   Chapter 4, Problem 42P

Chapter
Section
Textbook Problem

A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck’s flatbed is 0.350, and the coefficient of kinetic friction is 0.320. (a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck’s flatbed? (b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?

(a)

To determine
The maximum acceleration of the truck.

Explanation

Given Info: Mass of the crate is 45 kg. The co-efficient of static and kinetic friction are μs=0.350 and μk=0.320 respectively.

The free body diagram is given below.

In the above diagram,

F=(Fs)maxN=mg (I)

• Fs is the force of static friction.
• N is the normal force.
• g is the acceleration due to gravity.
• F is the horizontal force.
• mt is the total mass.

From Newton’s second law,

F=ma (II)

The force of static friction is,

Fs=μsN (III)

Substitute Equations (II) and (III) in (I).

ma=μsN

On Re-arranging,

a=μsNm

Substitute mg for N to get a

(b)

To determine
The acceleration of the crate relative to the ground.

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