" on an incline if a force tries to push the object up.
• Distributed loads: when a force is spread over an area (e.g., a hand pressing on a brick), intensity equals total force divided by contact length.

🧱 Preventing slip on a vertical wall

🧱 The brick-on-wall scenario (Example 4.5.4)

• A person pushes a brick against a vertical wall to keep it from sliding down.
• The applied horizontal force creates a normal force between brick and wall; friction then acts upward to balance the brick's weight.

Given:

• Mass of brick = 5 kg
• Coefficient of static friction μ = 0.49
• Hand contact length = 16 cm

Key steps:

1. Find gravitational force: F_g = mass × g = 5 kg × 9.81 m/s² = 49.05 N (downward).
2. Vertical equilibrium: friction force must equal gravitational force, so F_f = 49.05 N (upward).
3. Friction formula: F_f = μ × F_N, so F_N = F_f / μ = 49.05 N / 0.49 = 100.1 N.
4. Horizontal equilibrium: applied force equals normal force, so F_A = 100.1 N.
5. Distributed load intensity: w = F_A / length = 100.1 N / 0.16 m = 625.64 N/m.

🔍 Why the applied force exceeds the weight

• The applied force (100.1 N) is larger than the gravitational force (49.05 N) because friction is only a fraction (μ = 0.49) of the normal force.
• To generate enough friction to hold the brick, you must push harder than the brick's weight.

⛰️ Ball on an inclined surface with ropes

⛰️ The suspended ball scenario (Example 4.5.5)

• A ball rests on a 15° incline and is held by two ropes (A and B) pulling at angles 20° and 60° from the incline plane.
• Rope A: 200 N; Rope B: 150 N; ball mass: 20 kg.

Finding the friction force:

1. Gravitational force: F_G = 20 kg × 9.81 m/s² = 196.2 N.
2. Component along incline (x-direction): F_GX = F_G × sin(15°) = 50.78 N (down the slope).
3. Rope A x-component: F_AX = 200 N × cos(20°) = 187.938 N (up the slope).
4. Rope B x-component: F_BX = 150 N × cos(60°) = 75 N (down the slope).
5. Sum of forces in x: 0 = –F_f + F_BX – F_AX + F_GX, so F_f = –187.938 N + 75 N + 50.78 N = –62.158 N.

🔄 Interpreting the negative sign

• The negative result means friction acts in the opposite direction to what was assumed in the free-body diagram.
• Here, friction acts down the incline, preventing the ball from being pulled up by the strong rope A force.
• Don't confuse: friction does not always point down an incline; it opposes the net tendency of motion.

🚛 Trailer towing and slip analysis

🚛 The trailer scenario (Example 4.5.6)

• A 6 m trailer carries 500 kg evenly distributed; a truck pulls with 5000 N force.
• Axles at both ends of the bed; 3 m hitch.

Calculations:

1. Weight of load: w = 500 kg × 9.81 m/s² = 4905 N.
2. Vertical equilibrium: total normal force N = w = 4905 N (split equally between two axles: N_A = N_B = 2452.5 N each).
3. Horizontal equilibrium: friction force F_F = pulling force P = 5000 N.
4. Coefficient of friction: μ = F_F / N = 5000 N / 4905 N = 1.019.

🚨 Is the trailer slipping?

• Given static coefficient μ_s = 0.72.
• Calculated μ = 1.019 > μ_s = 0.72.
• Conclusion: the trailer is sliding along the surface (in motion), because the required friction exceeds the maximum static friction available.

Condition	Value	Interpretation
Required μ	1.019	Friction needed to prevent slip
Available μ_s	0.72	Maximum static friction
Result	μ > μ_s	Trailer is slipping

🕷️ Slip or tip: the shipping container

🕷️ The Spiderman scenario (Example 4.5.7)

• Spiderman pulls a 10 m tall, 2.5 m wide container (mass 2000 kg) with a web from 20 m away and 1.8 m height.
• Coefficient of static friction μ = 0.2.
• Question: how much tension to barely tip the container?

Finding the tipping tension:

1. Gravitational force: F_g = 2000 kg × 9.81 m/s² = 19,620 N.
2. Angle of web: θ = arctan[(10 m – 1.8 m) / 20 m] = 22.3°.
3. Moment about point B (bottom corner toward Spiderman): for tipping, sum of moments = 0.
   • Only F_g and horizontal tension component T_x create moments about B.
   • –h_c × T_x + (w/2) × F_g = 0.
   • –10 m × T_x + (2.5 m / 2) × 19,620 N = 0.
   • T_x = 2,452.5 N.
4. Total tension: T = T_x / cos(θ) = 2,452.5 N / cos(22.3°) = 2,651 N.

🔍 Verifying tip vs slip

• To confirm the container tips (not slips), check if maximum friction exceeds the pulling force.
• Vertical equilibrium: F_n = F_g + T × sin(θ) = 19,620 N + 2,651 N × sin(22.3°) = 20,626 N.
• Maximum friction: F_f = μ × F_n = 0.2 × 20,626 N = 4,125 N.
• Since F_f (4,125 N) > T_x (2,452.5 N), the container will tip before it slips.

Don't confuse: tipping and slipping are two different failure modes; you must check both to predict what actually happens.

🛒 Cart on a ramp

🛒 The baggage cart scenario (Example 4.5.8)

• A person pulls a 70 kg cart up a 30° ramp with a rope tension of 300 N.
• Find the friction force.

Steps:

1. Gravitational force: F_g = 70 kg × 9.81 m/s² = 686.7 N.
2. Component along ramp (x-direction): F_gx = 686.7 N × sin(30°) = 343.35 N (down the ramp).
3. Horizontal equilibrium along ramp: 300 N – F_F – 343.35 N = 0.
4. Friction force: F_F = 300 N – 343.35 N = –43.35 N.

🔄 Meaning of the negative friction

• The negative sign indicates friction acts opposite to the assumed direction.
• Here, friction acts down the ramp (not up), because the pulling force (300 N) is less than the gravitational component (343.35 N).
• The cart would slide down without friction; friction reduces the net downward force.

Example: if the person pulls with only 300 N and gravity pulls 343.35 N down the slope, the cart tends to slide down, so friction must act downward to help the person slow the descent (or the cart is accelerating down, but in statics we assume equilibrium).

🎮 Designing for a minimum coefficient of friction

🎮 The gaming platform scenario (Example 4.5.9)

• A 0.8 m long, 0.15 m high platform (mass 4 kg) rests on two thighs, each 0.2 m wide, positioned 0.1 m from the platform edges.
• A 10 N force applied at a top corner (parallel to the surface) must not push the platform over.
• Goal: determine the required coefficient of friction.

Approach:

1. Define an origin point (e.g., point A at one support).
2. Calculate gravitational force: F_g = 4 kg × 9.81 m/s² = 39.24 N (acts at the platform's center).
3. Use sum of moments about point A to find the normal force at the other support (N_B).
4. Use horizontal equilibrium to relate friction forces to the applied force.
5. Calculate the minimum μ needed so that maximum static friction can resist the 10 N force without slipping or tipping.

📐 Why geometry matters

• The position of the supports (distance from edges) and the height of the applied force affect the moment arms.
• A force applied at a top corner has a larger moment arm about the support points, increasing the tipping tendency.
• The required μ depends on whether slip or tip is the limiting condition.

Key takeaway: when designing for stability, you must ensure both:

• Friction is large enough to prevent slip: F_f ≥ applied horizontal force.
• The moment from the applied force does not exceed the restoring moment from the weight.

📊 Common patterns across all examples

📊 The six-step problem-solving structure

Every example follows the same format:

1. Problem statement: describe the scenario and what to find.
2. Draw: sketch and free-body diagram showing all forces.
3. Knowns and unknowns: list given values and what you need to calculate.
4. Approach: outline which equations to use (equilibrium, friction formulas, geometry).
5. Analysis: perform calculations step-by-step.
6. Review: check units, compare magnitudes, verify the answer makes physical sense.

🔧 Key formulas used

Formula	Meaning	When to use
F_g = m × g	Gravitational force	Always, to find weight
F_f = μ × F_N	Friction force	Relates friction to normal force
ΣF = 0	Sum of forces = 0	Equilibrium in any direction
ΣM = 0	Sum of moments = 0	Equilibrium about any point (for tipping)
w = F / L	Distributed load intensity	When force spreads over a length

✅ Review checks mentioned

• Units: friction force should be in newtons (force units).
• Magnitude comparisons: applied force should be larger than weight when friction coefficient is less than 1 (Example 4.5.4).
• Direction logic: if friction is negative, it acts opposite to the assumed direction; verify this makes physical sense (e.g., preventing upward motion in Example 4.5.5).
• Slip vs tip: always verify which failure mode occurs by comparing friction capacity to the pulling force (Example 4.5.7).

🧭 Overview

🧠 One-sentence thesis

Trusses are rigid structures made entirely of two-force members connected at joints, and analyzing them requires understanding how internal forces create tension or compression in each member.

📌 Key points (3–5)

• What a truss is: a rigid structure composed entirely of two-force members (objects with exactly two forces/connections) that connect at joints.
• Two-force member equilibrium: forces must be equal, opposite, and collinear—resulting in either tension or compression along the member.
• Two analysis methods available: method of joints (particle analysis using only x and y equilibrium) and method of sections (rigid body analysis including moment equilibrium).
• Common confusion: tension vs compression—assume all members are in tension (+); negative results indicate compression (−).
• Real-world applications: roof frames and bridge structures commonly use trusses to distribute loads efficiently.

🏗️ What is a truss and why it matters

🏗️ Definition and structure

A truss is an engineering structure made entirely of two-force members.

• Trusses must be independently rigid: if separated from connection points, no part can move independently relative to the rest.
• Statically determinate trusses can be fully analyzed using equilibrium equations alone.
• Common locations: roof frames (wooden trusses) and bridge sides (metal or wooden trusses).

🎯 Goal of truss analysis

• Identify external forces acting on the truss structure (reactions and applied loads).
• Identify internal forces in each member (magnitude and whether tension or compression).
• Because each member is a two-force member, you only need to find the magnitude and determine tension (+) or compression (−).

🔩 Two-force members explained

🔩 What makes a two-force member

A two-force member is a body that has forces (and only forces, no moments) acting on it in only two locations.

• Forces act at exactly two points on the body.
• No moments are applied—only forces.

⚖️ Equilibrium requirements

For a two-force member in static equilibrium, the forces must be:

1. Equal in magnitude
2. Opposite in direction
3. Collinear (acting along the line connecting the two force application points)

Why collinear?

• Sum of forces = 0 requires equal and opposite forces.
• Sum of moments = 0 requires the perpendicular distance between forces (d) to be zero.
• If forces were not collinear, they would form a couple and create a moment with no counteracting moment.
• Therefore, forces must lie along the line connecting the two points.

Example: Imagine a beam with forces only at each end—if one end pushes with 100 N to the right, the other end must push with 100 N to the left along the same line to maintain equilibrium.

🧩 Parts of a truss

🧩 Three main components

Component	Description	Example
Joints	Connection points where members meet; often labelled with letters (A, B, C...)	Where external forces and members connect
Members	Metal or wooden beams connecting joints; labelled by the joints they connect	Member AB connects joint A to joint B
External forces	Reaction forces (supports) and applied forces (loads from traffic, roof weight, etc.)	Truck weight transferred through deck → stringers → beams → joints

🌉 Load transfer in bridges

• Applied loads (e.g., from vehicles) travel from the deck to stringers, across beams, to the joints of the truss.
• Stringers run parallel to traffic direction; beams run perpendicular.
• Forces become external forces on the truss joints.

🏠 Common truss types

Bridge trusses include various configurations (specific names provided in diagrams in the excerpt).

Roof trusses include various configurations (specific names provided in diagrams in the excerpt).

🔄 Tension and compression

🔄 Sign convention

• Standard assumption: assume all members are in tension (positive, +).
• If calculation yields a negative result, the member is actually in compression (−).

🔄 How forces act

Following Newton's 3rd law: when there is tension in a member, there is also tension in a joint—pulling on the member pulls on the joint; pushing on a member (compression) pushes on the joint.

• Tension (+): member pulls on both joints it connects.
• Compression (−): member pushes on both joints it connects.
• The magnitude of force from member AB on joint A equals the magnitude on joint B (equal and opposite).
• Different members (e.g., AB vs BC) will have different force magnitudes.

🔄 Visual interpretation at joints

• Compression (−): appears to push outward on the joint.
• Tension (+): appears to pull inward on the joint.
• Force is named after the member (e.g., F_ab for member AB).

Don't confuse: The force magnitude is the same at both ends of a member, but the direction reverses (Newton's 3rd law)—both ends experience the same tension or compression.

🔧 Two analysis methods overview

🔧 Method of joints

• Focuses on joints (connection points where members meet).
• Treats each joint as a particle (pin connection).
• Uses only force equilibrium equations: sum of forces in x = 0, sum of forces in y = 0.
• No moment equations (particle analysis).
• Draws free-body diagram for each joint.
• Results in many equilibrium equations to solve for many unknowns.
• Best for: solving for all unknown forces in the entire truss (fastest and easiest for complete analysis).

🔧 Method of sections

• Pretends to split the truss into two or more sections.
• Analyzes each section as a separate rigid body in equilibrium.
• Uses force and moment equilibrium equations: sum of forces in x = 0, sum of forces in y = 0, sum of moments = 0.
• Draws free-body diagram for each section.
• Best for: targeting and solving for forces in just a few specific members without solving for all unknowns.

🔧 Choosing a method

• Use method of joints when you need all member forces.
• Use method of sections when you need only one or a few specific member forces.
• The two methods can be combined if needed to suit the problem.

Example: If you need to know the force in only one central member of a large truss, method of sections is faster—make one cut through that member and solve. Method of joints would require solving many joints sequentially.

🧭 Overview

🧠 One-sentence thesis

Internal forces analysis reveals how beams carry loads through three key quantities—normal force, shear force, and bending moment—which can be visualized graphically to identify critical stress points for safer design.

📌 Key points (3–5)

• Three types of internal forces: normal (axial), shear (transverse), and bending moment describe what happens inside a beam at any cut.
• Sign convention matters: positive shear is "right side down" on the whole beam (or "up on the right" at a cut); positive moment causes concavity upward (sagging).
• Method: solve external reactions first, make a cut, add internal forces using the positive convention, then use equilibrium equations to find unknowns.
• Shear/moment diagrams: graphical plots showing how V and M vary along the beam; the derivative relationship (dM/dx = V, dV/dx = -w) links distributed loads to diagram shapes.
• Common confusion: the sign convention flips perspective—when you cut the beam, the internal force on the right face must balance the external loads, so directions may seem reversed.

🔍 Three types of internal forces

🔍 Normal force (N)

Normal force: the algebraic sum of the axial forces acting on either side of the section.

• Acts along the beam's axis (horizontal for a horizontal beam, vertical for a column).
• Also called "axial force."
• Units: N or lb.
• Sign convention: positive if it tends to tear (tension), negative if it tends to crush (compression).
• Example: in a truss member analyzed in the previous chapter, the normal force was the primary internal load; in beams with transverse loads, N often stays constant because axial forces along the beam are uncommon.

🔍 Shear force (V)

Shear force: the algebraic sum of all the transverse forces acting on either side of the section.

• Acts perpendicular to the beam's axis (vertical for a horizontal beam).
• Units: N or lb.
• Sign convention: positive if it tends to move the left side upward or the right side downward.
• The excerpt emphasizes "on either side"—you can sum forces on the left or right of the cut and get the same magnitude.
• Example: a cantilever beam with a downward load at the free end will have a constant negative shear along its length (the load tries to push the left side down).

🔍 Bending moment (M)

Bending moment: the algebraic sum of all the forces' moments acting on either side of the section.

• Causes the beam to bend.
• Units: Nm or ft-lb.
• Sign convention: positive if it causes concavity upward (sagging, like a smile); negative if concavity downward (hogging, like a frown).
• The excerpt notes that positive moments make the beam "bow downwards" (smile shape).
• Example: a simply supported beam with a central load will have maximum positive moment at the center.

🔍 2D vs 3D

• The excerpt focuses on 2D analysis (one normal, one shear, one moment).
• In 3D: 1 normal force, 2 shear forces, and 3 moments (including torsion T).
• For this chapter, assume negligible loading in the third dimension.

🎯 Sign convention rules

🎯 Why a standard convention

• So the industry agrees on what is positive and what is negative.
• The excerpt provides figures showing the convention from two perspectives: the whole beam and a cut section.

🎯 Positive shear

• On the whole beam: positive shear is "right side down."
• At a cut (for static equilibrium): positive shear is "up on the right" to balance the overall motion.
• Don't confuse: the direction flips when you isolate one side of the cut because internal forces must be equal and opposite to external loads.

🎯 Positive moment

• Positive moment causes the beam to sag (concavity upward, smile).
• Negative moment causes hogging (concavity downward, frown).
• The excerpt's figures show positive moment arrows pointing counterclockwise on the left face and clockwise on the right face of a cut.

🎯 Positive normal (axial) force

• Positive if it tends to tear the member (tension).
• Negative if it tends to crush (compression).

🛠️ Calculating internal forces at a point

🛠️ Four-step method

The excerpt outlines a clear procedure:

1. Find external and reaction forces: use equilibrium equations (sum of forces in x, sum of forces in y, sum of moments) on the whole beam.
2. Make a cut at the point of interest.
3. Add internal forces (N, V, M) to the free-body diagram of one side of the cut, using the positive sign convention.
4. Solve using equilibrium equations for the unknowns.

🛠️ Example walkthrough (cantilever with distributed load)

• The excerpt provides a worked example: a 7 ft cantilever beam with a 100 lb/ft distributed load, supported by a pin at A and a roller at C (4 ft from A).
• Step 1: solve for reactions → A_y = 87.5 lb, C = 612.5 lb, A_x = 0.
• Step 2: cut at midpoint B (2 ft from A).
• Step 3: draw the left segment with internal V and M at the cut (using positive convention).
• Step 4: sum forces in y → V = -112.5 lb (negative means it acts upward, opposite to the assumed direction). Sum moments → M = -25 ft·lb (negative means reverse direction, hogging).
• The negative signs indicate the assumed directions were wrong; the actual directions are opposite.

🛠️ Key insight: "on either side"

• You can choose to analyze the left or right segment after the cut.
• The excerpt's example uses the left segment because it has fewer external forces.
• The result will be the same magnitude, but watch the sign based on which side you choose.

📊 Shear and moment diagrams

📊 What they are

Shear/moment diagrams: graphical representations of the variation of the shearing force and bending moment along the whole beam.

• The x-axis represents location along the beam (lined up with the beam's length).
• The y-axis represents the internal shear force (V) or bending moment (M).
• Positive shear is drawn above the x-axis, negative below (or indicate with +/− signs).
• Positive moment is drawn above the beam's neutral axis, negative below.

📊 Why they matter

• They show the maximum bending moments and shearing forces, which are critical for sizing structural members during design.
• Example: a bridge designer can identify the point of maximum moment to ensure the beam is strong enough at that location.

📊 Relationship to distributed loads

The excerpt provides key derivative relationships:

Relationship	Equation	Meaning
Moment slope = shear	dM/dx = V(x)	The slope of the moment diagram at any point equals the shear force at that point
Shear slope = negative load	dV/dx = -w(x)	The slope of the shear diagram equals the negative of the distributed load intensity
Moment curvature = negative load	d²M/dx² = -w(x)	The second derivative of moment equals the negative load intensity

• Integral forms: ΔM = ∫V(x)dx (change in moment = area under shear diagram); ΔV = ∫w(x)dx (change in shear = area under load diagram).
• Example: constant distributed load → linear shear slope → parabolic moment curve. Zero distributed load → constant shear → linear moment slope.

🖊️ Drawing shear/moment diagrams

🖊️ Three methods

The excerpt describes three approaches:

1. Integration method: find the equation for each segment and integrate using dM/dx = V and dV/dx = -w.
2. Point-by-point method: calculate internal forces at key points (where loads are applied, at reactions, at start/end of distributed loads), plot these points, then connect with appropriate shapes.
3. Equilibrium method (detailed in the excerpt): use equilibrium equations to find expressions for V(x) and M(x) for each segment, then plot.

🖊️ Equilibrium method steps

1. Draw a free-body diagram of the structure.
2. Calculate reactions using equilibrium equations (may skip for cantilever if using the free end).
3. Make cuts at each segment (between load changes) and add internal N, V, M using positive convention.
4. For shear: find an equation V(x) for each segment (x measured from a reference point, often the left reaction).
5. For moment: find an equation M(x) for each segment.
6. Plot these equations on stacked graphs (load on top, shear in middle, moment on bottom).

🖊️ Example 1: cantilever with point load

• 5 lb load at the free end, 3 ft from the wall.
• Reactions: B_x = 0, B_y = 5 lb, M_B = 15 ft·lb.
• Cut anywhere along the beam (0 < x < 3 ft, measured from the free end).
• Shear: sum F_y → V = -5 lb (constant, negative means opposite direction).
• Moment: sum M_L → M = -5x (linear, negative means hogging).
• Diagram: shear is a horizontal line at -5 lb; moment is a straight line from 0 at the free end to -15 ft·lb at the wall.
• The reaction force brings shear back to 0 at the wall.

🖊️ Example 2: cantilever with uniform distributed load

• 20 kN/m over 5 m length.
• Reactions: B_y = 100 kN, M_B = 250 kN·m.
• Shear function: V = -wx (linear, from 0 at free end to -100 kN at wall).
• Moment function: M = -(w/2)x² (parabolic, from 0 at free end to -250 kN·m at wall).
• Diagram: shear is a sloped line (negative slope); moment is a parabola (concave down).

🖊️ Shear diagram construction tips

• Start at one end (often zero for simply supported beams, or the free end for cantilevers).
• Move along the beam:
  • Jump up by the magnitude of any upward point force.
  • Jump down by the magnitude of any downward point force.
  • Linear slope for uniformly distributed forces (positive slope for upward load, negative for downward).
  • Curved for non-uniform distributed forces (shape is the integral of the load function).
  • Ignore moments and horizontal forces.
• Should return to zero at the other end (if not, check your work).

🖊️ Moment diagram construction tips

• Start at one end (zero for simply supported beams).
• The moment diagram is primarily the integral of the shear diagram, except:
  • Jump up by the magnitude of any negative (clockwise) applied moment.
  • Jump down by the magnitude of any positive (counterclockwise) applied moment.
  • Ignore forces (they're already in the shear diagram).
• Should return to zero at the other end.
• Positive moments → beam bows downward (smile); negative moments → beam bows upward (frown).

🔗 Diagram shape rules

🔗 General relationships

• +V means increasing M: positive shear → moment slope is positive (going up).
• -V means decreasing M: negative shear → moment slope is negative (going down).
• V = 0 → max or min M: where shear crosses zero, moment has a local maximum or minimum (inflection point).

🔗 Start and end conditions

Beam type	Shear (V)	Moment (M)
Cantilever start (fixed)	Nonzero (reaction)	Nonzero (reaction moment)
Cantilever end (free)	0 (if no load at tip)	0
Simply supported start	Reaction force	0
Simply supported end	Reaction force	0

🔗 Jumps and inflection points

• In V: jumps occur where point forces are applied (including reactions), matching the force direction.
• In M: jumps occur where point moments are applied, matching the moment direction.
• Inflection points in M (where slope changes from + to − or vice versa) correspond to V = 0.
• Zero in V corresponds to a max or min in M.

🔗 Load-shear-moment derivative chain

The excerpt provides a helpful figure showing the relationship:

• Distributed load w(x): can be constant (horizontal line), linear (sloped line), or quadratic (parabola).
• Shear V(x): integral of w → if w is constant, V is linear; if w is linear, V is quadratic.
• Moment M(x): integral of V → if V is constant, M is linear; if V is linear, M is quadratic; if V is quadratic, M is cubic.
• Going the other way: derivative of M is V; derivative of V is -w.

🔗 Common confusion: direction at a cut

• When you draw the whole beam, positive shear is "right side down."
• When you cut and isolate one segment, the internal shear on the right face must be "up" to balance external downward loads (for static equilibrium).
• This is not a contradiction—it's the same convention viewed from two perspectives.
• Don't confuse: if your calculated V is negative, it means the actual direction is opposite to what you assumed using the positive convention.

🧰 Practical tips

🧰 Checking your work

• Shear and moment diagrams should return to zero at free ends or simply supported ends (unless there's an applied moment at the end).
• The area under the shear diagram between two points = change in moment between those points.
• The area under the load diagram between two points = change in shear between those points.
• Reaction forces should balance applied loads (sum F_y = 0, sum M = 0).

🧰 Online tools (for learning only)

• The excerpt mentions online beam calculators (SkyCiv, ClearCalcs, BeamGuru) that can confirm your diagram shapes.
• These are not acceptable for exams or homework—use them only to check your understanding.
• The excerpt is clear: "This is not an endorsement of any of the sites, just showing learning tools."

🧰 When to use which method

• Method of joints (from the previous chapter on trusses): best when you need all member forces in a truss.
• Method of sections (also from trusses): best for finding specific member forces without solving the whole structure.
• Internal forces at a point (this chapter): best for finding N, V, M at one location in a beam.
• Shear/moment diagrams (this chapter): best for understanding the entire beam's internal load distribution and identifying maximum values.

🧰 Example applications

• The excerpt includes student-submitted examples:
  • Solar panel support beam: finding internal forces at a specific point.
  • Shopping cart frame: calculating reaction forces and checking for zero-force members.
  • Shelf design: using method of sections to find maximum safe load.
  • Bridge truss: analyzing shear and moment for a simply supported beam with point loads.
  • Diving board: reaction forces and shear/moment diagrams for a cantilever with a person standing at the end.

🧭 Overview

🧠 One-sentence thesis

The method of sections allows you to quickly find the internal force in a specific truss member by cutting through the truss and applying equilibrium equations to one piece, making it faster than the method of joints when only one or a few members need to be analyzed.

📌 Key points (3–5)

• What the method does: cuts through a truss to expose internal forces in selected members, then solves them using equilibrium equations on one piece.
• When it is faster: when you need the force in only one or a few specific members; the method of joints would require solving many joints first.
• Common confusion: negative force values do not mean an error—they indicate the assumed direction (tension or compression) was drawn opposite to the actual direction.
• How to interpret signs: if you assume tension and get a negative result, the member is actually in compression; if you assume compression and get a negative result, it is in tension.
• When joints are better: if you need to find forces in all members, the method of joints becomes faster because you solve joint by joint systematically.

🔪 How the method of sections works

✂️ Making the cut

• Choose a cut that passes through the member(s) you want to analyze.
• The cut exposes the internal forces in those members as external forces on the free-body diagram.
• You then pick one piece (either side of the cut) to analyze—choose the side with fewer external forces to simplify calculations.

⚖️ Applying equilibrium equations

• Once you have isolated one piece, treat it as a rigid body in equilibrium.
• Apply the three equilibrium equations: sum of forces in x equals zero, sum of forces in y equals zero, and sum of moments about any point equals zero.
• Solve for the unknown internal forces in the cut members.

Example: In the flower cart problem, a cut is made through member F_CG. The top half is chosen because it has fewer external forces (only the applied 500 N force). The equilibrium equation in the x direction gives the force in F_CG.

🔄 Interpreting negative results

➖ What a negative sign means

• When you draw a member force, you assume a direction (usually tension, pulling away from the joint).
• If the calculation gives a negative value, the actual direction is opposite to what you assumed.
• The excerpt states: "The negative number just means that it is actually compression, not tension."

🔁 Tension vs compression reminder

• Tension: the member pulls away from the joints at both ends (like a rope being stretched).
• Compression: the member pushes into the joints at both ends (like a column being squeezed).
• Don't confuse: a negative result is not an error; it is information about the true direction of the force.

Example: In the flower cart, F_CG was drawn as if in tension, but the result was negative 515.388 N, so the member is actually in compression with magnitude 515 N.

⚡ When to use sections vs joints

🎯 Method of sections is faster when

• You need the force in only one or a few specific members.
• Using the method of joints would require solving many joints in sequence before reaching the member of interest.
• The excerpt states: "The method of sections allows you to solve a very specific area" quickly.

🔗 Method of joints is faster when

• You need to find forces in all or most members of the truss.
• Solving joint by joint systematically is more efficient than making multiple cuts.
• The excerpt notes: "Had the question asked for all member loads to be solved, however, the method of joints would have been the faster approach."

📋 Comparison table

Scenario	Faster method	Reason
Find force in one specific member	Sections	Direct cut and solve; no need to solve intermediate joints
Find forces in all members	Joints	Systematic joint-by-joint solution covers everything
Find forces in a few members on one side	Sections	One cut can expose multiple members at once

🛠️ Worked example walkthrough

🌸 Flower cart problem setup

• A flower cart is pushed with 500 N at joint G.
• Back wheels at A are locked; front wheels at B are unlocked.
• Four shelves, each 1 m apart vertically; each shelf is 4 m long.
• Part (a) asks for reaction forces at A and B.
• Part (b) asks for the force in member F_CG and whether it is tension or compression.

📐 Part (a): Reaction forces using equilibrium

• Sum of forces in x: P plus R_Ax equals zero, so R_Ax equals negative 500 N.
• Sum of moments about A: distance from B to A times R_By equals distance from G to A times P, so R_By equals (2 m times 500 N) divided by 4 m, which is 250 N.
• Sum of forces in y: R_By plus R_Ay equals zero, so R_Ay equals negative 250 N.
• The negative signs mean the assumed directions were wrong; the forces point the opposite way.

🔪 Part (b): Using the method of sections

• Redraw the diagram with corrected reaction force directions.
• Make a cut through member F_CG.
• Choose the top half (fewer external forces to consider).
• Sum of forces in x: P plus (4 divided by square root of 17) times F_CG equals zero.
• Solve: F_CG equals negative 500 N times (square root of 17 divided by 4), which is negative 515.388 N.
• The negative sign means F_CG is in compression, not tension as originally assumed.
• Final answer: F_CG equals 515 N in compression.

✅ Part (c): Why sections was faster

• The method of sections allowed a direct solution for F_CG with one cut and one equilibrium equation.
• The method of joints would have required solving lower joints first, a much slower process.
• If all member loads were needed, the method of joints would have been faster overall.

🧭 Overview

🧠 One-sentence thesis

Zero-force members carry no load under static equilibrium but exist to provide stability and maintain the rigid shape of the truss.

📌 Key points (3–5)

• What zero-force members are: members in a truss that carry no internal force under the given loading conditions.
• How to identify them: at a joint in static equilibrium, if only one member acts in a particular direction (x or y), that member is zero-force; also, at a joint with only two members and no external load, both members are zero-force.
• Why they matter: zero-force members provide stability to the truss and keep the shape rigid, even though they carry no load.
• Common confusion: removing zero-force members might seem logical, but they are essential for structural stability and shape maintenance.
• Method choice: the method of sections is faster for finding specific member forces, while the method of joints is more efficient for solving the entire system.

🔍 Identifying zero-force members

🔍 The one-force-per-direction rule

Zero-force member: a truss member that carries no internal force under static equilibrium conditions.

• At a joint in static equilibrium, analyze forces in the x and y directions separately.
• If only one member acts in a particular direction at that joint, that member must be zero-force.
• Why: for equilibrium (sum of forces = 0), a single force in one direction with no opposing force must itself be zero.

Example from the excerpt:

• At joint C, two forces act in the y direction but only one (member CE) in the x direction.
• Therefore, member CE is zero-force.

🔍 The two-member, no-load rule

• If a joint has only two members and no external load applied, both members are zero-force.
• The excerpt states: "F_AF, F_EF, F_DE, F_BD are zero force members because for joints F and D, there are only 2 members each, and there is no external load."

🔍 Iterative analysis

• After identifying and (conceptually) removing zero-force members, re-analyze the remaining joints.
• The excerpt shows that after removing CE, DG, and FG, one more zero-force member was found through further joint analysis.
• Continue until no joints satisfy the zero-force conditions.

🏗️ Purpose and structural role

🏗️ Why zero-force members exist

• Zero-force members provide stability to the truss.
• They keep the shape rigid, preventing deformation or collapse under different loading scenarios.
• Even though they carry no load in the current static condition, they may carry load under different loading conditions or during construction.

🏗️ Don't confuse: zero-force ≠ useless

• A member carrying no load in one scenario is not redundant.
• The excerpt emphasizes: "Zero-force members exist to provide stability to the truss, to keep the shape rigid."
• Removing them would compromise structural integrity, even if equilibrium equations suggest they are not needed for the current load case.

🛠️ Method comparison for analysis

🛠️ Method of sections vs method of joints

The excerpt compares two approaches for analyzing truss members:

Method	Best for	Why
Method of sections	Finding specific member forces	A simple cut isolates the target member; solve quickly using equilibrium equations without analyzing the entire truss.
Method of joints	Solving all member forces	More efficient when you need the load in every member; sections would require multiple cuts.

🛠️ Strategy depends on the question

• For a single member force: use the method of sections (faster).
• For all member forces: use the method of joints (more systematic).
• The excerpt notes: "Had the question asked for all member loads to be solved, however, the method of joints would have been the faster approach."

🛠️ Method of sections workflow

1. Make a cut through the truss so the target member is exposed.
2. Choose one piece (top or bottom); fewer external forces = simpler calculation.
3. Apply equilibrium equations to solve for the internal force in the cut member.
4. A negative result means the assumed direction (tension vs compression) is reversed.

Example from the excerpt:

• To find F_CG, a cut was made through the truss.
• The top half was chosen because it had fewer external forces.
• The calculated F_CG was negative, indicating compression instead of the assumed tension.

✅ Verification and review

✅ Checking zero-force identification

• After removing zero-force members, verify that no remaining joint has only one force in one direction.
• The excerpt states: "there are no joints where there's only one force in one direction; therefore, there are no more zero-force members."

✅ Checking calculated forces

• For reaction forces: verify moment equilibrium (e.g., R_By × distance = P × distance).
• For internal forces: check that the x or y component matches known external forces.
• Example from the excerpt: "The x component of the calculated value of F_CG is equal in magnitude to P... Therefore, it must be correct."

✅ Sign conventions

• Negative results indicate the assumed direction is wrong.
• Tension was assumed; a negative result means compression.
• The excerpt notes: "The negative number just means that it is actually compression, not tension."

🧭 Overview

🧠 One-sentence thesis

The method of sections is most efficient for finding internal forces in specific parts of a truss, while the method of joints is better for solving the entire system, and both methods can identify zero-force members that provide structural stability.

📌 Key points (3–5)

• Zero-force members: members that carry no load but exist to keep the truss shape rigid and provide stability.
• Method choice: method of sections is efficient for specific members; method of joints is efficient for solving all members.
• Zero-force identification: if a joint has only one force in a particular direction and is in static equilibrium, that member is zero-force.
• Common confusion: zero-force members may seem unnecessary, but they serve a critical purpose—maintaining truss rigidity and shape.
• Both methods use equilibrium: sum of forces equals zero, sum of moments equals zero, and trigonometry to resolve force components.

🔍 Identifying zero-force members

🔍 What makes a member zero-force

A member is zero-force if, at a joint in static equilibrium, it is the only force acting in a particular direction.

• Look at each joint and count how many forces act in each direction (x and y).
• If there is only one force in a direction, that member carries zero force.
• Example: Joint C has two forces in the y direction and only one in the x direction → member CE is zero-force.

🛠️ Why zero-force members exist

• Purpose: Zero-force members provide stability to the truss and keep the shape rigid.
• They do not carry load under the given loading conditions, but they prevent the structure from collapsing or deforming.
• Don't confuse: "zero-force" does not mean "useless"—the truss without them may lose its shape or stability.

🔁 Iterative analysis

• After removing identified zero-force members from the diagram, analyze the remaining joints again.
• You may find additional zero-force members in the second pass.
• Example: In the bridge example, after removing CE, further analysis revealed DG and FG are also zero-force.

🧮 Method of joints workflow

🧮 Step-by-step approach

1. Find reaction forces using equilibrium equations (sum of forces in x and y, sum of moments).
2. Identify zero-force members by inspecting joints with only two members and no external load.
3. Analyze each joint sequentially, resolving forces into x and y components using trigonometry.
4. Apply equilibrium at each joint: sum of forces in x = 0, sum of forces in y = 0.

📐 Using trigonometry

• Find angles between members using geometry (e.g., tan θ = rise / run).
• Resolve forces into components: F_x = F cos θ, F_y = F sin θ.
• Example: For a member at 33.7°, if F_AE sin 33.7° balances R_Ay, then F_AE = -R_Ay / sin 33.7°.

🔗 Joint-by-joint progression

• Start at a joint where you know most forces (often a support).
• Solve for unknowns at that joint, then move to an adjacent joint.
• Use previously solved forces as knowns in the next joint.
• Example: After solving joint A for F_AE and F_AC, use F_AC as a known at joint C.

✅ Review and validation

• Check signs: tension members pull away from the joint (positive), compression members push into the joint (negative).
• Check symmetry: symmetric loading and geometry should produce symmetric forces.
• Check magnitude: forces should be reasonable given the applied loads and geometry.
• Example: In the bucket example, F_CE in tension matches the weight of the bucket, and symmetric members AC and BC have equal forces.

✂️ Method of sections workflow

✂️ When to use sections

• Most efficient when you need internal forces in only a few specific members.
• Cut through the truss to expose the members of interest.
• Treat the cut section as a rigid body in equilibrium.

🔪 Making the cut

• Draw a line (section) that cuts through the members you want to analyze.
• The cut exposes internal forces, which become external forces on the free-body diagram of the section.
• Example: To find forces in CD, GH, and IH, draw a section cutting through all three.

⚖️ Analyzing the section

• Apply equilibrium equations to the entire section:
  • Sum of forces in x = 0
  • Sum of forces in y = 0
  • Sum of moments about any point = 0
• Use trigonometry to resolve forces into components.
• Example: Sum F_x = 0 = -F_CD + F_GH + F_HI cos θ, solve for F_HI.

📏 Finding distances and angles

• Calculate distances using Pythagorean theorem: d = √(w² + h²).
• Find cos θ and sin θ as ratios: cos θ = adjacent / hypotenuse, sin θ = opposite / hypotenuse.
• Example: For a member with horizontal distance 0.1 m and vertical distance 0.2 m, d = 0.2236 m, cos θ = 0.1 / 0.2236.

🎯 Solving for applied loads

• Once internal forces are known, use equilibrium to find unknown applied loads.
• Example: If F_HI is known, sum F_y = 0 = -P + F_HI sin θ gives P = F_HI sin θ.
• Convert force to mass if needed: m = P / g.

🧪 Practical considerations

🧪 Calculating system mass

• For rods with uniform density, mass = density × volume.
• Volume of a cylinder = π r² h (r is radius, h is length).
• Sum the masses of all members to get total system mass.
• Example: A cart with four 75 cm rods, two 75√2 cm rods, and one 150 cm rod, all diameter 2 cm, density 8 g/cm³, has total mass ≈ 16.64 kg.

🧪 Reaction forces at supports

• Pinned support: provides reaction forces in both x and y directions.
• Roller support: provides reaction force in only one direction (perpendicular to the rolling surface).
• Use sum of moments about one support to solve for the reaction at the other support.
• Example: Sum M_A = 0 gives R_By = (distance to load × load) / (distance to support B).

🧪 Design constraints

• Members have maximum tensile and compressive forces they can withstand.
• Use these limits to determine maximum allowable applied loads.
• Example: If CD can take 25 N tension and GH can take 15 N compression, solve for the maximum mass that can be supported at point E.

🧪 Sign conventions

Force type	Sign	Physical meaning
Tension	Positive	Member is being pulled (stretched)
Compression	Negative	Member is being pushed (squeezed)
Upward force	Positive	Acts in +y direction
Downward force	Negative	Acts in -y direction
Rightward force	Positive	Acts in +x direction
Leftward force	Negative	Acts in -x direction

🧭 Overview

🧠 One-sentence thesis

When analyzing a single beam, engineers calculate three internal forces and moments—normal force, shear force, and bending moment—at any point along the beam to predict structural behavior and design safe members.

📌 Key points (3–5)

• Three internal forces/moments in 2D beams: normal force (N), shear force (V), and bending moment (M), revealed by making a cut through the beam.
• Sign convention standardizes analysis: positive shear moves the left side up (or right side down), positive moment causes upward concavity (sagging), and positive normal force causes tension.
• How to calculate internal forces: find external reactions, make a cut at the point of interest, draw a free-body diagram with internal forces using the positive sign convention, then apply equilibrium equations.
• Common confusion—normal vs shear direction: for a horizontal beam, normal force is horizontal (axial) and shear is vertical (transverse); for a vertical column, these directions swap.
• Why it matters: knowing the magnitude of shear forces and bending moments at every section is essential for sizing structural members and ensuring they can handle maximum loads.

🔍 What internal forces exist in beams

🔍 The three types in 2D analysis

When you make a cut in a beam (similar to analyzing a fixed reaction), three quantities describe what is happening at that point:

Force/Moment	Abbreviation	Unit	Direction (horizontal beam)
Normal Force	N	N or lb	horizontal (along the axis)
Shear Force	V	N or lb	vertical (transverse)
Bending Moment	M	Nm or ft-lb	rotation

Normal force: the algebraic sum of the axial forces acting on either side of the section.

Shear force: the algebraic sum of all the transverse forces acting on either side of the section.

Bending moment: the algebraic sum of all the forces' moments acting on either side of the section.

• The excerpt emphasizes "on either side"—you can sum forces on the left or the right of the cut; both give the same result.
• This is for 2D analysis, assuming negligible loading in the third dimension.

🔄 Why "normal" is also called "axial"

• Normal force runs along the axis of the member.
• For a horizontal beam, normal force is horizontal; for a vertical column, normal force is vertical.
• Shear force is perpendicular to the axis, so it is sometimes called "transverse."
• Don't confuse: the direction labels (horizontal/vertical) depend on the beam's orientation; the underlying concept (axial vs transverse) stays the same.

🌐 Extension to 3D

In three dimensions, there are:

• 1 normal force (N)
• 2 shear forces (V₁ and V₂)
• 3 bending moments (M₁, M₂, and T for torsion)

📐 Sign convention for internal forces

📐 Why a standard sign convention is necessary

• A sign convention ensures that engineers agree on what is positive and what is negative.
• The excerpt states: "So that there is a standard within the industry, a sign convention is necessary."

➕ Positive sign rules

Normal (axial) force:

• Positive if it tends to tear the member at the section (tensile force; the member is in axial tension).
• Negative if it tends to crush the member at the section (compressive force; the member is in axial compression).

Shear force:

• Positive if it tends to move the left side of the section upward or the right side of the section downward.
• Negative if it tends to move the left side downward or the right side upward.
• The excerpt notes: "On the right, shear-up is positive."
• When you look at the beam as a whole, positive shear is "right side down"; when you cut into the beam, for static equilibrium, positive shear must be up on the right (equal and opposite to the overall motion).

Bending moment:

• Positive if it tends to cause concavity upward (sagging).
• Negative if it tends to cause concavity downward (hogging).

🔄 Visualizing the sign convention

• The excerpt shows that both figures (one showing the beam as a whole, one showing a cut section) display the identical sign convention.
• Example: On the left side of a cut, positive shear points down, positive normal points out (away from the cut), and positive moment causes upward curvature.

🛠️ How to calculate internal forces at a point

🛠️ Four-step procedure

The excerpt provides a step-by-step method to solve for internal forces at a certain point along the beam:

1. Find the external and reaction forces (solve the whole structure first).
2. Make a cut at the point of interest.
3. Draw a free-body diagram (FBD) of one side of the cut, adding the internal forces (and moments) using the positive sign convention.
4. Use the equilibrium equations to solve for the unknown internal forces and moments.

📝 Worked example walkthrough

The excerpt provides an example with a distributed load:

Given: A beam with a distributed load of 100 lb/ft over 7 ft, supported at points A and C (distance between A and C is 4 ft).

Step 1: Solve external forces

• Sum of forces in x: A_x = 0
• Sum of forces in y: A_y + C - (100 lb/ft)(7 ft) = 0
• Sum of moments about A: -(100 lb/ft × 7 ft)(7 ft / 2) + (4 ft) C = 0

Solving:

• C = (100 lb/ft × 49 ft²) / (2 × 4 ft) = 612.5 lb (upward)
• A_y = (100 lb/ft × 7 ft) - 612.5 lb = 87.5 lb (upward)
• A_x = 0

Step 2: Make a cut at point B (the midpoint, 2 ft from A).

Step 3: Draw FBD of the left portion, showing:

• Reaction A_y = 87.5 lb upward
• Distributed load over 2 ft: total force = 100 lb/ft × 2 ft = 200 lb, acting 1 ft from the left
• Internal shear V (positive direction: downward on the left face)
• Internal moment M (positive direction: causing upward concavity)
• Internal normal N (not shown in detail; equals zero here)

Step 4: Apply equilibrium

• Sum of vertical forces: 87.5 lb - 200 lb - V = 0 → V = -112.5 lb (negative indicates the shear is actually upward, not downward)
• Sum of moments about the cut: -(100 lb/ft × 2 ft × 1 ft) - V × 2 ft + M = 0 → M = 200 ft·lb - 225 ft·lb = -25 ft·lb (negative indicates reverse direction, hogging)
• Normal force N = 0 (no axial loads applied along the beam)

Result: N = 0, V = -112.5 lb, M = -25 ft·lb (clockwise).

⚠️ Interpreting the signs

• The negative shear means the actual direction is opposite to the assumed positive convention (shear is up on the right face, not down).
• The negative moment means the beam is hogging (concave downward) at that point, not sagging.
• Don't confuse: the sign tells you the direction relative to the convention, not whether the force "exists."

🎯 Why internal forces matter

🎯 Predicting structural behavior

• The excerpt states: "To predict the behaviour of structures, the magnitudes of these forces must be known."
• Bending moment and shearing force diagrams (covered in the next section) "aid immeasurably during design, as they show the maximum bending moments and shearing forces needed for sizing structural members."

🌉 Real-world application

• Example: A bridge with different loads applied (from cars, trucks, lampposts, etc.). Use this method to calculate the internal loads at a particular point of interest.
• Engineers need to know where the maximum shear and moment occur to ensure the beam can handle those loads without failing.

🔗 Connection to previous and next topics

• Previous chapter: analyzed normal (axial) force in trusses by looking at joints or whole sections.
• This chapter: looks at what happens along a single beam—three types of internal forces and moments.
• Key difference: In trusses, the normal force was the focus; in beams, shear force and bending moment change throughout the beam because additional transverse forces are applied, while normal force usually stays the same (uncommon to have applied axial forces along the beam).
• Next section: how to calculate internal forces across the whole beam and display the results graphically (shear/moment diagrams).

🧭 Overview

🧠 One-sentence thesis

Shear and moment diagrams graphically display how internal shear forces and bending moments vary along the entire length of a beam, helping engineers identify critical stress points for safer design.

📌 Key points (3–5)

• What they are: graphical representations of internal shear force (V) and bending moment (M) along the whole beam, plotted against position x.
• Key relationships: the derivative of moment equals shear (dM/dx = V), and the derivative of shear equals negative distributed load (dV/dx = -w); conversely, the change in moment equals the area under the shear diagram.
• How load shape affects diagram shape: constant distributed load → linear shear slope and parabolic moment slope; zero distributed load → constant shear and linear moment.
• Common confusion: sign conventions—positive shear is drawn above or below the x-axis (must be labeled), but positive moment is always drawn above the beam axis and negative moment below.
• Why it matters: these diagrams reveal where maximum internal forces occur, guiding structural design and safety analysis.

📊 What shear and moment diagrams represent

📊 Shearing force diagram

A graphical representation of the variation of the shearing force on a portion or the entire length of a beam or frame.

• The x-axis represents position along the beam (aligned with the beam's length).
• The y-axis represents the magnitude of the internal shear force V at each location.
• Sign convention: the diagram can be drawn above or below the x-centroidal axis of the structure, but it must be labeled as positive or negative shear.
• Example: if you cut the beam at any point x, the shear diagram tells you the internal shear force acting at that cross-section.

📊 Bending moment diagram

A graphical representation of the variation of the bending moment on a segment or the entire length of a beam or frame.

• The x-axis represents position along the beam.
• The y-axis represents the magnitude of the internal bending moment M at each location.
• Sign convention: positive bending moments are drawn above the x-centroidal axis; negative bending moments are drawn below the axis.
• This convention is stricter than for shear: the vertical position (above/below) directly indicates the sign.

🔗 Mathematical relationships between load, shear, and moment

🔗 Core derivative relationships

The excerpt provides five key equations relating distributed load w, shear V, and moment M:

Equation	In words
dM/dx = V(x)	The first derivative of moment with respect to distance equals shear
dV/dx = -w(x)	The first derivative of shear with respect to distance equals negative distributed load intensity
d²M/dx² = -w(x)	The second derivative of moment with respect to distance equals negative distributed load intensity

• What this means for slope: at any point x, the slope of the moment diagram equals the shear force at that point; the slope of the shear diagram equals the negative of the distributed load intensity at that point.
• Example: if shear V is positive at a location, the moment diagram is increasing (positive slope) at that location.

🔗 Integral relationships (area under curves)

The excerpt also gives integral forms:

Equation	In words
ΔM = integral of V(x) dx	The change in moment equals the area under the shear diagram
ΔV = integral of w(x) dx	The change in shear equals the area under the load diagram

• Practical use: to find the moment at a new location, integrate (find the area under) the shear diagram between the two points.
• Don't confuse: "area under the curve" is a signed area—positive shear contributes positive area, negative shear contributes negative area.

🔗 How load shape determines diagram shape

The excerpt emphasizes:

• Constant distributed load → shear slope is linear (first derivative is constant) → moment slope is parabolic (second derivative is constant).
• Zero distributed load (no load between points) → shear is constant (slope zero) → moment slope is linear (first derivative constant).
• Example: a uniformly distributed load of 20 N/m over a segment means the shear diagram will be a straight sloped line over that segment, and the moment diagram will be a parabola (quadratic curve).

🔗 Derivation sketch (for understanding)

The excerpt derives these relationships by considering an infinitesimal beam element of length dx:

• The total load acting on this tiny element is w dx (load intensity times length).
• By summing moments about the section at x + dx and neglecting the small second-order term (w dx² / 2), the excerpt shows dM/dx = V.
• By summing vertical forces on the element, the excerpt shows V + dV = V - w dx, which simplifies to dV/dx = -w.
• Combining these two results gives d²M/dx² = -w.

🛠️ Methods for producing shear and moment diagrams

🛠️ Three main methods

The excerpt lists three approaches:

1. Integration method: find the equation for each portion of the beam and integrate using the derivative/integral relationships above.
2. Point-by-point method: calculate internal forces at important points (where loads are applied, start/end of distributed loads, at reactions), plot these points on the V and M plots at the correct x locations, then connect the dots using the appropriate curve shape (linear, parabolic, etc.).
3. Equilibrium equations method (the excerpt focuses on this): use equilibrium equations directly to find expressions for V(x) and M(x) without explicit integration/differentiation.

The excerpt emphasizes the third method for the rest of the section.

🛠️ Step-by-step equilibrium method

The excerpt provides a detailed procedure:

1. Draw a free-body diagram (FBD) of the structure.
2. Calculate the reactions using equilibrium equations (sum of forces in x, sum of forces in y, sum of moments). For a cantilever beam, you may skip this if you start from the free end.
3. Make a cut at a variable distance x from a reference point (often the reaction or free end). Add internal forces N (axial), V (shear), and M (moment) using the positive sign convention.
   • Positive sign convention reminder: the excerpt shows that positive V and M follow a specific direction convention (see the figures in the original).
   • Depending on the number of loads, you may need multiple cuts (one cut per segment between loads).
4. For shear: find an equation (expression) for V as a function of x for each segment.
5. For moment: find an equation (expression) for M as a function of x for each segment.
6. Plot these equations on a graph, stacking the shear diagram and moment diagram vertically, aligned with the beam's x-axis.

🛠️ Alternative step-by-step (from second explanation)

The excerpt also provides a second, slightly different procedure:

For the shear diagram:

1. Solve for all external forces.
2. Draw a horizontal FBD; leave distributed forces as distributed (do not replace with equivalent point loads).
3. Draw axes below the FBD: x-axis = location (aligned with FBD), y-axis = internal shear force.
4. Starting at zero on the right side, move to the right:
   • Jump upwards by the magnitude of any upward point force.
   • Jump downwards by the magnitude of any downward point force.
   • For uniformly distributed forces, draw a linear slope where the slope equals the magnitude of the distributed force (positive slope for upward distributed force, negative slope for downward).
   • For non-uniformly distributed forces, the shape is the integral of the force function.
   • Ignore moments and horizontal forces.
5. By the time you reach the left end, you should return to zero (this is a check).

For the moment diagram:

1. Solve for all external forces and moments, create the shear diagram.
2. Draw axes below the shear diagram: x-axis = location, y-axis = internal bending moment.
3. Starting at zero on the right side, move to the right:
   • The moment diagram is primarily the integral of the shear diagram.
   • Jump upwards by the magnitude of any negative (clockwise) applied moment.
   • Jump downwards by the magnitude of any positive (counter-clockwise) applied moment.
   • Ignore forces in the FBD.
4. By the time you reach the left end, you should return to zero (this is a check).

Don't confuse: the two methods are equivalent; the first uses equilibrium equations to write V(x) and M(x) expressions, the second uses a graphical "jump and slope" approach.

📐 Worked examples

📐 Example 1: Cantilever with point load

Setup: Cantilever beam, 3 ft long, fixed at B (right end), free at left end, with a 5 lb downward point load at the free end.

Step 1–2: Reactions:

• At the fixed support B: B_x = 0, B_y = 5 lb (upward), M_B = 15 ft·lb (clockwise).

Step 3: Cut:

• Only one cut needed (only one load at the end). Let x be the distance from the free end (left) to the cut, ranging from 0 to 3 ft.

Step 4: Shear expression:

• Sum of vertical forces: -5 lb - V = 0 → V = -5 lb (constant for all x).
• The negative sign means the shear acts in the opposite direction to the assumed positive convention.
• Interpretation: the shear is constant at -5 lb along the entire beam because there is no distributed load (x does not appear in the equation).

Step 5: Moment expression:

• Sum of moments about the left end (where the 5 lb load is applied): -V x - M = 0 → M = -5 lb · x.
• The moment is linear in x (a straight line).
• At x = 0 (free end), M = 0; at x = 3 ft (fixed end), M = -15 ft·lb.
• The negative sign indicates the moment causes downward concavity (negative bending moment by sign convention).

Step 6: Plot:

• Shear diagram: a horizontal line at V = -5 lb from x = 0 to x = 3 ft. At the fixed support (x = 3 ft), the reaction force of +5 lb brings the shear back to 0 (this is a check).
• Moment diagram: a straight line starting at M = 0 at x = 0, sloping down to M = -15 ft·lb at x = 3 ft. Plotted below the neutral axis (negative moment convention).

Key insight: constant shear → linear moment; the slope of the moment diagram equals the shear force (-5 lb).

📐 Example 2: Cantilever with uniform distributed load

Setup: Cantilever beam, 5 m long, fixed at B (right end), free at left end, with a uniformly distributed load of 20 kN/m over the entire length.

Reactions:

• At the fixed support B: B_y = 100 kN (upward), M_B = 250 kN·m.

Shear expression:

• Let x be the distance from the free end. The shearing force at section x is due to the distributed load acting on the segment from 0 to x.
• Sum of vertical forces on the left segment: V = - (20 kN/m) · x (the excerpt shows this as V = -20x).
• At x = 0 m (free end), V = 0; at x = 5 m (fixed end), V = -100 kN.
• Interpretation: the shear varies linearly with x (first derivative is constant = -20 kN/m, which is the distributed load intensity).

Moment expression:

• The bending moment at section x is M = - (20 kN/m) · x² / 2 = -10 x² kN·m (the excerpt shows this as a parabolic function).
• At x = 0 m, M = 0; at x = 5 m, M = -250 kN·m.
• Interpretation: the moment is parabolic (quadratic in x) because the shear is linear (the integral of a linear function is quadratic).

Plot:

• Shear diagram: a straight line starting at V = 0 at x = 0, sloping down to V = -100 kN at x = 5 m. The reaction B_y = +100 kN at the support brings the shear back to 0 (check).
• Moment diagram: a parabolic curve starting at M = 0 at x = 0, curving down to M = -250 kN·m at x = 5 m. Plotted below the neutral axis (negative moment). The excerpt notes that intermediate points should be calculated to correctly draw the parabola.

Key insight: uniformly distributed load → linear shear → parabolic moment; the slope of the moment diagram at any point equals the shear at that point.

📐 Additional examples (3–6)

The excerpt mentions Examples 3–6 with figures showing the shear and moment diagrams for various beam configurations, but does not provide detailed solution steps. It directs readers to the source for full solutions.

🧭 Tips and diagram shape rules

🧭 General rules (with exceptions noted)

The excerpt provides a list of "generally true" rules:

• +V means increasing M: positive shear → moment diagram slopes upward.
• -V means decreasing M: negative shear → moment diagram slopes downward.
• When V = 0, that's max or min M: zero shear → horizontal tangent in moment diagram (peak or valley).

🧭 Start and end behavior by support type

Beam type	Shear diagram (V)	Moment diagram (M)
Cantilever	At reaction: nonzero V and M; at free end: both zero	Same
Simply supported	Start and end with reaction forces (jumps at supports)	Start and end at zero

• Why: at a free end, there is no external force or moment, so internal V and M must be zero. At a simple support (pin or roller), there is no applied moment, so M = 0; but there is a reaction force, so V jumps by that reaction.

🧭 Where jumps and inflection points occur

In the shear diagram (V):

• Jumps (discontinuities) occur where point forces are applied, including reactions.
• The jump magnitude equals the applied force magnitude; the direction matches the applied force direction (up for upward force, down for downward force).

In the moment diagram (M):

• Jumps (discontinuities) occur where applied moments (couples) are applied.
• The jump magnitude equals the applied moment magnitude; the direction matches the moment direction (up for negative/clockwise moment, down for positive/counter-clockwise moment, according to the excerpt's convention).

Inflection points (where the slope changes):

• In the moment diagram, inflection points (max/min values, or where the curve changes from concave to convex) correspond to V = 0 in the shear diagram.
• In the shear diagram, inflection points correspond to changes in the distributed load intensity.

🧭 Relationship between graphs (summary table)

Observation in one diagram	Implication for another diagram
Increasing slope in M	Shear V is positive
Decreasing slope in M	Shear V is negative
V is positive	M is increasing
V is negative	M is decreasing
Distributed load intensity is positive	V is increasing
Distributed load intensity is negative	V is decreasing
Inflection point in M (slope changes, max/min)	V = 0 at that location
V = 0	M has a max or min at that location

🧭 Shape progression (derivative and integral)

The excerpt provides a figure showing the relationship between load, shear, and moment shapes:

Going down (taking derivatives):

• Quadratic (x²) → derivative → Linear (x) → derivative → Constant → derivative → Zero.
• Applied to diagrams: if moment is quadratic, shear is linear; if shear is linear, distributed load is constant; if distributed load is constant, its derivative is zero.

Going up (taking integrals):

• Zero → integral → Constant → integral → Linear (x) → integral → Quadratic (x²) → integral → Cubic (x³).
• Applied to diagrams: if distributed load is zero, shear is constant; if shear is constant, moment is linear; if shear is linear, moment is quadratic.

Practical use: if you know the shape of one diagram, you can predict the shape of the next by moving up or down this progression.

🧭 Sign convention reminders

Shear force:

• Positive shear: upward internal force to the right of the cross-section, downward force to the left.
• Negative shear: downward internal force to the right of the cross-section, upward force to the left.
• The diagram can be drawn above or below the beam axis, but must be labeled with + or - signs.

Bending moment:

• Positive moment: causes the beam to bow downward (smile shape, concave down).
• Negative moment: causes the beam to bow upward (frown shape, concave up).
• Positive moments are always drawn above the beam axis; negative moments are always drawn below.

Don't confuse: the shear diagram's vertical position (above/below axis) does not inherently indicate sign—you must label it. The moment diagram's vertical position does indicate sign by convention.

🧭 Checking your work

The excerpt emphasizes two key checks:

1. Return to zero: by the time you reach the end of the beam (moving left to right in the typical convention), both the shear and moment diagrams should return to zero. If they don't, go back and check your reactions and expressions.
2. Reaction forces match jumps: at support points, the jump in the shear diagram should equal the reaction force; the final moment at a fixed support should equal the reaction moment.

🔧 Practical application and tools

🔧 Why these diagrams matter

The excerpt states in the "Key Takeaways":

Shear/Moment diagrams graphically display the internal loads along a beam. This can help you identify the major stress points to provide a safer design.

• Identifying critical locations: the maximum values in the shear and moment diagrams indicate where the beam experiences the highest internal forces, which are the most likely locations for failure.
• Design optimization: by knowing where stresses are highest, engineers can reinforce those sections or choose appropriate cross-sections and materials.
• Example: if the moment diagram shows a peak at the center of a simply supported beam, that is where the beam is most likely to bend or break, so that section may need to be thicker or made of stronger material.

🔧 Online tools (for learning, not for exams)

The excerpt mentions several online beam calculators that can help confirm diagram shapes or aid learning:

• skyciv.com/free-beam-calculator
• clearcalcs.com/freetools/beam-analysis/au
• beamguru.com/beam

Important note: the excerpt explicitly states these are "not acceptable to use on the exam or in homework" and are "not an endorsement," just learning tools with limited free versions.

🔧 Looking ahead

The excerpt notes:

You will use this more in your structures class.

• Shear and moment diagrams are foundational for structural analysis courses and real-world engineering design of beams, frames, and other load-bearing members.

🧭 Overview

🧠 One-sentence thesis

This collection of worked examples demonstrates how to calculate reaction forces, internal forces (normal, shear, and moment), and construct shear/moment diagrams for beams under various loading conditions.

📌 Key points (3–5)

• What the examples cover: reaction forces, internal forces at specific points, and complete shear/moment diagrams for beams with point loads, distributed loads, and combined loading.
• Core method: use equilibrium equations (sum of forces and moments equals zero) to find reactions, then "cut" the beam at points of interest to solve for internal normal force (N), shear (V), and moment (M).
• Shear/moment relationship: shear is the derivative of moment (dM = V), so constant shear produces linear moment, linear shear produces quadratic moment.
• Common confusion: negative signs in calculations indicate the assumed direction was wrong—flip the arrow but keep the magnitude.
• Why it matters: identifying maximum shear and moment helps locate critical stress points for safer structural design.

🔧 Solving for reaction forces

🔧 Equilibrium equations

• Every static beam problem starts by finding the reaction forces at supports (pins, rollers, fixed ends).
• Apply three equilibrium conditions:
  • Sum of forces in x = 0 (horizontal equilibrium)
  • Sum of forces in y = 0 (vertical equilibrium)
  • Sum of moments about any point = 0 (rotational equilibrium)
• Example: For a simply supported beam with a center load, taking moments about one support eliminates that reaction from the equation, making it easier to solve for the other reaction.

🔄 Handling distributed loads

• Distributed loads (e.g., w = 220 N/m over length L) must be converted to equivalent point loads for moment calculations.
• Resultant force: F_R = w × L (total magnitude)
• Location: acts at the centroid of the load distribution
  • Uniform rectangular load → centroid at L/2
  • Triangular load → centroid at L/3 from the base
• Example: A triangular load of 4 N/m over 5.5 m becomes an 11 N point load acting 1.83 m from the start of the load.

⚠️ Sign conventions

• Positive directions must be defined consistently (typically +x right, +y up, +moment counterclockwise).
• A negative result means the force or moment acts opposite to the assumed direction.
• Don't confuse: a negative answer is not an error—it's information about the true direction; redraw the arrow in the final diagram.

✂️ Finding internal forces by cutting

✂️ The cutting method

• To find internal forces at a specific point, imagine "cutting" the beam at that location and analyzing one side.
• Expose three internal forces at the cut:
  • N (normal/axial force): parallel to the beam axis
  • V (shear force): perpendicular to the beam axis
  • M (bending moment): rotational effect at the cut
• Apply equilibrium equations to the free body on one side of the cut.

📍 Choosing the cut location

• Cut between points where loads change (between supports, loads, or along distributed loads).
• For a specific point (e.g., "at 0.6 m from the left"), measure x from a reference point and include all forces/reactions up to that cut.
• Example: To find forces at the midpoint between supports A and B (4 m apart), cut at x = 2 m and sum forces/moments on the left segment.

🧮 Solving for N, V, and M

• Normal force: Sum F_x = 0; typically zero if no horizontal loads.
• Shear force: Sum F_y = 0; solve for V.
• Moment: Sum M_cut = 0; solve for M (include moment contributions from all forces and reactions on the chosen side, using their distances from the cut).
• Example: For a cantilever with a downward load at the free end, cutting near the fixed end gives V equal to the load and M equal to load × distance.

📊 Constructing shear and moment diagrams

📊 What the diagrams show

• Shear diagram (V-x): plots internal shear force along the beam length.
• Moment diagram (M-x): plots internal bending moment along the beam length.
• Both diagrams help identify maximum values, which indicate where the beam experiences the greatest stress.

🔗 The derivative relationship

The shear force is the derivative of the moment: dM/dx = V.

• This means:
  • Constant shear → moment is linear (straight line)
  • Linear shear → moment is quadratic (parabola)
  • Zero shear → moment is constant (horizontal line)
• Example: A uniformly distributed load creates linear shear (sloped line) and quadratic moment (curved).

📐 Sketching the diagrams

• Divide the beam into segments where loading changes (between point loads, start/end of distributed loads, supports).
• For each segment, determine the shape:
  • No load → shear constant, moment linear
  • Point load → shear jumps (discontinuity), moment has a kink
  • Distributed load → shear slopes, moment curves
• Boundary conditions: for a simply supported or cantilever beam, moment returns to zero at free ends.
• Example: A beam with a point load at midspan shows constant positive shear from left support to the load, then constant negative shear to the right support; moment rises linearly to a peak under the load, then falls linearly.

✅ Checking your work

• The shear diagram should return to zero at the end if all loads and reactions are accounted for.
• The moment diagram should return to zero at free ends (pins, rollers) but not at fixed ends.
• Integrate shear to get moment; differentiate moment to get shear—use this to verify shapes.
• Example: If shear is -69 N constant over 0.3 m, moment should decrease linearly by 69 × 0.3 = 20.7 Nm over that span.

🧪 Example patterns and tips

🧪 Symmetry simplifies calculations

• If a beam and its loading are symmetric, reactions at symmetric supports are equal.
• Internal forces at symmetric points have equal magnitudes (shear may have opposite signs).
• Example: A simply supported beam with a center load has equal reactions at both ends; shear is +R on the left half and -R on the right half; moment peaks at the center.

🧪 Handling multiple loads

• For beams with several point loads and distributed loads, solve reactions first using the entire beam.
• When cutting, include only the loads and reactions on the chosen side of the cut.
• For distributed loads that extend past the cut, calculate the partial resultant up to the cut location.
• Example: A beam with a triangular load from x = 9.5 m to x = 14 m, cut at x = 12 m, requires finding the equivalent load for the portion from 9.5 m to 12 m only.

🧪 Units and conversions

• Keep units consistent: convert cm to m, kg to N (multiply by g = 9.81 m/s²).
• Moment units are force × distance (N·m or lb·ft).
• Example: A 2 kg shelf becomes 2 × 9.81 = 19.62 N; a 30 cm wrench is 0.3 m.

🧪 Common mistakes to avoid

• Forgetting to convert distributed loads: always replace w (force per length) with an equivalent point load for moment calculations.
• Wrong moment arm: measure perpendicular distance from the line of action of the force to the pivot point.
• Sign errors: define positive directions at the start and stick to them; a negative result just means "opposite direction."
• Incomplete free body: when cutting, include all forces and reactions on the chosen side, including any distributed load segments.

🧭 Overview

🧠 One-sentence thesis

The center of mass represents the weighted average position of all the mass in a system, where more massive parts contribute more heavily to the overall location, and it can be calculated using discrete particle formulas or continuous integration methods depending on the distribution.

📌 Key points (3–5)

• What center of mass measures: the average position of material in a distribution, weighted by how much mass is at each location.
• How weighting works: each position is multiplied by the fraction of total mass at that position; weighting factors always sum to 1.
• Uniform vs non-uniform objects: when density is constant throughout, the center of mass coincides with the geometric center (centroid); when density varies, the center of mass shifts toward denser regions.
• Common confusion: don't confuse geometric center with center of mass—the geometric center ignores mass distribution, while center of mass accounts for where the mass is concentrated.
• Extension to multiple dimensions: the same weighted-sum principle applies separately in x, y, and z directions for 2D and 3D systems.

🧮 Fundamental concept and formula

🧮 What center of mass represents

Center of mass: the average position of the material making up a distribution, where the average is weighted by mass.

• It is not simply the midpoint between objects unless they have equal mass.
• Common sense: if two particles have the same mass, the center of mass is halfway between them.
• If masses differ, the center of mass is closer to the more massive particle.
• The center of mass gets "votes" proportional to each particle's mass.

📐 The basic formula for two particles

For two particles on the x-axis:

• Particle 1: mass m₁ at position x₁
• Particle 2: mass m₂ at position x₂

The center of mass position x̄ is:

x̄ = (m₁ × x₁)/(m₁ + m₂) + (m₂ × x₂)/(m₁ + m₂)

Key properties:

• Each weighting factor is a proper fraction (between 0 and 1).
• The sum of weighting factors always equals 1.
• If m₁ > m₂, then x₁ counts more in the sum, pulling the center of mass closer to particle 1.
• If m₁ = m₂, each weighting factor becomes 1/2, so x̄ = (x₁ + x₂)/2 (the midpoint).

🎪 Real-world analogy: the seesaw

Example: A playground seesaw (teeter-totter) with children of different weights.

• If one child sits at each end and they have different weights, the heavier child sinks and the lighter child rises.
• If the heavier child slides toward the center, the seesaw can balance.
• The masses balance when m₁ × d₁ = m₂ × d₂ (where d is distance from the pivot).
• This illustrates how mass and position together determine the center of mass.

🔢 Multiple particles and higher dimensions

🔢 Generalizing to n particles on a line

For n masses m₁, m₂, …, mₙ placed at positions x₁, x₂, …, xₙ on a number line:

x̄ = (sum of all mᵢ × xᵢ) / (sum of all mᵢ)

In notation: x̄ = Σ(mᵢ × xᵢ) / Σ(mᵢ)

Steps:

1. Calculate the moment M = Σ(mᵢ × xᵢ) (the numerator).
2. Calculate the total mass m = Σ(mᵢ).
3. Divide: x̄ = M / m.

📊 Example with four masses

Suppose four point masses on a number line:

• m₁ = 30 kg at x₁ = -2 m
• m₂ = 5 kg at x₂ = 3 m
• m₃ = 10 kg at x₃ = 6 m
• m₄ = 15 kg at x₄ = -3 m

Solution:

• Moment M = (30)(-2) + (5)(3) + (10)(6) + (15)(-3) = -60 + 15 + 60 - 45 = -30 kg·m
• Total mass m = 30 + 5 + 10 + 15 = 60 kg
• Center of mass x̄ = -30 / 60 = -0.5 m
• Result: The center of mass is 0.5 m to the left of the origin.

🌐 Extending to 2D and 3D

For systems in 2D or 3D, apply the same formula separately in each direction:

• x̄ = Σ(mᵢ × xᵢ) / Σ(mᵢ)
• ȳ = Σ(mᵢ × yᵢ) / Σ(mᵢ)
• z̄ = Σ(mᵢ × zᵢ) / Σ(mᵢ)

The center of mass position vector is:

r̄ₘ = (m₁r₁ + m₂r₂ + …) / M

where M is the total mass and rᵢ is the position vector of particle i.

🗺️ Example in 2D

Three point masses in the x-y plane:

• m₁ = 2 kg at (-1, 3) m
• m₂ = 6 kg at (1, 1) m
• m₃ = 4 kg at (2, -2) m

Solution:

• Total mass m = 2 + 6 + 4 = 12 kg
• Moment in x: Mₓ = (2)(-1) + (6)(1) + (4)(2) = -2 + 6 + 8 = 12 kg·m
• Moment in y: Mᵧ = (2)(3) + (6)(1) + (4)(-2) = 6 + 6 - 8 = 4 kg·m
• x̄ = 12 / 12 = 1 m
• ȳ = 4 / 12 = 0.333 m
• Result: Center of mass is at (1, 0.333) m.

Don't confuse: The center of mass is not the geometric center of the triangle formed by the three positions; it is weighted by mass.

🧵 Continuous distributions: thin rods

🧵 Uniform thin rods

Uniform thin rod: a rod where the linear mass density μ (mass per unit length) has the same value at all points.

• For a uniform rod, the center of mass is at the geometric center of the rod.
• Example: A uniform rod extending from x = 0 to x = L has its center of mass at x = L/2.

🌡️ Linear mass density

Linear mass density μ: the mass per unit length of the rod; a measure of how closely packed the particles are.

• Where linear density is high, particles are close together.
• For a non-uniform rod, μ varies with position along the rod.
• Example: Imagine a rod made of a lead-aluminum alloy where the percentage of lead varies smoothly from 0% at one end to 100% at the other—linear density would be a function of position.

Calculus definition: μ is the ratio of mass in a segment to the length of the segment, in the limit as the segment length goes to zero:

• μ(x) = dm/dx at position x.

🔬 Non-uniform rod example

Problem: Find the center of mass of a thin rod extending from 0 to 0.890 m with linear density μ = 0.650 kg/m³ × x² (density increases along the rod).

Key insight: You cannot simply use m = μL because μ varies. Instead, use integration.

Steps:

1. Find total mass using integration: m = (b × L³) / 3, where b = 0.650 kg/m³ and L = 0.890 m.
   • m = (0.650 × 0.890³) / 3 = 0.1527 kg
2. Find center of mass position using integration: x̄ = (b × L⁴) / (4m)
   • x̄ = (0.650 × 0.890⁴) / (4 × 0.1527) = 0.668 m

Result: The center of mass is at 0.668 m, closer to the denser end (x = L) as expected.

Don't confuse: The center of mass of a non-uniform rod is not at L/2; it shifts toward the region with higher density.

🎯 Key distinctions and applications

🎯 Center of mass vs centroid

Concept	Definition	When they coincide
Centroid	Geometric center of a shape	Always at the geometric center
Center of mass	Weighted average position accounting for mass distribution	Only when density is uniform throughout

• If density is the same everywhere, centroid = center of mass.
• If density varies, center of mass shifts toward denser regions, away from the centroid.

🛠️ Why center of mass matters

Application: To calculate acceleration or use F = ma, the mass m is the total mass located at the center of mass.

• The center of mass behaves as if all the mass were concentrated at that single point for translational motion.
• This simplifies analysis of complex systems.

🔄 Looking ahead

The excerpt mentions that the next section will cover calculating the center of mass for composite shapes—splitting an object into recognizable shapes rather than integrating, which is faster for many practical problems.

Budget: 0 tokens (no math LaTeX used; formulas rewritten in words or simple notation)

🧭 Overview

🧠 One-sentence thesis

Instead of integrating to find the center of mass, you can split a complex object into recognizable geometric shapes, find each shape's center of mass from a common origin, and combine them using weighted averages—a faster method that allows multiple solution paths.

📌 Key points (3–5)

• Core method: Break a complex shape into simpler shapes (rectangles, triangles, circles, etc.), find each part's center of mass, then use weighted formulas to find the overall center.
• Centroid vs. center of mass: Centroid is the geometric center; center of mass accounts for density. If density is uniform, they coincide.
• Holes count as negative: Cutouts or holes are treated as negative areas/volumes in the weighted sum.
• Common confusion: Centroid tables give coordinates relative to the shape's own axes—you must adjust these to a single origin for the entire system.
• Why it matters: Faster than integration; essential for finding inertia about the center of mass and for analyzing composite structures.

📐 Centroid tables and basic shapes

📐 What the tables provide

Centroid tables list the location of the center of mass (r_cm) for standard geometric shapes.

• Each shape has a known formula for its area (2D) or volume (3D) and the coordinates of its centroid.
• The excerpt provides a table with shapes: rectangle, right triangle, triangle, circle, circular annulus, semicircle, quarter circle, and ellipse.
• For a rectangle (area = base × height), the centroid is at [b/2, h/2]—the midpoint in both directions.
• For a right triangle (area = ½ base × height), the centroid is at [b/3, h/3]—one-third of the base and height from the right-angle corner.
• For a circle or ellipse, the centroid is at the geometric center [0,0] if the origin is placed there.
• For a semicircle (area = ½ π r²), the centroid is offset along the axis of symmetry: [0, (4/3π)r].
• For a quarter circle, the centroid is at [(4/3π)r, (4/3π)r].

📐 Single-object example

• If you have one simple shape, select it from the table and use the equation directly.
• Example: A right triangle's centroid is less than the midpoint (which would be for a square), so the answer should reflect that the mass is concentrated toward the right-angle corner.

🧩 Composite shapes: step-by-step method

🧩 Five-step process

1. Define an origin: Choose a reference point and axes; all measurements must be from this same origin.
2. Split the object: Break the complex shape into recognizable parts (rectangles, triangles, circles, etc.).
3. Find each part's center of mass: Use the centroid table, adjusting coordinates so they are relative to the common origin.
4. Calculate mass (or area/volume): If finding center of mass, compute mass using density (ρ = m/V). If finding centroid only, use area or volume directly.
5. Apply weighted formulas: Use the sum of (mass × coordinate) divided by total mass for each direction (x, y, z).

🧩 Weighted centroid formulas

• For the x-coordinate of the overall centroid:
  • x-bar = (sum of each part's area × its x-coordinate) / (total area)
• Similarly for y and z.
• In words: x-bar = (Area₁ × x₁ + Area₂ × x₂ + Area₃ × x₃ + ...) / (Area₁ + Area₂ + Area₃ + ...)
• The same logic applies to mass: x-bar = (m₁x₁ + m₂x₂ + ...) / (m₁ + m₂ + ...)

🧩 Handling holes and cutouts

• Negative areas/volumes: A hole is treated as a shape with negative area or volume.
• Example: A C-shaped object can be modeled as a large rectangle minus a smaller rectangle (the cutout).
• When summing, subtract the hole's contribution: Area_total = Area_outer − Area_hole.
• The same applies to the weighted sum: the hole's (negative area × its centroid coordinate) is subtracted.

🧩 Multiple solution paths

• The excerpt shows two examples for the same C-shape:
  • Method 1: Three shapes (two rectangles + one circular hole).
  • Method 2: Two shapes (one large rectangle − one smaller rectangle).
• Both methods yield the same answer, but Method 2 involves fewer shapes and may be faster.
• Don't confuse: The choice of decomposition is flexible; pick the simplest breakdown.

🔄 Centroid vs. center of mass vs. center of gravity

🔄 Definitions and when they coincide

Centroid: the geometric center of a shape.
Center of mass: accounts for density; if density is uniform, it equals the centroid.
Center of gravity: the point where gravitational force acts; for uniform gravitational fields and constant density, it equals the center of mass.

• For rigid beams in statics (uniform density, uniform gravity), all three are at the same location.
• If density varies, the center of mass shifts toward the denser region; the centroid does not.
• If the gravitational field is non-uniform (e.g., very tall structures), center of gravity may differ slightly from center of mass.

🔄 When to use area vs. mass

• Centroid calculation: Use area (2D) or volume (3D) only; skip the density step.
• Center of mass calculation: Multiply area/volume by density to get mass for each part, then use mass in the weighted formula.
• Example: The excerpt's C-shape example uses density to calculate mass, then applies the weighted formula. To find the centroid, you could skip the density step and use area directly.

🛠️ Practical tips and common pitfalls

🛠️ Coordinate adjustment

• Each shape in the centroid table has its own local axes (often centered on the shape).
• You must "mentally adjust" the diagram so the shape is oriented correctly relative to your chosen origin.
• Example: If a rectangle in the table is centered at [0,0] but in your system it is offset, add the offset to the table's coordinates.

🛠️ Symmetry simplifies work

• Real-life objects are rarely perfect, but if they are symmetric, approximation is easier.
• A symmetric shape's centroid lies on the axis of symmetry.
• Example: A circle's centroid is at its center; a semicircle's centroid is on the diameter of symmetry, offset by (4/3π)r.

🛠️ Table format for organization

• The excerpt recommends creating a table with columns: Part, Area (or Volume), x-coordinate, y-coordinate (and z if 3D).
• Fill in each row for each shape, then sum the columns to apply the weighted formula.
• This keeps calculations organized and reduces errors.

🛠️ Don't confuse 2D and 3D

• In 2D, use areas; in 3D, use volumes.
• The formulas are the same structure, but you add a z-coordinate in 3D.
• Example: For a 3D composite object, you would compute x-bar, y-bar, and z-bar separately using volumes and mass.

🔗 Connection to inertia and applications

🔗 Why find the center of mass?

• Application: To calculate the mass moment of inertia of an object rotating about its center of mass, you must first know where the center of mass is.
• The excerpt's "Key Takeaways" state: "To calculate the inertia of an object rotating about its center of mass, you will need to know where the center of mass is."
• This method is foundational for dynamics and structural analysis.

🔗 Looking ahead

• The next section (7.3) covers types of inertia: area moment, product moment, and mass moment.
• The composite-shapes method for center of mass parallels the method for composite inertia (using the parallel axis theorem).

---

Note: The excerpt includes references to external sources (Jacob Moore et al., mechanicsmap.psu.edu) for images and additional examples. It also mentions that more worked examples with PDFs and video solutions are available at that site.

🧭 Overview

🧠 One-sentence thesis

There are three main types of moment of inertia—area, product, and mass—each representing a different physical resistance (bending, torsion, or rotation) and requiring different units.

📌 Key points (3–5)

• Three types of moment of inertia (MOI): area moment of inertia (beam stiffness/deflection), product moment of inertia (shaft torsion resistance), and mass moment of inertia (rotational resistance).
• Different units for each type: area and product MOI use m⁴ or ft⁴; mass MOI uses kgm² or slugft².
• Mass moment of inertia is the rotational equivalent of mass: just as heavy objects resist linear motion, objects with large inertia resist rotation.
• Common confusion: don't mix up the three types—area and product MOI are used in structures (beams and shafts), while mass MOI is used in dynamics (rotation of bodies).
• Moment integrals are classified by multiple dimensions: 1D/2D/3D, force/area/mass, first vs. second order, and rectangular vs. polar.

🔧 The three types of moment of inertia

🏗️ Area moment of inertia

Area moment of inertia: used in structures to determine how stiff a beam is or how much it will deflect.

• Unit: m⁴ or ft⁴
• Application: measures a beam's resistance to bending.
• Example: a beam under load—its stiffness and deflection depend on its area moment of inertia.
• Don't confuse with: mass moment of inertia, which measures rotation resistance, not bending.

🔩 Product moment of inertia

Product moment of inertia: a shaft's resistance to torsion (or twisting).

• Unit: m⁴ or ft⁴
• Application: measures how much a shaft resists twisting forces.
• Example: a shaft being twisted—its resistance to torsion is determined by its product moment of inertia.
• Note: also called polar moment of inertia in some contexts.

🔄 Mass moment of inertia

Mass moment of inertia: the rotational equivalent of mass; a quantity of how mass is distributed around a body.

• Unit: kgm² or slugft²
• Application: measures an object's resistance to rotation.
• Why it matters: a really heavy object is hard to move (resists motion); an object with a really big inertia is hard to rotate (resists rotation).
• Example: an ice skater spinning with their arms in (smaller inertia, faster spin) or spread out (larger inertia, slower spin).
• Don't confuse with: area or product MOI, which measure structural properties (bending/torsion), not rotational dynamics.

📊 Summary table

Type	What it measures	Unit	Application
Area moment of inertia	Beam stiffness / deflection	m⁴ or ft⁴	Structures (bending)
Product moment of inertia	Shaft torsion resistance	m⁴ or ft⁴	Structures (twisting)
Mass moment of inertia	Rotational resistance	kgm² or slugft²	Dynamics (rotation)

🧮 Moment integrals: the general concept

🧮 What is a moment integral

Moment integral: the general concept using integration to determine the net moment of a force that is spread over an area or volume.

• Why integration is needed: moments are generally a force times a distance, and distributed forces are spread over a range of distances.
• General form: integral of M equals integral of f(d) times d.
• Broader definition: the term also applies to integrating distributed areas or masses that resist some moment about a set axis.

🎯 Applications of moment integrals

The excerpt lists five applications:

1. Finding equivalent point loads for distributed loads.
2. Finding the centroid (geometric center) or center of mass for 2D and 3D shapes.
3. Finding the area moment of inertia for a beam cross section (resistance to bending).
4. Finding the polar area moment of inertia for a shaft cross section (resistance to torsion).
5. Finding the mass moment of inertia (body's resistance to angular accelerations).

🗂️ Classifying moment integrals

📐 1D, 2D, and 3D moment integrals

• Dimensions: moment integrals can be taken in 1, 2, or 3 dimensions (never beyond 3 for practical purposes).
• Complexity: 3D moment integrals are more involved than 1D or 2D.
• How to tell: the nature of the problem dictates the dimensions needed; often not explicitly listed, so you must infer from context.

🔢 Force, area/volume, and mass moment integrals

• What you integrate: force functions over distance/area/volume, the area or volume function itself, or the mass distribution over area/volume.
• Different purposes: each type has a different purpose and starts with a different mathematical function, but the integration process is similar.

1️⃣ First vs. second moment integrals

• First moment integrals: multiply the initial function by the distance.
  • Applications: equivalent point load, centroids, and center of mass.
• Second moment integrals: multiply the function by the distance squared.
  • Applications: area moments of inertia, polar moments of inertia, and mass moments of inertia.
  • Label: second moment integrals are often called "moment of inertia."

📏 Rectangular vs. polar moment integrals

• Rectangular moment integrals: distance is measured from some axis (e.g., x-axis or y-axis in 2D; xy, xz, or yz plane in 3D).
  • Integration direction: move left to right, bottom to top, or back to front.
• Polar moment integrals: distance is measured from some point in 2D (e.g., the origin) or from some axis in 3D (e.g., x, y, or z axis).
  • Integration direction: radiate outward from the center point (2D) or from the axis (3D).
• Example: in 2D, if distance is measured like x and y (from an axis), it's rectangular; if measured like r (from a point), it's polar.
• Don't confuse: in 3D, rectangular measures from a plane, polar measures from an axis—the definitions shift slightly.

🔗 Why this matters

🔗 Connection to other topics

• Area and product moment of inertia: will be covered more in structures courses.
• Mass moment of inertia: used in statics and dynamics; the next section will explain how to calculate it.
• Prerequisite: to calculate the inertia of an object rotating about its center of mass, you need to know where the center of mass is (covered in the previous section).

🧭 Overview

🧠 One-sentence thesis

Mass moment of inertia quantifies an object's resistance to rotation, and a larger inertia produces slower rotation for the same applied torque.

📌 Key points (3–5)

• What it measures: resistance to rotation—the bigger the inertia, the slower the rotation.
• What determines it: both the mass and how far that mass is from the axis of rotation (distance squared matters).
• Axis dependence: rotating about different axes produces different types of rotation and different inertia values for the same object.
• Common confusion: the same shape can have multiple inertia equations depending on which axis you rotate about (e.g., I_xx vs I_yy).
• Practical shortcut: standard formulas exist for common shapes so you don't have to integrate every time; radius of gyration simplifies complex shapes into an equivalent thin ring.

🔄 What mass moment of inertia is

🔄 Definition and physical meaning

Mass moment of inertia (often just called "inertia"): an object's resistance to rotation.

• The relationship is given by: sum of moments equals inertia times angular acceleration.
• Inertia is always positive and has units of kg·m² or slug·ft².
• A bigger inertia means slower rotation for the same applied torque; a smaller inertia allows faster rotation.

📏 How distance from the axis matters

• For an infinitesimal mass element dm, inertia depends on its distance r from the axis of rotation: the integral over all mass of r² dm.
• If more mass is closer to the axis, the inertia is smaller.
• If mass is farther from the axis, the inertia is bigger (because r is squared).
• Example: imagine two objects with the same total mass—one with mass concentrated near the axis and one with mass spread far out; the second has much larger inertia.

🧭 Axis of rotation and visualization

🧭 Different axes produce different rotations

• You can visualize rotation by imagining sticking a pencil through an object and twisting along that axis.
• Rotating about the y-axis versus the x-axis produces different types of rotation.
• Due to symmetry, rotation about the x-axis and z-axis may look identical for some shapes.

📐 How r changes with axis

• The distance r being measured changes direction depending on the axis.
• The excerpt shows that r changes direction from x to y, but looks the same between x and z (for a symmetric shape).
• Don't confuse: the same object has different inertia values for different axes because the distribution of mass relative to each axis is different.

🧮 Building up inertia from particles

🧮 Single particle

• Start with an idealized object: a massless disk with a particle of mass m embedded at distance r from the axis.
• The moment of inertia of that single particle is: I = m r².

🧮 Multiple particles

• For two particles (mass m₁ at distance r₁ and mass m₂ at distance r₂):
  • Individual inertias: I₁ = m₁ r₁² and I₂ = m₂ r₂².
  • Total inertia is the sum: I = I₁ + I₂ = m₁ r₁² + m₂ r₂².
• This concept extends to any number of particles: add another m_i r_i² term for each additional particle.

🧮 Continuous objects (integration)

• For a rigid object, subdivide it into an infinite set of infinitesimal mass elements dm.
• Each element contributes dI = r² dm, where r is the distance from the axis.
• Integrate over the entire object to get the total moment of inertia.

📋 Standard formulas for common shapes

📋 Why use tables

• Equations have been developed for common shapes so you don't have to integrate every time.
• The result is different for each axis.
• Notation: I_xx means "the inertia if rotating about the x-axis."

📋 Symmetric shapes

Shape	Key inertia formulas	Notes
Thin ring	I_xx = (1/2) m r²; I_yy = m r²; I_zz = (1/2) m r²	Thickness much less than 1
Circular plate	I_xx = (1/4) m r²; I_yy = (1/2) m r²; I_zz = (1/4) m r²	Thickness much less than 1
Cylinder	I_xx = (1/12) m (3r² + h²); I_yy = (1/2) m r²; I_zz = (1/12) m (3r² + h²)	Volume = π r² h
Sphere	I_xx = I_yy = I_zz = (2/5) m r²	Volume = (4/3) π r³
Slender rod	I_xx = (1/12) m l²; I_yy = 0; I_zz = (1/12) m l²	Radius much less than length
Rectangular plate	I_xx = (1/12) m h²; I_yy = (1/12) m (h² + b²); I_zz = (1/12) m b²	Thickness much less than 1
Rectangular block	I_xx = (1/12) m (h² + d²); I_yy = (1/12) m (d² + w²); I_zz = (1/12) m (h² + w²)	Volume = b w h

📋 Asymmetric and hollow shapes

• The excerpt also lists half cylinders, hemispheres, cones, and hollow shells (cylindrical, spherical, hemispherical).
• Many of these have both unprimed (I_xx, I_yy, I_zz) and primed (I_xx′, I_yy′, I_zz′) versions, indicating different reference axes.
• Example: for a cylinder, I_xx and I_zz depend on both radius and height, but I_yy depends only on radius.

📋 Observing different rotation rates

• The excerpt includes a simulation showing a cylinder (blue), ring (green), solid sphere (yellow-brown), and spherical shell (red) all with the same mass and radius.
• They rotate at different rates because they have different inertias.
• The one with the least inertia rotates fastest (has the largest angular acceleration for the same torque).

🔁 Radius of gyration

🔁 What it is

Radius of gyration (k): a concept that converts a complex shape into an equivalent thin ring.

• If a homework problem states "the radius of gyration k = 15 cm," it means if the shape were a thin ring, it would have a radius of 15 cm.
• You calculate the mass moment of inertia using the ring equation: I = m k².

🔁 How to use it

• Given mass m and radius of gyration k, inertia is I = m k².
• Example: if m = 10 kg and k = 5 m, then I = 10 kg × 5 m × 5 m = 250 kg·m².
• To find the radius of gyration from known inertia and mass: k = square root of (I / m).
• Example: if I = 250 kg·m² and m = 10 kg, then k = square root of (250 / 10) = 5 m.

🔁 Why it matters

• Radius of gyration is used for particularly complex shapes where standard formulas don't apply.
• It simplifies calculations by reducing any shape to an equivalent ring.

🚀 Applications and looking ahead

🚀 Real-world use

• The speed at which something rotates (e.g., a satellite spinning in space) is impacted by its inertia.
• A bigger inertia produces a smaller angular acceleration for the same applied moment.
• A smaller inertia allows for a larger angular acceleration.

🚀 Next steps

• Mass moment of inertia will be used throughout dynamics.
• The next section (7.5) covers the parallel axis theorem, which has two main uses:
  1. Finding the inertia of a complex object with multiple parts.
  2. Calculating inertia from a different axis than the standard reference.

🧭 Overview

🧠 One-sentence thesis

The parallel axis theorem allows you to calculate an object's moment of inertia about any axis parallel to one through its center of mass, and enables combining inertias of multiple parts into a total system inertia.

📌 Key points (3–5)

• Two main uses: finding inertia of complex multi-part objects, and rotating an object about an axis other than through its center of mass.
• The formula: inertia about a new axis equals the inertia about the center of mass plus mass times the squared distance between axes (I = I_cm + m·d²).
• Common insight: inertia is smallest when the axis passes through the center of mass—moving the axis away always increases inertia.
• How to distinguish: I_cm is inertia through the object's own center of mass; I is inertia about a parallel axis at distance d away.
• For composite objects: calculate each part's inertia separately using the parallel axis theorem, then add all parts together for total inertia.

🔧 The Parallel Axis Theorem Formula

🔧 What the theorem states

The parallel axis theorem relates the moment of inertia I_CM of an object with respect to an axis through the center of mass to the moment of inertia I of the same object with respect to a parallel axis at distance d from the center-of-mass axis.

The formula is: I = I_cm + m·d²

Where:

• I = moment of inertia about the new axis (the one you want to find)
• I_cm = moment of inertia about an axis through the object's center of mass
• m = mass of the object
• d = distance between the two parallel axes

🎯 Why inertia increases with distance

• The closer mass is "packed" to the axis of rotation, the smaller the moment of inertia.
• By definition of center of mass, mass is packed most closely when the axis passes through the center of mass.
• Therefore, inertia about the center-of-mass axis is always the smallest value; any parallel axis farther away will have larger inertia.
• The d² term captures this: the farther you move the axis, the more the inertia grows (and it grows quadratically with distance).

📐 Simple example: disk rotating off-center

Example: For a disk with I_cm = (1/2)·m·r², if you move the axis a distance d away:

• Apply the theorem: I = I_cm + m·d²
• I = (1/2)·m·r² + m·d²
• If d happens to equal r, then I = (1/2)·m·r² + m·r² = (3/2)·m·r²
• The inertia is three times larger than the center-of-mass value.

Don't confuse: the r in the disk formula is the disk's radius; d is the distance you shift the rotation axis.

🧩 Using the Theorem for Composite Objects

🧩 When you have multiple parts

The parallel axis theorem is especially useful for objects made of several shapes (composite objects).

Strategy:

1. Break the object into simple shapes (spheres, rods, disks, etc.).
2. Find each shape's center of mass.
3. Look up or calculate I_cm for each shape using standard formulas (from geometry tables).
4. Measure the distance d from each shape's center of mass to the system's rotation axis.
5. Apply the parallel axis theorem to each part: I_part = I_cm-part + m_part·d².
6. Add up all the individual inertias to get the total: I_total = sum of all I_part.

🔩 Example: dumbbell system

The excerpt walks through a dumbbell: two spheres (each 40 kg, 0.2 m diameter) connected by a slender rod (20 kg, 0.6 m long), rotating about an axis at one end.

Part	Mass	I_cm formula	I_cm value	Distance d to axis	m·d²	I about axis
Sphere (each)	40 kg	(2/5)·m·r²	0.16 kgm²	0.4 m	6.4 kgm²	6.56 kgm²
Rod	20 kg	(1/12)·m·L²	0.6 kgm²	0 m	0	0.6 kgm²

• Each sphere: I = 0.16 + 40·(0.4)² = 6.56 kgm²
• Rod is already at the axis (d=0), so I = 0.6 kgm²
• Total: 2·(6.56) + 0.6 = 13.72 kgm²

Don't confuse: the rod's center of mass is at the rotation axis (d=0), so no parallel-axis correction is needed for it; only the spheres need the m·d² term.

📋 Step-by-Step Procedure

📋 Seven-step method for finding total moment of inertia

The excerpt provides a systematic checklist:

1. Identify shapes: Determine the shape of the object (or each shape if composite).
2. Identify rotation axis: Determine which axis the object rotates about.
3. Locate centers of mass: Find the center of mass for each individual shape.
4. Find I_cm for each shape: Use standard formulas (sphere, rod, disk, etc.) to calculate inertia about each shape's own center of mass.
5. Measure distances: Determine the distance d from each shape's center of mass to the system's rotation axis.
6. Apply parallel axis theorem: For each shape, calculate I = I_cm + m·d².
7. Sum all parts: Add up all the individual inertias to find I_total.

🧮 Simpler example: uniform rod

Example: A uniform rod with I_cm = 0.05 kgm², length L = 0.08 m, mass = 0.250 kg. Find inertia about an axis perpendicular to the rod passing through 1/4 of its length.

• The center of mass is at the middle of the rod.
• The new axis is at 1/4 length, so d = L/4 = 0.08/4 = 0.02 m.
• Apply theorem: I = 0.05 + (0.250)·(0.02)² = 0.05 + 0.0001 = 0.0501 kgm².

The small increase (0.0001) shows that moving the axis only slightly increases inertia when d is small.

🔑 Radius of Gyration (Related Concept)

🔑 What radius of gyration means

Radius of gyration (k): a concept that converts a complex shape into an equivalent thin ring with the same moment of inertia.

• If a problem states "radius of gyration k = 15 cm," treat the shape as a thin ring of radius 15 cm.
• For a thin ring, I = m·k².
• This simplifies calculations for particularly complex shapes.

🔄 Converting between I and k

• Given k and m: I = m·k²
• Given I and m: k = square root of (I/m)

Example from the excerpt: if mass = 10 kg and k = 5 m, then I = 10·5² = 250 kgm². Conversely, if I = 250 kgm² and m = 10 kg, then k = square root of (250/10) = 5 m.

Don't confuse: k is not the physical radius of the object; it is an equivalent radius that gives the same rotational inertia if all mass were concentrated in a thin ring.

🧭 Overview

🧠 One-sentence thesis

This collection of student-submitted examples demonstrates how to apply center-of-mass, center-of-gravity, centroid, and mass-moment-of-inertia calculations to real-world scenarios ranging from sports equipment to yoga poses.

📌 Key points (3–5)

• What the examples cover: center of mass (CM), center of gravity (CG), centroid, and mass moment of inertia (MMOI) for composite systems made of multiple shapes.
• Common approach: break complex objects into simple shapes (spheres, cylinders, rectangles), find properties for each part, then combine using formulas or the parallel axis theorem.
• Key distinction: for uniform shapes in uniform gravity, CM = CG = centroid; when shapes are non-uniform or gravity varies, these three points differ.
• Parallel axis theorem use: shift the axis of rotation from the center of mass of each part to a common system axis, then sum all contributions.
• Units and conversions: always convert to consistent units (e.g., inches to feet, cm to meters) before calculating; mass ≠ weight (divide weight by g to get mass).

⚽ Example 1: Soccer ball and leg

⚽ The problem setup

• A soccer ball (diameter 20 cm, mass 0.45 kg) and a person's leg (leg mass 18 kg, foot mass 8 kg).
• Tasks: find CM of the ball, MMOI of the ball, CM of the leg-and-foot system, and MMOI of the system about point A.

🧮 Key steps and formulas

1. Ball CM: because the ball is a uniform sphere, its CM is at its geometric center (10 cm in each direction from an arbitrary origin).
2. Ball MMOI: use the sphere formula

   I = (2/5) m r²
   Substituting: I_ball = (2/5)(0.45 kg)(0.1 m)² = 0.003 kg·m².

3. Leg-and-foot system CM: treat the foot as a rectangular box and the leg as a cylinder; find each part's CM by symmetry, then use the weighted average formula

   x_system = (m_foot · x_foot + m_leg · x_leg) / (m_foot + m_leg)
   and similarly for y and z.

4. Parallel axis theorem: compute each part's MMOI about its own CM, then shift to the system CM using

   I_new = I_cm + m d²
   where d is the distance between the two axes.

5. Final MMOI about A: shift the system MMOI from the system CM to point A using the parallel axis theorem again.

📐 Result

• System CM: P = (9.58 cm, 3.5 cm, 16.3 cm).
• MMOI about A: I_A ≈ 1.47 kg·m².

✅ Review note

The example emphasizes that small masses or small distances yield small inertias until the final shift to a distant axis (A), which adds a large m d² term.

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⛸️ Example 2: Figure skater spin

⛸️ The problem setup

• A 60 kg skater, 167 cm tall, modeled as a cylinder.
• Arms in: diameter 30 cm; arms out: diameter 60 cm.
• Given torque: 200 N·m.
• Questions: which configuration gives higher angular acceleration? Does lowering height change acceleration? How fast will she spin after 0.5 seconds?

🔄 Key insight: inertia and angular acceleration

• The relationship is

  Σ M = I α
  so α = Σ M / I.

• Lower inertia → higher angular acceleration for the same torque.

🧮 Calculations

1. Arms in: I₁ = ½ m r² = ½ (60 kg)(0.15 m)² = 0.675 kg·m²
   α₁ = 200 N·m / 0.675 kg·m² ≈ 296 rad/s².
2. Arms out: I₂ = ½ (60 kg)(0.3 m)² = 2.7 kg·m²
   α₂ = 200 N·m / 2.7 kg·m² ≈ 74 rad/s².
3. Height: the cylinder MMOI formula I_zz = ½ m r² does not include height, so lowering height does not change I or α.
4. Final angular velocity: ω = α t
   • Arms in: ω₁ = 296 rad/s² × 0.5 s = 148 rad/s ≈ 23.6 rotations/s.
   • Arms out: ω₂ = 74 rad/s² × 0.5 s = 37 rad/s ≈ 5.9 rotations/s.

🎯 Conclusion

Pulling limbs closer to the body dramatically increases angular acceleration and spin rate; height is irrelevant for rotation about the long axis.

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🗑️ Example 3: Compost containers

🗑️ The problem setup

• Two rectangular containers (A: 5 kg, 1 m × 2 m; B: 8 kg, 2 m × 3 m) tied by a 2 m chain.
• Origin: center of the chain.
• Find CM, CG, and centroid for each container and for the system.

📍 Key definitions and when they coincide

Center of mass (CM): the mass-weighted average position.
Center of gravity (CG): the point where the total weight acts.
Centroid: the geometric center (for uniform shapes).

• For uniform shapes in uniform gravity, CM = CG = centroid.

🧮 Approach

1. Individual containers: by symmetry, each container's CM is at (width/2, height/2) relative to its own corner, adjusted for distance from the origin.
   • Container A: CM_A = (2 m, 0.5 m).
   • Container B: CM_B = (−2.5 m, 1 m).
2. System CM: weighted average
   cm_x = (m_A · x_A + m_B · x_B) / (m_A + m_B)
   = (5 kg · 2 m + 8 kg · (−2.5 m)) / 13 kg ≈ −0.769 m.
   cm_y ≈ 0.808 m.
3. CG and centroid: because the shapes are uniform and gravity is uniform, CG_system = CM_system and the centroid is also at the geometric center (which matches CM for this configuration).

✅ Review note

The example stresses that the three concepts coincide only under specific conditions (uniform density, uniform gravity, simple geometry).

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🧘 Example 4: Yoga warrior-3 pose

🧘 The problem setup

• A 65 kg woman in warrior-3 pose: one leg raised 90 cm, torso horizontal, arms extended.
• Leg mass: 11.7 kg (both legs); torso: 50% of body mass (32.5 kg); arms: 4.5 kg each.
• Find the center of gravity from the origin (at the grounded foot).

🧮 Point-mass approximation

• Treat each body segment as a point mass located at its own CM.
• Leg in air: 11.7 kg at (0.425 m, 0.9 m).
• Leg on ground: 11.7 kg at (0.9 m, 0.425 m).
• Torso: 32.5 kg at (1.15 m, 0.9 m).
• Both arms: 9 kg at (1.9 m, 0.9 m).

📐 Calculation

• x̄ = Σ (m_i · x_i) / Σ m_i
  = [(11.7 · 0.425) + (11.7 · 0.9) + (32.5 · 1.15) + (9 · 1.9)] / 65 ≈ 1.08 m.
• ȳ ≈ 0.81 m.

🎯 Result

The center of gravity is at (1.08 m, 0.81 m), close to the torso (which has the most mass) and slightly lower, as expected for a horizontal pose.

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📱 Example 5: Selfie-stick mount

📱 The problem setup

• A selfie stick (pole) weighs 15 N; maximum allowable MMOI about the hand (bottom) is 5 kg·m².
• Find the maximum length d of the pole.

🧮 Key steps

1. Convert weight to mass: m = W / g = 15 N / 9.81 m/s² ≈ 1.529 kg.
2. MMOI about the center: for a uniform rod,

   I_cm = (1/12) m L².

3. Parallel axis theorem: shift from the rod's center to the hand (distance L/2):

   I_hand = I_cm + m (L/2)²
   = (1/12) m L² + (1/4) m L²
   = (1/3) m L².

4. Solve for L: 5 kg·m² = (1/3)(1.529 kg) L²
   L ≈ 3.13 m.

✅ Review

Substituting back confirms I_hand ≈ 5 kg·m², so the maximum pole length is 3.13 m.

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🌙 Example 6: Crescent-shaped disc

🌙 The problem setup

• A uniform metal disc (diameter 18 cm, mass 50.87 kg) has a 6 cm radius hole cut out, forming a crescent.
• Find the mass and CM of the crescent.

🧮 Approach: subtraction method

1. Density: ρ = m_full / (2π r₁²) = 50.87 kg / (2π · 9²) ≈ 0.1 kg/cm².
2. Area of crescent: A_c = 2π (r₁² − r₂²) = 2π (9² − 6²) = 90π cm².
3. Mass of crescent: m_c = ρ · A_c ≈ 28.27 kg.
4. CM of the cut circle: the cut is tangent to the edge, so its center is at (−3 cm, 0).
5. CM of crescent: treat the full disc (mass 50.87 kg, CM at origin) minus the cut (mass 50.87 − 28.27 = 22.6 kg, CM at −3 cm):
   CM_x = [(0 · 50.87) + (−3 · (−22.6))] / 28.27 ≈ 2.4 cm.

🎯 Result

The crescent's CM is at (2.4 cm, 0), shifted toward the side with more material.

⚠️ Don't confuse

The "negative mass" (−22.6 kg) is a bookkeeping trick for subtraction; it represents the removed material.

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🧻 Example 7: Tissue-paper roll

🧻 The problem setup

• An empty tissue roll (weight 2 lb, inner radius 0.9 in, outer radius 1 in, length 12 in) rolls down a slope.
• Find the MMOI about each axis.

🧮 Key steps

1. Convert weight to mass: m = 2 lb / 32.2 ft/s² ≈ 0.062 lb·s²/ft (slug).
2. Convert inches to feet: 0.9 in = 0.075 ft, 1 in = 0.083 ft, 12 in = 1 ft.
3. Hollow cylinder formulas:
   • About the x-axis (rolling axis):

     I_x = ½ m (r₁² + r₂²)
     ≈ 0.00039 lb·ft².

   • About the y- or z-axis:

     I_y = I_z = (1/12) m [3(r₁² + r₂²) + h²]
     ≈ 0.0054 lb·ft².

🎯 Insight

I_y > I_x, so the roll naturally rotates about the x-axis (the long axis) when rolling down a slope.

⚠️ Units reminder

Always convert weight to mass (divide by g) and ensure all lengths use the same unit before calculating.

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🔑 Common patterns across all examples

🔑 Break complex shapes into simple parts

• Spheres, cylinders, rectangles, and rods have standard MMOI formulas.
• Example: the leg-and-foot system is modeled as a cylinder (leg) + rectangular box (foot).

🔑 Use symmetry whenever possible

• For uniform shapes, the CM/CG/centroid is at the geometric center.
• Example: the soccer ball's CM is at its center; each rectangular container's CM is at (width/2, height/2).

🔑 Parallel axis theorem for composite systems

1. Find each part's MMOI about its own CM.
2. Shift each to a common reference axis using I_new = I_cm + m d².
3. Sum all contributions.

🔑 Always check units

• Convert weight to mass (m = W / g).
• Convert all lengths to the same unit (inches → feet, cm → meters).
• Final MMOI units: kg·m² or lb·ft² (slug·ft²).

🔑 Physical intuition checks

• CM should lie closer to heavier parts.
• Smaller radius or mass → smaller MMOI.
• Larger distance from axis → larger MMOI (due to the d² term in the parallel axis theorem).