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Question

A 1400 N crate is being pushed across a level floor at a constant speed by a force F of 300 N at an angle of 20° below the horizontal a) What is the coefficient of kinetic friction between the crate and the floor?

(b) If the 300 N force is instead pulling the block at an angle of 20° above the horizontal as shown in Figure P4.31b, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a).
m/s2

Answer 1

For both A the crate is moving at constant velocity, so there is no acceleration. That means that all forces are balanced on the crate.

What are the forces?

Graivity directly downward
1400 N
Force F, 300N pushing downward at an angle of 20 degrees
and friction = kN
where k is the coefficient of kinetic friction
and N is the normal force

N=1400+sin(20)*300
so friction is
k*1502.6 N

The friction must exactly balance the horizontal force since we have constant velocity
Cos(20)*300=k*1502.6
k=cos(20)*300/1502.6
=.188

b) Now that we know k, reevaluate the forces on the crate under the conditions stated:
since f=m*a
a=
(cos(20)*F-k*(mg-sin(20)*F))/m
a=
9.8*(cos(20)*300-.188*(1400-sin(20)*300))/1400
a=.266 m/s^2

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