   Chapter 4, Problem 29P

Chapter
Section
Textbook Problem

A dockworker loading crates on a ship finds that a 20.0-kg crate, initially at rest on a horizontal surface, requires a 75.0-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 60.0 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

To determine
The co-efficient of static and kinetic friction.

Explanation

Section 1:

To determine: The co-efficient of static friction.

The co-efficient of static friction is 0.383.

Explanation:

Given Info: Mass of the crate is 20.0 kg. When the crate is at rest, the force required is 75.0 N.

At equilibrium, when the crate is on the verge of moving, the maximum static friction is,

(Fs)max=μsN                                          (I)

• μs is co-efficient of static friction.
• N is the normal force.

Normal force is equal to the weight of the crate and the applied force equals the friction force.

N=mg                                                   (II)

• m is the mass of the crate.
• g is the acceleration due to gravity.

Substitute Equation (II) in (I).

(Fs)max=μsmg

On Re-arranging,

μs=(Fs)maxmg                                           (III)

Substitute 75 N for (Fs)max , 20.0 kg for m and 9.8ms2 for g in Equation (III) to get μs

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