Lesson 4: Newton's Laws — Applied Problems

Week 4 — Solving real-world mechanics problems

🎯 Learning Objectives

📐 Problem-Solving Strategy with Newton's Laws

Follow these steps for every Newton's Law problem:

  1. Identify the system — What object are you analyzing?
  2. Draw a free-body diagram — Show all forces on the object
  3. Choose coordinates — Usually align x with the direction of motion
  4. Break forces into components — Use sin/cos for angled forces
  5. Apply Newton's Second Law — ΣFx = max, ΣFy = may
  6. Solve — Solve the equations for the unknown(s)
  7. Check — Does the answer make sense? Units? Magnitude?

🔧 Special Cases

Case 1: Atwood Machine (Two Masses on a Pulley)

Two masses m₁ and m₂ hang from a massless string over a frictionless pulley.

Acceleration: a = (m₂ − m₁)g / (m₁ + m₂)
Tension: T = 2m₁m₂g / (m₁ + m₂)
📝 Worked Example: Atwood Machine

m₁ = 3 kg, m₂ = 5 kg. Find the acceleration and tension.

a = (5 − 3)(9.8)/(3 + 5) = 19.6/8 = 2.45 m/s²
T = 2(3)(5)(9.8)/(8) = 294/8 = 36.75 N

✅ The heavier mass accelerates down at 2.45 m/s², tension = 36.75 N

Case 2: Blocks on a Ramp (Connected by a String)

Block A sits on a ramp; block B hangs freely, connected by a string over a pulley at the ramp's top.

📝 Worked Example:

Block A (10 kg) sits on a 30° ramp (μk = 0.2). It's connected to block B (8 kg) hanging off the top. Find acceleration.

Step 1 — Forces on Block A:

  • Weight parallel: m₁g sin θ = 10 × 9.8 × 0.5 = 49 N
  • Weight perpendicular: m₁g cos θ = 10 × 9.8 × 0.866 = 84.9 N
  • Normal: N = 84.9 N
  • Friction: fk = 0.2 × 84.9 = 17.0 N

Step 2 — Forces on Block B:

  • Weight: m₂g = 8 × 9.8 = 78.4 N down

Step 3 — Set up equations (assuming B moves down, A moves up the ramp):

For A: T − m₁g sin θ − fk = m₁a
For B: m₂g − T = m₂a

Step 4 — Add equations:

m₂g − m₁g sin θ − fk = (m₁ + m₂)a
78.4 − 49 − 17.0 = (10 + 8)a
12.4 = 18a
a = 12.4/18 = 0.69 m/s²

✅ a = 0.69 m/s² (B moves down, A slides up the ramp)

Case 3: Circular Motion

When an object moves in a circle at constant speed, it experiences centripetal acceleration toward the center:

a₍ = v²/r = ω²r
F₍ = mv²/r
💡 Key Point:

Centripetal force is not a new force — it's the label for whatever force is pulling the object inward. For a car turning, it's friction. For a satellite, it's gravity. For a ball on a string, it's tension.

📝 Worked Example: Car on a Curve

A 1200-kg car rounds a 50-m radius curve at 20 m/s. What is the minimum coefficient of friction needed to avoid skidding?

Step 1 — Centripetal force needed:

F₍ = mv²/r = (1200)(20²)/50 = 48000/50 = 9600 N

Step 2 — Maximum friction available:

Ffriction,max = μsN = μsmg = μs(1200)(9.8) = 11760μs

Step 3 — Set equal:

11760μs = 9600
μs = 9600/11760 = 0.82

✅ Minimum μs = 0.82 (typical rubber-on-dry-asphalt ≈ 0.8-1.0)

📐 Banked Curves

On a banked curve (tilted road), part of the normal force provides the centripetal force:

No-friction ideal speed: v = √(rg tan θ)

✏️ Practice Problems

Problem 1: Two masses, m₁ = 4 kg and m₂ = 6 kg, hang from a pulley. What is the acceleration?

Problem 2: A 2000-kg car turns a 100-m radius curve at 25 m/s. What centripetal force is needed?

Problem 3: A 5-kg block on a 37° ramp (μs = 0.5). Will it slide? (sin 37° = 0.6, cos 37° = 0.8)

✅ Check Your Understanding

Q1: If you're in an elevator accelerating upward, does the normal force from the floor feel larger or smaller than usual?

Click to reveal answer

Larger. The normal force must both support your weight AND provide upward acceleration: N = mg + ma. That's why you feel heavier.

Q2 (mini-problem): A 3-kg block is pulled up a 40° ramp at constant speed by a rope parallel to the ramp. μk = 0.15. What tension is needed?

Click to reveal answer

Constant speed → a = 0 → T = parallel component + friction

T = mg sin θ + μk mg cos θ = 3(9.8)(0.643) + 0.15 × 3(9.8)(0.766) = 18.92 + 3.38 = 22.3 N