Let's solve the problem step-by-step. **Given:** - \( m_A = 200.0\, \text{kg} \) - \( m_B = 350.0\, \text{kg} \) - \( m_C = 600.0\, \text{kg} \) - \( \theta_A = 25.0^\circ \) - \( \theta_C = 50.0^\circ \) - \( \mu = 0.2500 \) (all surfaces) - Assume ideal pulleys and cords. We are asked for: 1. The **acceleration** of the system, and 2. The **tension** in each connecting cord (\(T_1, T_2\)). --- ### **Step 1: Draw Free-Body Diagrams** #### **Block A (Up the incline)** - Gravity down the incline: \( m_A g \sin\theta_A \) - Normal force: \( m_A g \cos\theta_A \) - Friction: \( f_{A} = \mu m_A g \cos\theta_A \) - Tension \( T_1 \) up the incline #### **Block B (Horizontal)** - Two tensions: \( T_1 \) (to the left), \( T_2 \) (to the right) - Friction: \( f_{B} = \mu m_B g \) - No incline force #### **Block C (Down the incline)** - Gravity down the incline: \( m_C g \sin\theta_C \) - Normal force: \( m_C g \cos\theta_C \) - Friction: \( f_{C} = \mu m_C g \cos\theta_C \) - Tension \( T_2 \) up the incline --- ### **Step 2: Write Equations of Motion** Let acceleration of the system be \( a \), all blocks move as one system. #### **Block A:** \[ T_1 - m_A g \sin\theta_A - f_A = m_A a \] #### **Block B:** \[ T_2 - T_1 - f_B = m_B a \] #### **Block C:** \[ m_C g \sin\theta_C - f_C - T_2 = m_C a \] Where: - \( f_A = \mu m_A g \cos\theta_A \) - \( f_B = \mu m_B g \) - \( f_C = \mu m_C g \cos\theta_C \) --- ### **Step 3: Substitute Values** Let \( g = 9.8\, \text{m/s}^2 \). #### **Calculate Friction Forces:** - \( f_A = 0.25 \times 200 \times 9.8 \times \cos(25^\circ) \) - \( f_B = 0.25 \times 350 \times 9.8 \) - \( f_C = 0.25 \times 600 \times 9.8 \times \cos(50^\circ) \) Calculate cosines: - \( \cos(25^\circ) = 0.9063 \) - \( \cos(50^\circ) = 0.6428 \) So: - \( f_A = 0.25 \times 200 \times 9.8 \times 0.9063 = 444.27\, \text{N} \) - \( f_B = 0.25 \times 350 \times 9.8 = 857.5\, \text{N} \) - \( f_C = 0.25 \times 600 \times 9.8 \times 0.6428 = 946.52\, \text{N} \) --- ### **Step 4: Substitute Forces into Equations** #### **Block A:** - \( m_A g \sin(25^\circ) = 200 \times 9.8 \times 0.4226 = 828.3\, \text{N} \) - \( T_1 - 828.3 - 444.27 = 200a \) - \( T_1 - 1272.57 = 200a \) - \( T_1 = 200a + 1272.57 \) #### **Block B:** - \( T_2 - T_1 - 857.5 = 350a \) - \( T_2 = 350a + T_1 + 857.5 \) #### **Block C:** - \( m_C g \sin(50^\circ) = 600 \times 9.8 \times 0.7660 = 4509.5\, \text{N} \) - \( 4509.5 - 946.52 - T_2 = 600a \) - \( 3562.98 - T_2 = 600a \) - \( T_2 = 3562.98 - 600a \) --- ### **Step 5: Solve the System of Equations** From above: 1. \( T_1 = 200a + 1272.57 \) ... (i) 2. \( T_2 = 350a + T_1 + 857.5 \) ... (ii) 3. \( T_2 = 3562.98 - 600a \) ... (iii) Equate (ii) and (iii): \[ 350a + T_1 + 857.5 = 3562.98 - 600a \] \[ 350a + T_1 + 857.5 + 600a = 3562.98 \] \[ 950a + T_1 = 3562.98 - 857.5 = 2705.48 \] Now substitute \( T_1 \) from (i): \[ 950a + (200a + 1272.57) = 2705.48 \] \[ 1150a + 1272.57 = 2705.48 \] \[ 1150a = 2705.48 - 1272.57 = 1432.91 \] \[ a = \frac{1432.91}{1150} = 1.2469\, \text{m/s}^2 \] --- ### **Step 6: Find Tensions** #### **T1:** \[ T_1 = 200a + 1272.57 = 200 \times 1.2469 + 1272.57 = 249 + 1272.57 = 1521.57\, \text{N} \] #### **T2:** \[ T_2 = 3562.98 - 600a = 3562.98 - 600 \times 1.2469 = 3562.98 - 748.14 = 2814.84\, \text{N} \] --- ## **Final Answers** ### **Acceleration of the system:** \[ \boxed{1.25\, \text{m/s}^2} \] ### **Tensions:** - \( \boxed{T_1 = 1522\, \text{N}} \) - \( \boxed{T_2 = 2815\, \text{N}} \) *(rounded to three significant figures)* Let me know if you need the step-by-step with all numbers plugged in!