Let's break the problem into parts as requested. --- ## **Given Data** - Spring constant, \( k = 1850\ \text{N/m} \) - Uncompressed spring length, \( l = 2.50\ \text{m} \) - Spring is compressed to \( 96.0\% \) of \( l \): \( l_{\text{compressed}} = 0.96 \times 2.50 = 2.40\ \text{m} \) - Coefficient of static friction, \( \mu_s = 0.150 \) - Mass of crate, \( m = 23.0\ \text{kg} \) - Angle of wedge, \( \theta = 11.0^\circ \) - **Neglect the mass of the wedge.** - **Assume the crate only moves vertically (in the y direction).** --- ## **PART A: Normal Force \( N_C \) Exerted by the Crate on the Wedge** ### **Free-Body Diagram (Vertical - y direction)** For the crate: - **Upward forces:** Spring force \( F_{spring} \) and normal force from the wedge \( N_C \sin{\theta} \) - **Downward forces:** Weight \( mg \) - **Friction and horizontal forces do not affect vertical equilibrium here.** ### **Step 1: Calculate the Spring Force** \[ \Delta x = l - l_{\text{compressed}} = 2.50\ \text{m} - 2.40\ \text{m} = 0.10\ \text{m} \] \[ F_{spring} = k \Delta x = 1850 \times 0.10 = 185\ \text{N} \] ### **Step 2: Vertical Equilibrium for the Crate** \[ \sum F_y = 0: \quad F_{spring} + N_C \sin{\theta} = mg \] \[ N_C \sin{\theta} = mg - F_{spring} \] \[ N_C = \frac{mg - F_{spring}}{\sin{\theta}} \] #### Plug in numbers: - \( m = 23.0\ \text{kg} \) - \( g = 9.81\ \text{m/s}^2 \) - \( F_{spring} = 185\ \text{N} \) - \( \theta = 11.0^\circ \) - \( \sin{11.0^\circ} = 0.1908 \) \[ N_C = \frac{(23.0)(9.81) - 185}{0.1908} \] \[ N_C = \frac{225.63 - 185}{0.1908} \] \[ N_C = \frac{40.63}{0.1908} \] \[ N_C \approx 213\ \text{N} \] --- ## **PART B: Minimum Force \( P \) to Move the Crate Up** For the crate to **just begin to move upward**, friction must be overcome on both the wedge and the crate. ### **Crate Free Body Diagram (Horizontal - x direction):** - **To move up, the wedge must overcome friction between itself and the crate, and between itself and the ground.** - The force \( P \) is applied horizontally to the wedge. #### **Forces on the Wedge:** - Applied force \( P \) to the left. - Friction between wedge and ground: \( f_{A} \). - Force from crate: \( N_C \) at angle \(\theta\) upward and right. - Friction between crate and wedge: \( f_{B} \). Let's sum forces **on the wedge in the x-direction** (to the left is positive for \( P \)): \[ P - N_C \cos{\theta} - f_A = 0 \] But, let's account for all friction: - **Friction between wedge and crate:** \( f_B = \mu_s N_C \) - **Friction between wedge and ground:** \( f_A = \mu_s N_A \) - \( N_A \) is the normal force between wedge and ground. #### **Step 1: Find \( N_A \)** The wedge supports both the vertical component of \( N_C \) and the spring force. \[ N_A = N_C \sin{\theta} + F_{spring} \] (We'll ignore wedge mass as given.) \[ N_A = 213 \times 0.1908 + 185 = 40.63 + 185 = 225.63\ \text{N} \] #### **Step 2: Find Friction Forces** \[ f_B = \mu_s N_C = 0.150 \times 213 = 31.95\ \text{N} \] \[ f_A = \mu_s N_A = 0.150 \times 225.63 = 33.84\ \text{N} \] #### **Step 3: Sum Forces on Wedge (x-direction)** Friction forces resist motion. The crate pushes on the wedge to the right (horizontal component of \( N_C \)). Friction at both contact points resists motion left. \[ P = N_C \cos{\theta} + f_A + f_B \] \[ N_C \cos{\theta} = 213 \times \cos{11.0^\circ} = 213 \times 0.9816 = 209.1\ \text{N} \] So, \[ P = 209.1 + 33.84 + 31.95 = 274.89\ \text{N} \] ## **Final Answers** ### **(A) Normal force \( N_C \):** \[ \boxed{213\ \text{N}} \] ### **(B) Minimum force \( P \):** \[ \boxed{275\ \text{N}} \] (Rounded to three significant figures.) --- **Let me know if you need a step-by-step free-body diagram or more details on the force analysis!**