How to Solve: Newton’s Laws & Free-Body Diagrams (Inclines, Pulleys)

How to Solve: Newton’s Laws & Free-Body Diagrams (Inclines, Pulleys)

Introduction

"Mastering free-body diagrams on inclines and pulleys unlocks 10–15% of your AP Physics 1/Mechanics exam score—and the same skills let you predict whether a car will skid on a hill or how much weight a crane can lift. One diagram, three equations, full credit."

What You Need To Know First

1. Newton’s 2nd Law: Fₙₑₜ = ma (net force = mass × acceleration).
2. Trigonometry: sin θ (opposite/hypotenuse) and cos θ (adjacent/hypotenuse) for inclines.
3. Tension: A pulling force in ropes/strings, same magnitude throughout (ideal pulleys).

Key Vocabulary

Formulas To Know

Step-by-Step Method

Step 1: Identify the System

• Circle one object to analyze (e.g., a block on a ramp, a mass on a pulley).
• Ignore everything else for now.

Step 2: Draw the Free-Body Diagram (FBD)

• Axes: Choose x and y directions. For inclines, align x parallel to the ramp, y perpendicular.
• Forces:
• Weight (mg): Always straight down.
• Normal Force (N): Perpendicular to the surface.
• Friction (f): Parallel to the surface, opposes motion.
• Tension (T): Along the rope, away from the object.
• Label all forces with arrows. Write magnitudes (e.g., mg, N, T) next to arrows.

Step 3: Split Forces into Components (Inclines Only)

• Weight (mg):
• Parallel to ramp: mg sin θ (downhill).
• Perpendicular to ramp: mg cos θ (into ramp).
• Normal Force (N): N = mg cos θ (if no vertical acceleration).
• Friction (f): f = μN (if moving) or f ≤ μₛN (if stationary).

Step 4: Write Newton’s 2nd Law Equations

• x-direction (parallel to ramp/pulley): ΣFₓ = maₓ
• Example (incline): mg sin θ – f = ma
• y-direction (perpendicular to ramp): ΣFᵧ = maᵧ
• Example (incline): N – mg cos θ = 0 (no vertical acceleration).
• For pulleys: Write Fₙₑₜ = ma for each mass, then solve the system.

Step 5: Solve for Unknowns

• Plug in known values (e.g., m, θ, μ).
• Solve for a, T, or N using algebra.
• Check units: kg, m/s², N.

Step 6: Verify Reasonableness

• Does the answer make sense?
• Acceleration a should be less than g (9.8 m/s²).
• Tension T should be between the weights of the two masses (for pulleys).
• Normal force N should be less than mg on an incline.

Worked Examples

Example 1 – Basic: Block on a Frictionless Incline

Problem: A 5 kg block slides down a 30° frictionless ramp. Find its acceleration.

Step 1: System = the 5 kg block. Step 2: FBD: - mg down. - N perpendicular to ramp. - No friction (frictionless). Step 3: Components: - mg sin 30° = (5)(9.8)(0.5) = 24.5 N (parallel). - mg cos 30° = (5)(9.8)(0.866) = 42.4 N (perpendicular). Step 4: Equations: - x: mg sin θ = ma → 24.5 = 5a → a = 4.9 m/s². - y: N – mg cos θ = 0 → N = 42.4 N. Step 5: a = 4.9 m/s² (down the ramp). What we did and why: We split the weight into components, applied Newton’s 2nd Law in the x-direction, and solved for acceleration. The y-direction gave the normal force, but we didn’t need it for a.

Example 2 – Medium: Block on a Rough Incline

Problem: A 4 kg block is held at rest on a 20° ramp with μₛ = 0.3. Will it slide? If so, find its acceleration (μₖ = 0.2).

Step 1: System = the 4 kg block. Step 2: FBD: - mg down. - N perpendicular to ramp. - fₛ up the ramp (static friction). Step 3: Components: - mg sin 20° = (4)(9.8)(0.342) = 13.4 N (downhill). - mg cos 20° = (4)(9.8)(0.94) = 36.8 N (perpendicular). Step 4: Check if it slides: - Max static friction: fₛ,max = μₛN = 0.3 × 36.8 = 11.0 N. - mg sin θ = 13.4 N > 11.0 N → It slides. Step 5: Now use kinetic friction: - fₖ = μₖN = 0.2 × 36.8 = 7.36 N. - x: mg sin θ – fₖ = ma → 13.4 – 7.36 = 4a → a = 1.51 m/s². What we did and why: We first checked if static friction could hold the block. When it couldn’t, we switched to kinetic friction and solved for acceleration.

Example 3 – Exam-Style: Atwood’s Machine with Friction

Problem: Two blocks (m₁ = 3 kg, m₂ = 5 kg) are connected by a light string over a pulley. The 3 kg block is on a horizontal table (μₖ = 0.1). Find the acceleration of the system.

Step 1: System = m₁ (on table) and m₂ (hanging). Step 2: FBDs: - m₁: T right, fₖ left, N up, mg down. - m₂: T up, mg down. Step 3: Forces: - m₁: N = m₁g = 3 × 9.8 = 29.4 N. - fₖ = μₖN = 0.1 × 29.4 = 2.94 N. - m₂: mg = 5 × 9.8 = 49 N. Step 4: Equations: - m₁: T – fₖ = m₁a → T – 2.94 = 3a. - m₂: m₂g – T = m₂a → 49 – T = 5a. Step 5: Solve the system: - Add equations: (T – 2.94) + (49 – T) = 3a + 5a → 46.06 = 8a → a = 5.76 m/s². What we did and why: We wrote Fₙₑₜ = ma for both masses, combined the equations, and solved for a. Friction on m₁ reduced the net force.

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s the night-before cheat sheet for Newton’s Laws on inclines and pulleys:
1. One object at a time: Circle it, draw its FBD.
2. Inclines: Split mg into mg sin θ (downhill) and mg cos θ (into ramp). Align x parallel to the ramp.
3. Pulleys: Write Fₙₑₜ = ma for each mass. Tension T is the same on both sides (ideal pulley).
4. Friction: f = μN. Check if it’s static (holding) or kinetic (sliding).
5. Solve: Combine equations, plug in numbers, check units. You’ve got this—one diagram, three equations, full credit. Now go ace that exam!