Question

Suppose you have to move a heavy crate of weight 875 N by sliding it along a horizontal concrete floor. You push the crate to the right with a horizontal force of magnitude 300 N, but friction prevents the crate from sliding. What is the magnitude

Fp of the minimum force you need to exert on the crate to make it start sliding along the floor? Let the coefficient of static friction ?s between the crate and the floor be 0.56 and that of kinetic friction, ?k, be 0.47.

Answer 1

Given data

Weight of the crate is, mg = 875 N

Magnitude of applied force is, F = 300 N

Coefficient of static friction is, ?s = 0.56

Coefficient of kinetic friction is, ?k = 0.47

Let the minimum force required to move the crate be Fp

Solution:

The static frictional force is,

fs = ?s mg

= (0.56 ) (875 N)

= 490 N

For the crate to slide, we need to apply atlest Fp= 490 N

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The kinetic frictional force is,

fk = ?k mg

= (0.47 ) (875 N)

= 411.25 N

After the crate slides, the force required to keep the crate to move

with constant speed is 411.25 N