Let's solve the problem step by step. --- ## **Step 1: List the Given Values** - \(F_1 = 18.64 \, \text{N}\) - \(F_2 = 69.12 \, \text{N}\) - \(\theta_1 = 110.4^\circ\) (Angle for \(F_1\) from x-axis) - \(\theta_2 = 24.29^\circ\) (Angle for \(F_2\) from x-axis) - \(m = 10.6 \, \text{kg}\) --- ## **Step 2: Find the Y-Components of Each Force** The y-component of a force is given by \( F_y = F \sin(\theta) \). - For \(F_1\): \( F_{1y} = F_1 \sin(\theta_1) \) - For \(F_2\): \( F_{2y} = F_2 \sin(\theta_2) \) ### **Calculate Each:** 1. **\(F_{1y}\):** \[ F_{1y} = 18.64 \times \sin(110.4^\circ) \] \[ \sin(110.4^\circ) \approx .939 \] \[ F_{1y} = 18.64 \times .939 \approx 17.50\, \text{N} \] 2. **\(F_{2y}\):** \[ F_{2y} = 69.12 \times \sin(24.29^\circ) \] \[ \sin(24.29^\circ) \approx .411 \] \[ F_{2y} = 69.12 \times .411 \approx 28.43\, \text{N} \] --- ## **Step 3: Find the Net Y-Component of the Force** \[ F_{\text{net},y} = F_{1y} + F_{2y} \] \[ F_{\text{net},y} = 17.50 + 28.43 = 45.93\, \text{N} \] --- ## **Step 4: Calculate the Acceleration in the Y-Direction** Newton’s Second Law: \( F = ma \implies a = \frac{F}{m} \) \[ a_y = \frac{F_{\text{net},y}}{m} \] \[ a_y = \frac{45.93}{10.6} \approx 4.33\, \text{m/s}^2 \] --- ## **Final Answer** \[ \boxed{a_y \approx 4.33\, \text{m/s}^2} \] **The acceleration of the box in the y direction is approximately \(4.33\, \text{m/s}^2\).**