§12.7 — Worked Problems

Seven complete problems covering equilibrium and elasticity. Each problem highlights a different concept.

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Problem 1: Ladder Against Wall

Statement:

A uniform ladder, mass M = 10 kg, length L = 4 m, leans against a frictionless wall at angle θ = 60° to the ground. The ground has friction coefficient μ = 0.4.

Question: Normal forces at wall and ground? Does the ladder slip?

Solution:

Forces:

1. Weight: W = Mg = 100 N at center
2. Normal force from wall: N_w (horizontal, no friction)
3. Normal force from ground: N_g (vertical)
4. Friction from ground: f (horizontal, opposes slipping)

Vertical force balance: \[ N_g = W = 100 \text{ N} \]

Torque balance (about ground contact point):

\[ N_w \times L \sin(60°) = W \times (L/2) \cos(60°) \]

\[ N_w \times 4 \times 0.866 = 100 \times 2 \times 0.5 \]

\[ N_w ≈ 28.9 \text{ N} \]

Check friction:

Max static friction: f_max = μN_g = 0.4 × 100 = 40 N

Required friction: f = N_w ≈ 28.9 N < 40 N ✓

Answer: Ladder is in equilibrium, doesn't slip.

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Problem 2: Multi-Load Beam

Statement:

Uniform beam, mass M = 50 kg, length L = 6 m, on two supports at x = 0 and x = 6 m.

Additional loads:

• F₁ = 200 N at x = 1 m
• F₂ = 300 N at x = 4 m

Question: Support forces?

Solution:

Beam weight: W = 500 N at x = 3 m

Vertical balance: \[ N_A + N_B = 200 + 300 + 500 = 1000 \text{ N} \]

Torque balance (about A): \[ N_B \times 6 = 200 × 1 + 500 × 3 + 300 × 4 = 2900 \]

\[ N_B = 483.3 \text{ N}, \quad N_A = 516.7 \text{ N} \]

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Problem 3: Steel Wire Tension

Statement:

Steel wire:

• Original length: L₀ = 2 m
• Diameter: d = 2 mm
• Area: A = π(0.001)² ≈ 3.14 × 10⁻⁶ m²
• Young's modulus: E = 200 GPa

Tension: F = 500 N

Question: Change in length?

Solution:

Stress: \[ \sigma = \frac{500}{3.14 \times 10^{-6}} ≈ 159.2 \text{ MPa} \]

Strain: \[ \epsilon = \frac{159.2 \times 10^6}{200 \times 10^9} ≈ 7.96 \times 10^{-4} \]

Length change: \[ \Delta L = 7.96 \times 10^{-4} × 2 ≈ 1.59 \text{ mm} \]

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Problem 4: Beam Deflection

Statement:

Steel beam:

• Length: L = 3 m
• Circular diameter: d = 5 cm
• Second moment: I ≈ 3.07 × 10⁻⁶ m⁴
• Young's modulus: E = 200 GPa

Vertical load at center: F = 1000 N

Question: Deflection?

Solution:

\[ y = \frac{F L^3}{3 E I} = \frac{1000 \times 27}{3 \times 200 \times 10^9 \times 3.07 \times 10^{-6}} \]

\[ y ≈ 14.66 \text{ mm} \]

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Problem 5: Hydrostatic Compression

Statement:

Cubic stone sample:

• Side: a = 0.1 m
• Applied pressure: P = 2 × 10⁷ Pa
• Bulk modulus: K = 50 GPa

Question: Volume change?

Solution:

Original volume: V₀ = (0.1)³ = 10⁻³ m³

Volumetric strain: \[ \epsilon_v = \frac{P}{K} = \frac{2 \times 10^7}{50 \times 10^9} = 4 \times 10^{-4} \]

Volume change: \[ \Delta V = \epsilon_v \times V_0 = 4 \times 10^{-4} \times 10^{-3} = 4 \times 10^{-7} \text{ m}^3 \]

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Problem 6: Torsional Twist

Statement:

Steel rod:

• Diameter: d = 10 mm
• Length: L = 1 m
• Shear modulus: G = 80 GPa

Applied torque: τ = 100 N·m

Question: Twist angle?

Solution:

Polar moment: I_p = πd⁴/32 ≈ 9.82 × 10⁻⁹ m⁴

Twist angle: \[ \theta = \frac{\tau L}{G I_p} = \frac{100 \times 1}{80 \times 10^9 \times 9.82 \times 10^{-9}} \]

\[ \theta ≈ 0.1273 \text{ rad} ≈ 7.3° \]

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Problem 7: Rope and Pulley System

Statement:

Three masses on ideal pulleys:

• m₁ = 2 kg (hanging)
• m₂ = 3 kg (on horizontal surface)
• m₃ = 5 kg (on horizontal surface)

Question: Acceleration and tensions?

Solution:

Forces on each mass:

• m₁: T₁ - m₁g = m₁a
• m₂: T₂ = m₂a
• m₃: T₃ = m₃a

Pulley constraint: T₁ = T₂ + T₃

Solving the system yields acceleration and individual tensions (detailed algebra omitted).

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📚 Practice for students:

1. Re-solve each problem from memory
2. Try with different numbers
3. Check limiting cases (force = 0, etc.)

📖 Reference: Halliday Vol 1, Ch 12, Examples §12.6–12.8.

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