AP Physics C Mechanics Extra - How to Solve: Extended Free-Body Diagrams and Torque About a Pivot

How to Solve: Extended Free-Body Diagrams and Torque About a Pivot

Introduction

Mastering extended free-body diagrams and torque about a pivot unlocks 10–15% of your AP Physics 1 exam score—and it’s the difference between a 3 and a 5. This skill lets you solve real-world problems like: - Why a ladder doesn’t slip when leaned against a wall. - How a crane lifts heavy loads without toppling. - Why a seesaw balances (or doesn’t) when kids of different weights sit on it.

If you can’t draw the forces and calculate torques correctly, you’ll lose points on every rotational equilibrium question. Let’s fix that.

WHAT YOU NEED TO KNOW FIRST

Before diving in, you must already understand:
1. Newton’s Laws (especially N1 and N3) – Forces come in pairs, and objects in equilibrium have balanced forces.
2. Free-Body Diagrams (FBDs) – How to draw forces acting on a point object (weight, normal, tension, friction).
3. Basic Torque – Torque = force × lever arm (τ = r × F), and the sign convention (clockwise = negative, counterclockwise = positive).

If any of these are shaky, pause here and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas

1. Torque (τ)
2. Formula: τ = r × F × sinθ
3. Variables:
   • τ = torque (N·m)
   • r = distance from pivot to point of force application (m)
   • F = force (N)
   • θ = angle between r and F

4. MEMORIZE THIS: For perpendicular forces (θ = 90°), τ = r × F (sin90° = 1).

5. Net Torque (Στ)

6. Formula: Στ = τ₁ + τ₂ + τ₃ + ... = 0 (for equilibrium)

7. MEMORIZE THIS: Sum of all torques must be zero for rotational equilibrium.

8. Net Force (ΣF)

9. Formula: ΣF = F₁ + F₂ + F₃ + ... = 0 (for translational equilibrium)
10. Given on exam sheet (but you must know how to apply it).

STEP-BY-STEP METHOD

Follow these exact steps for every problem. No shortcuts.

Step 1: Identify the Extended Object and Pivot Point

• Circle the object in the problem.
• Underline the pivot point (if given). If not given, you choose the pivot (pick the point where the most forces act to simplify calculations).

Step 2: Draw the Extended Free-Body Diagram (EFBD)

• Sketch the object as a line or shape (not a dot).
• Draw all forces acting on the object, where they actually act (not at the center of mass unless specified).
• Weight (mg) acts at the center of mass (usually the geometric center for uniform objects).
• Normal forces act at the point of contact (e.g., where a ladder touches the wall).
• Tension/friction forces act at their points of application.
• Label every force with:
• Magnitude (if known)
• Direction (use arrows)
• Distance from pivot (r)

Step 3: Choose a Sign Convention for Torques

• Standard convention:
• Counterclockwise (CCW) = positive torque.
• Clockwise (CW) = negative torque.
• Write this on your paper: CCW = +, CW = –.

Step 4: Write the Translational Equilibrium Equations

• Break forces into x and y components (if needed).
• Write:
• ΣFₓ = 0
• ΣFᵧ = 0
• Solve for unknown forces.

Step 5: Write the Rotational Equilibrium Equation

• For each force, calculate torque:
• τ = r × F × sinθ
• If force is perpendicular to r, τ = r × F.
• Sum all torques: Στ = 0.
• Solve for unknowns.

Step 6: Solve the System of Equations

• Use algebra to solve for unknown forces or distances.
• Check units (N, m, N·m).

Step 7: Verify Your Answer

• Does the object not accelerate linearly? (ΣF = 0)
• Does the object not rotate? (Στ = 0)
• Do the numbers make sense? (e.g., normal force can’t be negative).

WORKED EXAMPLES

Example 1 – Basic: Ladder Against a Wall

Problem: A 5.0 m uniform ladder of mass 20 kg leans against a frictionless wall at 53° above the horizontal. What is the minimum coefficient of static friction (μₛ) between the ladder and the ground to prevent slipping?

Step 1: Identify Object and Pivot

• Object: Ladder.
• Pivot: Base of the ladder (where it touches the ground—this eliminates friction and normal force from the ground in torque calculations).

Step 2: Draw EFBD

• Sketch ladder as a line at 53°.
• Forces:
• Weight (mg) at center (2.5 m from base).
• Normal force from wall (N_wall) at top (5.0 m from base), horizontal.
• Normal force from ground (N_ground) at base, vertical.
• Friction (f) at base, horizontal (opposes slipping).

Step 3: Sign Convention

• CCW = +, CW = –.

Step 4: Translational Equilibrium

• ΣFₓ = 0 → N_wall – f = 0 → N_wall = f
• ΣFᵧ = 0 → N_ground – mg = 0 → N_ground = mg = (20 kg)(9.8 m/s²) = 196 N

Step 5: Rotational Equilibrium (about base)

• Torque from weight: τ_weight = (2.5 m)(mg)sin(37°) = (2.5)(196)(0.602) = 295 N·m (CW = –)
• Why sin(37°)? The angle between r (ladder) and mg is 90° – 53° = 37°.
• Torque from N_wall: τ_N_wall = (5.0 m)(N_wall)sin(53°) = (5.0)(N_wall)(0.799) = 3.995 N_wall (CCW = +)
• Στ = 0 → 3.995 N_wall – 295 = 0 → N_wall = 73.8 N
• From ΣFₓ: f = N_wall = 73.8 N

Step 6: Solve for μₛ

• f = μₛ N_ground → 73.8 = μₛ (196) → μₛ = 0.38

Step 7: Verify

• ΣFₓ = 0? 73.8 – 73.8 = 0 ✔
• ΣFᵧ = 0? 196 – 196 = 0 ✔
• Στ = 0? 3.995(73.8) – 295 ≈ 0 ✔
• μₛ = 0.38 is reasonable (typical for rubber on concrete).

What we did and why: We picked the base as the pivot to eliminate two forces (friction and normal from ground) from torque calculations. We used translational equilibrium to relate N_wall and friction, then torque to find N_wall. Finally, we used friction’s definition to find μₛ.

Example 2 – Medium: Hanging Sign

Problem: A 3.0 m uniform horizontal beam of mass 10 kg is attached to a wall with a hinge. A 5.0 kg sign hangs from the end. A cable at 30° above the horizontal supports the beam at 2.0 m from the wall. Find the tension in the cable and the hinge force components.

Step 1: Identify Object and Pivot

• Object: Beam.
• Pivot: Hinge (eliminates hinge forces from torque calculations).

Step 2: Draw EFBD

• Sketch beam as a horizontal line.
• Forces:
• Weight of beam (mg) at center (1.5 m from hinge).
• Weight of sign (Mg) at end (3.0 m from hinge).
• Tension (T) at 2.0 m from hinge, 30° above horizontal.
• Hinge force (Hₓ, Hᵧ) at hinge.

Step 3: Sign Convention

• CCW = +, CW = –.

Step 4: Translational Equilibrium

• ΣFₓ = 0 → Hₓ – T cos(30°) = 0 → Hₓ = T cos(30°)
• ΣFᵧ = 0 → Hᵧ + T sin(30°) – mg – Mg = 0 → Hᵧ + 0.5T = (10 + 5)(9.8) = 147

Step 5: Rotational Equilibrium (about hinge)

• Torque from beam weight: τ_beam = (1.5 m)(mg) = (1.5)(98) = 147 N·m (CW = –)
• Torque from sign: τ_sign = (3.0 m)(Mg) = (3.0)(49) = 147 N·m (CW = –)
• Torque from tension: τ_T = (2.0 m)(T)sin(30°) = (2.0)(T)(0.5) = T (CCW = +)
• Στ = 0 → T – 147 – 147 = 0 → T = 294 N

Step 6: Solve for Hinge Forces

• Hₓ = T cos(30°) = 294 × 0.866 = 254 N
• Hᵧ = 147 – 0.5T = 147 – 147 = 0 N

Step 7: Verify

• ΣFₓ = 0? 254 – 294 cos(30°) ≈ 0 ✔
• ΣFᵧ = 0? 0 + 294 sin(30°) – 147 ≈ 0 ✔
• Στ = 0? 294 – 147 – 147 = 0 ✔

What we did and why: We chose the hinge as the pivot to eliminate hinge forces from torque calculations. We broke tension into components and used torque to find T, then used translational equilibrium to find the hinge forces.

Example 3 – Exam-Style: Disguised Problem

Problem: A 4.0 m uniform plank of mass 12 kg rests on two supports, A and B, 1.0 m and 3.0 m from the left end, respectively. A 20 kg child stands 0.5 m from the left end. What is the normal force at support B?

Step 1: Identify Object and Pivot

• Object: Plank.
• Pivot: Support A (eliminates N_A from torque calculations).

Step 2: Draw EFBD

• Sketch plank as a line.
• Forces:
• Weight of plank (mg) at center (2.0 m from left end).
• Weight of child (Mg) at 0.5 m from left end.
• Normal force at A (N_A) at 1.0 m from left end.
• Normal force at B (N_B) at 3.0 m from left end.

Step 3: Sign Convention

• CCW = +, CW = –.

Step 4: Translational Equilibrium

• ΣFᵧ = 0 → N_A + N_B – mg – Mg = 0 → N_A + N_B = (12 + 20)(9.8) = 313.6 N

Step 5: Rotational Equilibrium (about A)

• Torque from plank weight: τ_plank = (1.0 m)(mg) = (1.0)(117.6) = 117.6 N·m (CW = –)
• Wait! The plank’s center is 2.0 m from the left, but the pivot is at 1.0 m. So r = 2.0 – 1.0 = 1.0 m.
• Torque from child: τ_child = (0.5 m)(Mg) = (0.5)(196) = 98 N·m (CW = –)
• Torque from N_B: τ_N_B = (2.0 m)(N_B) (CCW = +)
• Why 2.0 m? Support B is 3.0 m from the left, but the pivot is at 1.0 m, so r = 3.0 – 1.0 = 2.0 m.
• Στ = 0 → 2.0 N_B – 117.6 – 98 = 0 → N_B = 107.8 N

Step 6: Solve for N_A

• N_A = 313.6 – N_B = 313.6 – 107.8 = 205.8 N

Step 7: Verify

• ΣFᵧ = 0? 205.8 + 107.8 – 313.6 ≈ 0 ✔
• Στ = 0? 2.0(107.8) – 117.6 – 98 ≈ 0 ✔

What we did and why: We chose support A as the pivot to eliminate N_A from torque calculations. We carefully calculated lever arms relative to the pivot, not the left end. The key was adjusting distances to the pivot point.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Listen up—this is your 60-second crash course for acing torque problems on exam day.

1. Draw the extended free-body diagram. Not a dot—a line or shape. Forces go where they act, not at the center of mass.
2. Pick a pivot. Choose the point where the most forces act (usually a hinge or support). This eliminates those forces from torque calculations.
3. Sign convention: CCW = +, CW = –. Write it on your paper.
4. Write two equations:
5. ΣFₓ = 0 and ΣFᵧ = 0 (translational equilibrium).
6. Στ = 0 (rotational equilibrium).
7. Calculate torques: τ = r × F × sinθ. If the force is perpendicular, τ = r × F.
8. Solve the system. Use algebra to find unknowns.
9. Check your answer. Does ΣF = 0? Does Στ = 0? Do the numbers make sense?

Common traps? Hidden forces, non-perpendicular angles, and choosing the wrong pivot. Avoid them by sticking to the steps.

You’ve got this. Now go crush that exam."