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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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Most popular questions from this chapter

Four separate blocks are placed side by side in a left-toright row on a table. A horizontal force, acting toward the right, is applied to the block on the far left end of the row. Draw FBDs for (a) the second block on the left and for (b) the system of four blocks.
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