Let's break the problem into steps: #### **Given Data** - Ladder length: \( L = 5.5\, \text{m} \) - Wall height: \( h = 4.5\, \text{m} \) - Ladder mass: \( m_1 = 16\, \text{kg} \) - Worker mass: \( m_2 = 90\, \text{kg} \) - Angle with ground: \( \theta = 74^\circ \) - Coefficient of static friction: \( \mu_s = 0.44 \) - Gravity: \( g = 9.8\, \text{m/s}^2 \) --- ### **Step 1: Forces Acting on the Ladder** - Weight of ladder acts at its center: \( W_1 = m_1 g \) - Weight of worker acts at the top: \( W_2 = m_2 g \) - Normal force at the base: \( N \) - Friction at the base: \( f \) - Normal force at the wall (horizontal): \( F_W \) --- ### **Step 2: Force Equilibrium** #### **Vertical:** \[ N = W_1 + W_2 \] #### **Horizontal:** \[ F_W = f \] --- ### **Step 3: Torque (Moment) Equilibrium** **Take moments about the base:** - Weight of ladder acts at \( L/2 \) from base, downward at angle \( \theta \) - Weight of worker acts at the top at \( L \) from base, downward at angle \( \theta \) - Wall force acts at top at height \( h \), perpendicular to ladder \[ \sum \tau_\text{base} = 0 \] - **Torque due to ladder:** \( (m_1 g) \cdot \frac{L}{2} \cdot \cos\theta \) - **Torque due to worker:** \( (m_2 g) \cdot L \cdot \cos\theta \) - **Torque due to wall:** \( F_W \cdot h \) \[ (m_1 g) \frac{L}{2} \cos\theta + (m_2 g) L \cos\theta = F_W h \] \[ F_W = \frac{(m_1 g) \frac{L}{2} \cos\theta + (m_2 g) L \cos\theta}{h} \] --- ### **Step 4: Friction Force Needed** Since \( F_W = f \): \[ f = \frac{(m_1 g) \frac{L}{2} \cos\theta + (m_2 g) L \cos\theta}{h} \] --- ### **Step 5: Maximum Friction Available** \[ f_{\text{max}} = \mu_s N = \mu_s (m_1 + m_2) g \] --- ## **Plug in Numbers** - \( m_1 = 16\, \text{kg} \) - \( m_2 = 90\, \text{kg} \) - \( g = 9.8\, \text{m/s}^2 \) - \( L = 5.5\, \text{m} \) - \( h = 4.5\, \text{m} \) - \( \theta = 74^\circ \) - \( \mu_s = 0.44 \) #### **Cosine:** \[ \cos(74^\circ) \approx 0.275 \] #### **Calculate \( f \):** \[ f = \frac{(16 \cdot 9.8) \cdot \frac{5.5}{2} \cdot 0.275 + (90 \cdot 9.8) \cdot 5.5 \cdot 0.275}{4.5} \] \[ = \frac{(156.8) \cdot 2.75 \cdot 0.275 + (882) \cdot 5.5 \cdot 0.275}{4.5} \] \[ = \frac{(156.8 \cdot 2.75 \cdot 0.275) + (882 \cdot 5.5 \cdot 0.275)}{4.5} \] Calculate each term: - \( 156.8 \cdot 2.75 = 431.2 \) - \( 431.2 \cdot 0.275 = 118.58 \) - \( 882 \cdot 5.5 = 4851 \) - \( 4851 \cdot 0.275 = 1334.03 \) - Total numerator: \( 118.58 + 1334.03 = 1452.61 \) - Divide by \( 4.5 \): \[ f = \frac{1452.61}{4.5} \approx 323 \text{ N} \] --- #### **Calculate \( f_{\text{max}} \):** \[ f_{\text{max}} = 0.44 \times (16 + 90) \times 9.8 \] \[ = 0.44 \times 106 \times 9.8 \] \[ = 0.44 \times 1038.8 = 457.07 \text{ N} \] --- ## **Final Answers (Magnitudes Only):** **Friction force needed:** \[ \boxed{323 \text{ N}} \] **Maximum friction available:** \[ \boxed{457 \text{ N}} \]