   Chapter 4, Problem 28P

Chapter
Section
Textbook Problem

A block of mass 55.0 kg rests on a slope having an angle of elevation of 25.0°. If pushing downhill on the block with a force just exceeding 187 N and parallel to the slope is sufficient to cause the block to start moving, find the coefficient of static friction.

To determine
The co-efficient of static friction.

Explanation

Given Info: Mass of the block is 55.0 kg. Angle of inclination is 25° . Applied force is 187 N.

The free body diagram is given below.

From the above diagram,

Fsmgsinθ=F                             (I)

N=mgcosθ                                    (II)

• m is the mass of the student.
• g is the acceleration due to gravity.
• θ is the angle of inclination.
• Fs is the static friction force.
• F is the applied force.
• N is the normal force.

Formula for static friction force is,

Fs=μsN                                           (III)

• μs is the co-efficient of static friction.

Using Equations (I), (II) and (III), we get the expression for the co-efficient of static friction as,

μs=Fmgcosθ+tanθ

Substitute 187 N for F, 55

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