AP Physics 1 - How to Solve: Newton’s Laws & Free-Body Diagrams (Inclines, Pulleys)

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

Introduction

Mastering Newton’s Laws and free-body diagrams on inclines and pulleys unlocks 15-20% of your AP Physics 1 exam score—and it’s the foundation for every mechanics problem in C: Mechanics. If you can solve these, you can solve any force problem, from a block sliding down a ramp to a pulley lifting a crate.

WHAT YOU NEED TO KNOW FIRST

Before diving in, you must already understand:
1. Newton’s Three Laws (especially N2L: ΣF = ma)
2. Vector components (breaking forces into x- and y-directions)
3. Basic trigonometry (sin, cos, tan for angles)

If you’re shaky on these, stop now and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas

STEP-BY-STEP METHOD

Follow these exact steps for every Newton’s Laws problem:

Step 1: Identify the System

• Circle the object(s) you’re analyzing.
• If there’s a pulley, treat each mass separately unless told otherwise.

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

• One object per FBD.
• Label all forces with arrows pointing in the correct direction.
• Weight (W = mg) → Always straight down.
• Normal Force (N) → Perpendicular to the surface.
• Tension (T) → Along the rope, away from the object.
• Friction (f) → Parallel to the surface, opposing motion.
• For inclines:
• Break weight (mg) into parallel (mg sinθ) and perpendicular (mg cosθ) components.
• Normal force (N) = mg cosθ (if no vertical acceleration).

Step 3: Choose a Coordinate System

• For inclines: Align axes parallel and perpendicular to the ramp.
• For pulleys: Align axes along the string (usually vertical).
• Label + and – directions (e.g., up = +, down = –).

Step 4: Write Newton’s 2nd Law (ΣF = ma) for Each Direction

• X-direction (parallel to motion): ΣFₓ = maₓ
• Y-direction (perpendicular to motion): ΣFᵧ = maᵧ
• If no acceleration in a direction, ΣF = 0 (e.g., no vertical acceleration on a flat surface → N = mg).

Step 5: Solve for Unknowns

• Combine equations if needed (e.g., tension in a pulley system).
• Check units (N, kg, m/s²).
• Does the answer make sense? (e.g., acceleration can’t be negative if the object is speeding up).

Step 6: Answer the Question

• Re-read the question to ensure you answered what was asked (e.g., "Find tension" vs. "Find acceleration").

WORKED EXAMPLES

Example 1 – Basic Incline (No Friction)

Problem: A 5.0 kg block slides down a frictionless 30° incline. What is its acceleration?

Solution:
1. System: The block.
2. FBD: - Weight (mg) → down. - Normal force (N) → perpendicular to incline. - No friction (frictionless).
3. Coordinate System: - X-axis: Parallel to incline (down = +). - Y-axis: Perpendicular to incline (up = +).
4. Newton’s 2nd Law: - X-direction: ΣFₓ = mg sinθ = ma → (5.0 kg)(9.8 m/s²)(sin 30°) = (5.0 kg)a → 24.5 N = 5.0a → a = 4.9 m/s² - Y-direction: ΣFᵧ = N – mg cosθ = 0 (no vertical acceleration) → N = mg cosθ (not needed here).
5. Answer: The acceleration is 4.9 m/s² down the incline.

What we did and why: - We broke weight into components because the block moves parallel to the incline. - We ignored the y-direction because there’s no acceleration perpendicular to the ramp. - We used sinθ for the parallel component (opposite side of the triangle).

Example 2 – Medium Incline (With Friction)

Problem: A 10 kg block is on a 20° incline with μₖ = 0.20. What is its acceleration?

Solution:
1. System: The block.
2. FBD: - Weight (mg) → down. - Normal force (N) → perpendicular to incline. - Kinetic friction (fₖ) → up the incline (opposes motion).
3. Coordinate System: - X-axis: Parallel to incline (down = +). - Y-axis: Perpendicular to incline (up = +).
4. Newton’s 2nd Law: - Y-direction: ΣFᵧ = N – mg cosθ = 0 → N = mg cosθ = (10 kg)(9.8 m/s²)(cos 20°) = 92.1 N - X-direction: ΣFₓ = mg sinθ – fₖ = ma → fₖ = μₖN = (0.20)(92.1 N) = 18.4 N → (10 kg)(9.8 m/s²)(sin 20°) – 18.4 N = (10 kg)a → 33.5 N – 18.4 N = 10a → a = 1.51 m/s²
5. Answer: The acceleration is 1.51 m/s² down the incline.

What we did and why: - We calculated normal force first because friction depends on it. - We subtracted friction because it opposes motion. - We used cosθ for the perpendicular component (adjacent side of the triangle).

Example 3 – Exam-Style Pulley System

Problem: Two blocks (m₁ = 3.0 kg, m₂ = 5.0 kg) are connected by a light string over a frictionless pulley. Block 1 is on a horizontal surface (μₖ = 0.10), and Block 2 hangs vertically. Find the acceleration of the system and the tension in the string.

Solution:
1. System: Two separate FBDs (one for each block).
2. FBD for Block 1 (horizontal): - Weight (m₁g) → down. - Normal force (N₁) → up. - Tension (T) → right. - Friction (fₖ) → left.
3. FBD for Block 2 (vertical): - Weight (m₂g) → down. - Tension (T) → up.
4. Coordinate System: - Block 1: Right = +, Up = +. - Block 2: Down = + (to match Block 1’s motion).
5. Newton’s 2nd Law: - Block 1 (X-direction): ΣFₓ = T – fₖ = m₁a → fₖ = μₖN₁ = μₖm₁g = (0.10)(3.0 kg)(9.8 m/s²) = 2.94 N → T – 2.94 N = (3.0 kg)a - Block 2 (Y-direction): ΣFᵧ = m₂g – T = m₂a → (5.0 kg)(9.8 m/s²) – T = (5.0 kg)a → 49 N – T = 5.0a
6. Combine Equations: - From Block 1: T = 3.0a + 2.94 - Substitute into Block 2: 49 – (3.0a + 2.94) = 5.0a → 46.06 = 8.0a → a = 5.76 m/s² - Then, T = 3.0(5.76) + 2.94 = 20.2 N
7. Answer: - Acceleration = 5.76 m/s² (Block 2 moves down). - Tension = 20.2 N.

What we did and why: - We treated each block separately but linked them via tension. - We aligned the coordinate systems so both blocks accelerate in the same direction. - We solved two equations simultaneously to find acceleration and tension.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

Listen up—this is your last-minute lifeline.

1. Draw the FBD first. Every force, every arrow. No shortcuts.
2. Break weight into components on inclines: mg sinθ (parallel), mg cosθ (perpendicular).
3. Choose axes wisely: Parallel/perpendicular for inclines, vertical for pulleys.
4. Write ΣF = ma for each direction. If no acceleration, ΣF = 0.
5. Solve for unknowns. Combine equations if needed (like in pulleys).
6. Check units and reasonableness. Acceleration can’t be 100 m/s²—something’s wrong.

For pulleys: Tension is not the same as weight. Use T = mg ± ma. For inclines: Normal force is not mg—it’s mg cosθ.

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