# Acceleration of Box Along a Sloped Ceiling Let's break down and solve the problem step by step. --- ## **Step 1: Draw the Forces and Find the Net Force Along the Slope** **Forces acting on the box:** - Weight (\(mg\)), acts downward. - Normal force (\(N\)), acts perpendicular to the slope, upward. - Stabilizing force (\(F_{stab}\)), acts perpendicular to the surface, pressing the box against the ceiling. - Friction force (\(f_k\)), acts opposite to the direction of motion along the slope. **Decompose the weight:** - Parallel to slope: \(mg \sin\theta\) - Perpendicular to slope: \(mg \cos\theta\) **Normal force:** Since the box is under a sloped ceiling, both \(F_{stab}\) and the perpendicular component of weight contribute to the normal force: \[ N = F_{stab} + mg\cos\theta \] **Kinetic friction:** \[ f_k = \mu_k N = \mu_k (F_{stab} + mg\cos\theta) \] **Net force along the slope:** \[ F_{net} = mg\sin\theta - f_k \] --- ## **Step 2: Calculate the Acceleration** ### **Plug in the numbers:** Given: - \(m = 5\,\text{kg}\) - \(\theta = 30^\circ\) - \(F_{stab} = 100\,\text{N}\) - \(\mu_k = .2\) - \(g = 9.8\,\text{m/s}^2\) 1. **Calculate \(mg\sin\theta\) and \(mg\cos\theta\):** - \(mg\sin\theta = 5 \times 9.8 \times \sin 30^\circ = 5 \times 9.8 \times .5 = 24.5\,\text{N}\) - \(mg\cos\theta = 5 \times 9.8 \times \cos 30^\circ = 5 \times 9.8 \times .866 = 42.435\,\text{N}\) 2. **Normal force:** - \(N = 100 + 42.435 = 142.435\,\text{N}\) 3. **Friction force:** - \(f_k = .2 \times 142.435 = 28.487\,\text{N}\) 4. **Net force along the slope:** - \(F_{net} = 24.5 - 28.487 = -3.987\,\text{N}\) 5. **Acceleration:** - \(F = ma \implies a = \frac{F_{net}}{m} = \frac{-3.987}{5} = -.797\,\text{m/s}^2\) --- ## **Final Answer** \[ \boxed{a = -.80~\text{m/s}^2} \] (The negative sign means the box accelerates **up the slope**, i.e., opposite to the expected "downhill" direction, due to friction being stronger than the downslope component of gravity.) --- ## **Graphical Representation**  *Alt text: Free-body diagram of a box under a sloped ceiling, showing gravity, normal force, stabilizing force, and friction.* --- ### **Summary Table of Steps and Example Values** | Step | Expression | Example Value | |------|------------|--------------| | 1 | \(mg\sin\theta\) | 24.5 N | | 2 | \(mg\cos\theta\) | 42.435 N | | 3 | \(N = F_{stab} + mg\cos\theta\) | 142.435 N | | 4 | \(f_k = \mu_k N\) | 28.487 N | | 5 | \(F_{net} = mg\sin\theta - f_k\) | -3.987 N | | 6 | \(a = F_{net}/m\) | -.80 m/s² | --- **The box accelerates up the slope with \(.80\,\text{m/s}^2\).**