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Chapter 4: Problem 65

Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on the surface. The sanding block weighs \(2.0 \mathrm{N}\) and Brenda pushes on it with a force of $3.0 \mathrm{N}\( at an angle of \)30.0^{\circ}$ with respect to the vertical, and angled toward the wall. Draw an FBD for the sanding block as it moves straight up the wall at a constant speed. What is the coefficient of kinetic friction between the wall and the block?

Short Answer

Expert verified
Answer: The coefficient of kinetic friction between the sanding block and the wall is approximately 0.745.

Step by step solution

01

Draw a FBD for the sanding block

Identify the forces acting on the sanding block. There are four forces present: weight (W), normal force (N), friction force (f), and applied force (P). Draw an FBD for the sanding block, representing it as a point and vector arrows for each force acting on it. Step 2: Analyzing the forces acting on the block
02

Analyze the forces

Break down the applied force (P) into its vertical and horizontal components. The vertical component is \(P_y = P\sin{30.0^{\circ}} = 3.0\mathrm{N} * \frac{1}{2} = 1.5\mathrm{N}\). The horizontal component is \(P_x = P\cos{30.0^{\circ}} = 3.0\mathrm{N} * \frac{\sqrt{3}}{2} = 1.5\sqrt{3}\mathrm{N}\). Step 3: Applying the equations of motion for vertical forces
03

Equations of motion for vertical forces

Since the sanding block moves at a constant speed (which means constant velocity) up the wall, the net force in the vertical direction is zero. Thus, we can set up an equation: \(N - W - P_y = 0\). Substitute the given values for weight (2.0 N) and \(P_y\): \(N - 2.0\mathrm{N} - 1.5\mathrm{N} = 0\). Solving for N, we find that \(N = 3.5\mathrm{N}\). Step 4: Applying the equations of motion for horizontal forces
04

Equations of motion for horizontal forces

In the horizontal direction, we have two forces acting on the block: \(P_x\) and friction force \(f\). Since the block moves at a constant velocity, these two forces must balance each other. We can set up an equation: \(f = \mu_k * N = P_x\). Using the previously calculated values for N (\(3.5 \mathrm{N}\)) and \(P_x\) (\(1.5\sqrt{3}\mathrm{N}\)), we can find the coefficient of kinetic friction: \(\mu_k * 3.5\mathrm{N} = 1.5\sqrt{3}\mathrm{N}\). Step 5: Solving for the coefficient of kinetic friction
05

Solve for the coefficient of kinetic friction

Dividing by the normal force, we get: \(\mu_k = \frac{1.5\sqrt{3}\mathrm{N}}{3.5\mathrm{N}} = \frac{3\sqrt{3}}{7}\). Thus, the coefficient of kinetic friction between the wall and the block is \(\mu_k = \frac{3\sqrt{3}}{7} \approx 0.745\).

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