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A Ladder of length = 4 m leans against a frame house at an angle 6 = 74 degrees. Frictionless contacts occurs with a wall of height h = 3.5 m. The ladder has a mass m; — 18kg. The service worker who must climb the ladder has a mass of m; = 65 kg Assume the contact point with the ground has a coefficient of friction between the ladder footing and the ground of yz, = 0.29. Compute the friction force needed to keep the ladder in equilibrium when the worker is at the top of the ladder. What is the maximum friction that can be supported in this configuration? Note: Express answers as magnitudes only and ignore signs/directions. L / | [4

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Uploaded ImageA Ladder of length = 4 m leans against a frame house at an angle 6 = 74 degrees. Frictionless contacts occurs with a wall of height h = 3.5 m. The ladder has a mass m; — 18kg. The service worker who must climb the ladder has a mass of m; = 65 kg Assume the contact point with the ground has a coefficient of friction between the ladder footing and the ground of yz, = 0.29. Compute the friction force needed to keep the ladder in equilibrium when the worker is at the top of the ladder. What is the maximum friction that can be supported in this configuration? Note: Express answers as magnitudes only and ignore signs/directions. L / | [4

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Asked by: Test User 100096
Created at: 2025-05-06 07:21:20

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Let's break down the solution step by step: ## 1. **Given Data** - Length of ladder, \( L = 4\, \text{m} \) - Angle with ground, \( \theta = 74^\circ \) - Height of wall, \( h = 3.5\, \text{m} \) - Mass of ladder, \( m_1 = 18\, \text{kg} \) - Mass of worker, \( m_2 = 65\, \text{kg} \) - Coefficient of static friction, \( \mu_s = 0.29 \) - Gravity, \( g = 9.8\, \text{m/s}^2 \) --- ## 2. **Forces on the Ladder** - **Weight of ladder** at center: \( W_1 = m_1 g \) - **Weight of worker** at top: \( W_2 = m_2 g \) - **Normal force at ground**: \( N \) - **Friction force at ground**: \( f \) - **Normal force at wall (horizontal)**: \( F_w \) --- ## 3. **Equilibrium Conditions** ### (a) **Vertical Forces** \[ N = W_1 + W_2 \] ### (b) **Horizontal Forces** \[ f = F_w \] ### (c) **Torque about the ground point** Taking moments about the base (ground contact): - Perpendicular distance from base to ladder top: \( L \sin\theta \) - Perpendicular distance from base to ladder center: \( \frac{L}{2}\sin\theta \) - Wall force acts at height \( h \) (or \( L \sin\theta \)) #### **Moment equation:** \[ F_w \cdot h = W_1 \cdot \left(\frac{L}{2}\cos\theta\right) + W_2 \cdot (L \cos\theta) \] --- ## 4. **Plug in the Numbers** - \( W_1 = 18 \times 9.8 = 176.4\, \text{N} \) - \( W_2 = 65 \times 9.8 = 637\, \text{N} \) - \( N = 176.4 + 637 = 813.4\, \text{N} \) - \( \cos(74^\circ) \approx 0.275 \) - \( \sin(74^\circ) \approx 0.961 \) #### **Moment equation:** \[ F_w \cdot 3.5 = 176.4 \cdot \left(\frac{4}{2} \cdot 0.275\right) + 637 \cdot (4 \cdot 0.275) \] \[ F_w \cdot 3.5 = 176.4 \cdot 0.55 + 637 \cdot 1.1 \] \[ F_w \cdot 3.5 = 97.02 + 700.7 = 797.72 \] \[ F_w = \frac{797.72}{3.5} = 227.92\, \text{N} \] So, **required friction force** \( f = F_w = 227.92\, \text{N} \) --- ## 5. **Maximum Static Friction** \[ f_{\text{max}} = \mu_s N = 0.29 \times 813.4 = 235.89\, \text{N} \] --- ## 6. **Final Answers** ### - **Friction force needed:** \( \boxed{228\, \text{N}} \) (rounded to 3 sig. figs.) ### - **Maximum friction supported:** \( \boxed{236\, \text{N}} \) (rounded to 3 sig. figs.) --- **Let me know if you need a free-body diagram or further breakdown!**

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