Let's solve **both parts step-by-step**: --- ## **Part 1: Nurse Pushing Cart** ### **Given:** - Angle of force \( \theta = 45^\circ \) below horizontal - Mass of cart \( m = 28.0 \) kg - Force of friction \( f_k = 61 \) N - Cart moves at **constant velocity** (so acceleration \( a = 0 \)) ### **Required:** Find the **magnitude of force** \( F \) the nurse must exert. --- ### **Step 1: Draw a Free Body Diagram** - **Forces acting on cart:** - Weight: \( mg \) (down) - Normal force: \( N \) (up) - Friction: \( f_k \) (opposite to motion) - Push by nurse: \( F \) at \( 45^\circ \) below horizontal - **Decompose nurse's force:** - Horizontal: \( F_x = F\cos\theta \) - Vertical (down): \( F_y = F\sin\theta \) --- ### **Step 2: Vertical Force Balance** Since there's no acceleration vertically, \[ N + F_y = mg \] But \( F_y \) is *downward*, so: \[ N = mg - F\sin\theta \] --- ### **Step 3: Friction Force** Kinetic friction: \[ f_k = \text{given as } 61~\text{N} \] (So, you don't need to calculate using \(\mu_k N\), it's provided.) --- ### **Step 4: Horizontal Force Balance** Constant velocity \(\implies\) no net force in horizontal direction: \[ F_x = f_k \implies F\cos\theta = f_k \] \[ F = \frac{f_k}{\cos\theta} \] --- ### **Step 5: Plug in the Numbers** - \( f_k = 61~\text{N} \) - \( \theta = 45^\circ \) - \( \cos 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071 \) \[ F = \frac{61}{0.7071} \approx 86.3~\text{N} \] --- ### **Final Answer (Part 1):** \[ \boxed{86.3~\text{N}} \] The nurse must exert a force of **86.3 N** at \(45^\circ\) below the horizontal to move the cart at constant velocity. --- ## **Part 2: Maximum Force Before Crate Moves** ### **Given:** - Mass of crate \( m = 161 \) kg - Coefficient of static friction \( \mu_s = 0.5 \) - Coefficient of kinetic friction (for reference) \( \mu_k = 0.3 \) - Crate is on a wood floor (horizontal) - \( g = 9.8~\text{m/s}^2 \) ### **Required:** Maximum horizontal force you can exert **without moving the crate** (i.e., maximum static friction). --- ### **Step 1: Calculate Normal Force** On a flat surface, the normal force equals the weight: \[ N = mg = 161 \times 9.8 = 1577.8~\text{N} \] --- ### **Step 2: Calculate Maximum Static Friction** \[ f_{s,\text{max}} = \mu_s N = 0.5 \times 1577.8 = 788.9~\text{N} \] --- ### **Final Answer (Part 2):** \[ \boxed{789~\text{N}} \] (Rounded to 3 significant figures.) --- ## **Summary Table** | Part | Final Answer | Units | |------|-------------|-------| | 1 | 86.3 | N | | 2 | 789 | N | --- **Explanations included above. Let me know if you need further clarification!**