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In a laboratory experiment on friction, a 135-N block resting on a rough horizontal table is pulled by a horizontal wire. The pull gradually increases until the block begins to move and continues to increase thereafter. Figure E5.26 shows a graph of the friction force on this block as a function of the pull. (a) Identify the regions of the graph where static friction and kinetic friction occur. (b) Find the coefficients of static friction and kinetic friction between the block and the table. (c) Why does the graph slant upward at first but then level out? (d) What would the graph look like if a 135-N brick were placed on the block, and what would the coefficients of friction be?

Figure E5.26

Short Answer

Expert verified
  • the region for static friction is when pull is less than 75N and the region of kinetic friction is the region when the pull is greater than 75N.
  • The coefficient of static and kinetic friction are 0.55 and 0.37 respectively.
  • Coefficient of static friction increases with an increase in the pull, so the graph slants upward.
  • There is no effect of weight on the value of the coefficient of friction.

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The weight of the box is, W = 135N .
02

Significance of coefficient of friction.

The coefficient of friction is the ratio of the force which resist the one surface to slide over another surface to that of normal force is known as coefficient of friction.

03

Determination of static friction and kinetic friction region in graph.

Part (a)

We know that static friction is a self-adjustable force and that it will be equal to the applied force until the body starts to move. In the above diagram, part OA is the region of static friction that means the region in which the pull is less than 75N .

The kinetic friction on the other hand, remains constant with time. So, the region of kinetic friction is when pull is greater than 75N .

04

Determination of coefficient of static friction and kinetic friction.

Part (b)

Expression for the static frictional force can be expressed as,

Fr=μsR

HereFris the frictional force,μs is the coefficient of static friction, and the normal reaction on the box,R=W=135N.

Substitute 75 N for Fr , 135 N for in the above equation.

75N=μs135N=0.55

Hence, required coefficient of static friction is, 0.55.

Expression for the kinetic frictional force can be expressed as,

Fr=μkR

Here μk is the coefficient of kinetic friction.

Substitute 50 N for , 135 N for in the above equation.

50N=μs135N=0.37

Hence, required coefficient of kinetic friction is, 0.37.

05

Determination why graph slant upward at first but then level out.

Part (c)

When a body is pulled, a sliding tendency develops in it but the body does not move due to friction. As, the pull is increased this sliding tendency goes on increasing but the body still remains stationary. This means that with increase in sliding tendency the frictional force also increased. On further increasing the pull, a certain maximum limit is reached beyond which if the pull is increased, the object will start moving. This is what is shown with the graph slanting upward.

06

Determination of coefficient of friction when 135 N brick placed on the block.

Part (d)

We know the coefficient of friction doesn’t depend upon the weight, but it depends on the roughness of the material.

Hence, the coefficient of friction remain unchanged.

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