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Chapter 5: Problem 97

Block \(A\), with weight \(3w\), slides down an inclined plane \(S\) of slope angle 36.9\(^{\circ}\) at a constant speed while plank \(B\), with weight \(w\), rests on top of A. The plank is attached by a cord to the wall \(\textbf{(Fig. P5.97).}\) (a) Draw a diagram of all the forces acting on block \(A\). (b) If the coefficient of kinetic friction is the same between \(A\) and \(B\) and between \(S\) and \(A\), determine its value.

Short Answer

Expert verified
(a) Diagram includes forces: gravity, normal, friction on block A. (b) Coefficient of kinetic friction \(\mu_k \approx 0.53\).

Step by step solution

01

Understand the Forces

To solve this exercise, let's identify the forces acting on block \(A\). These include: (1) the gravitational force \(F_g = 3w\) acting downwards, (2) the normal force \(N_S\) from the surface of the inclined plane, acting perpendicular to the plane, and (3) the frictional force \(f_S\) between \(A\) and the inclined plane, acting up the slope. We also have the tension \(T\) in the cord due to plank \(B\) and the frictional force between \(A\) and \(B\), \(f_{A/B}\), acting down the plane.
02

Draw the Force Diagram for Block A

On a diagram of block \(A\), draw the following forces: - Gravitational force \(F_g = 3w\) pointing directly downward.- The normal force \(N_S\) acting perpendicular to the inclined plane.- Frictional force \(f_S\) acting parallel to the inclined plane, directed up the slope.- Frictional force \(f_{A/B}\) directed downwards along the slope due to the plank \(B\).Ensure to label all these forces accurately on the diagram.
03

Apply Newton's Second Law Along the Incline

Apply Newton's Second Law along the plane. Since block \(A\) slides at constant speed, the net force along the slope direction must be zero. Write the equation:\[3w \sin(36.9^{\circ}) - f_S - f_{A/B}= 0\]where \(f_S\) is the kinetic friction force between \(A\) and the slope, and \(f_{A/B}\) is the kinetic frictional force between \(A\) and \(B\), each given by \(\mu_k N_S\) and \(\mu_k N_{A/B}\), respectively.
04

Calculate the Normal Forces

Calculate the normal force \(N_S\). For \(N_S\), use the force balance perpendicular to the plane:\[N_S = 3w \cos(36.9^{\circ})\]Similarly, for \(N_{A/B}\) due to plank \(B\), it is just \(w\), the weight of the plank, as it rests on \(A\) without any inclination change.
05

Equate Frictional Forces and Solve for Coefficient

Substitute \(N_S\) and \(N_{A/B}\) into the equation from step 3:\[3w \sin(36.9^{\circ}) = \mu_k (3w \cos(36.9^{\circ}) + w)\]Divide through by \(w\) and solve for \(\mu_k\):\[3\sin(36.9^{\circ}) = \mu_k (3 \cos(36.9^{\circ}) + 1)\]Now solve for \(\mu_k\). Substitute the angle values \(\sin(36.9^{\circ}) \approx 0.6\) and \(\cos(36.9^{\circ}) \approx 0.8\):\[1.8 = \mu_k (2.4 + 1)\]\[\mu_k = \frac{1.8}{3.4} \approx 0.53\]
06

Verify Calculated Coefficient

Check the work by plugging back the \( \mu_k \) value into the context of the forces to ensure equilibrium. This verifies that the coefficient of kinetic friction \(\mu_k \approx 0.53\) holds the system in the necessary state of constant speed as specified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Force Diagrams
When dealing with physics problems, force diagrams, or free-body diagrams, help us visualize and analyze the forces acting on an object. Here, block A has several forces to consider. These forces include:
  • The gravitational force, which pulls block A downward with a force of magnitude equal to its weight, 3w.
  • The normal force, which is perpendicular to the inclined plane and balances part of the gravitational force.
  • The frictional force between block A and the inclined plane, which acts to oppose the motion of block A down the slope.
  • The tension in the cord, which influences the forces acting between block A and the plank B.

      • In our scenario, it's crucial to label these forces when drawing a force diagram. Doing so allows us to visually interpret how each force interacts with block A. By applying Newton's Second Law, we can set up the equations needed to solve the problem effectively.
Exploring Kinetic Friction
Kinetic friction occurs when two surfaces slide past each other. It opposes the motion and acts to slow down the moving object. The force of kinetic friction depends on two main factors:
  • The types of surfaces in contact, described by the coefficient of kinetic friction ( \( \mu_k \)).
  • The normal force, which is the force perpendicular to the surfaces in contact.

In our problem, there are two surfaces of contact involving kinetic friction: between block A and the inclined plane, and between block A and plank B. For both, the frictional force is calculated using the formula: \( f_k = \mu_k N \), where \( N \) is the normal force. The frictional force ( \( f_k \)) between the slope and block A, and between block A and B, will have an impact on the overall motion, allowing us to determine the necessary value of \( \mu_k \) for block A to slide consistently.
Inclined Plane Problems Simplified
Inclined plane problems often involve analyzing the forces acting on objects moving along a slope. Understanding the components of these forces is vital.
  • The weight of the object can be broken down into two components: one parallel to the plane which moves the object downwards, and another perpendicular to the plane.
  • The angle of the incline affects how these forces are divided. For instance, the perpendicular normal force is a result of the gravitational force component that does not aid the motion down the slope.
  • Newton's Second Law assists by ensuring the sum of forces along the slope equals the mass times acceleration. In the case of a constant velocity, this net force is zero.

For our specific exercise, understanding these concepts allows us to set up equations reflecting the process. By breaking the gravitational force of block A into components and balancing forces along the inclined slope, the strength of friction needed to keep the block moving at a constant speed can be analyzed and calculated.

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Most popular questions from this chapter

A light rope is attached to a block with mass 4.00 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass \(m\) is suspended from the other end. When the blocks are released, the tension in the rope is 15.0 N. (a) Draw two free-body diagrams: one for each block. (b) What is the acceleration of either block? (c) Find \(m\). (d) How does the tension compare to the weight of the hanging block?

A box with mass 10.0 kg moves on a ramp that is inclined at an angle of 55.0\(^\circ\) above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is \(\mu_k =\) 0.300. Calculate the magnitude of the acceleration of the box if you push on the box with a constant force \(F =\) 120.0 N that is parallel to the ramp surface and (a) directed down the ramp, moving the box down the ramp; (b) directed up the ramp, moving the box up the ramp.

When you do a chin-up, you raise your chin just over a bar (the chinning bar), supporting yourself with only your arms. Typically, the body below the arms is raised by about 30 cm in a time of 1.0 s, starting from rest. Assume that the entire body of a 680-N person doing chinups is raised by 30 cm, and that half the 1.0 s is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Draw a free-body diagram of the person's body, and use it to find the force his arms must exert on him during the accelerating part of the chin-up.

A box is sliding with a constant speed of 4.00 m/s in the \(+x\)-direction on a horizontal, frictionless surface. At \(x =\) 0 the box encounters a rough patch of the surface, and then the surface becomes even rougher. Between \(x =\) 0 and \(x =\) 2.00 m, the coefficient of kinetic friction between the box and the surface is 0.200; between x = 2.00 m and x = 4.00 m, it is 0.400. (a) What is the \(x\)-coordinate of the point where the box comes to rest? (b) How much time does it take the box to come to rest after it first encounters the rough patch at \(x =\) 0?

A 40.0-kg packing case is initially at rest on the floor of a 1500-kg pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed. Find the magnitude and direction of the friction force acting on the case (a) when the truck accelerates at 2.20 m/s\(^2\) northward and (b) when it accelerates at 3.40 m/s\(^2\) southward.
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