Question

A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at constant speed? Important: Assume that the pushing force is to the right.

Answer 1

a)F = 456 N

b)F = 246 N

Step-by-step explanation:

Given that

m = 60 kg

Coefficients of static friction μs=0.76

Coefficients of kinetic friction μk= 0.410

We know that friction is always oppose the motion of a body.It act in opposite to the motion of the body.

When body is in rest position then static friction will act and when body is in moving position then kinetic friction will act.

The maximum value Static friction Frs = μs .N N is the normal force.

The maximum value kinetic friction Frk = μk .N

Here N = mg

a)

When crate just start the motion then maximum value of static friction will act on the crate.

So

The maximum value Static friction Frs = μs .mg

Take
`g=10\ m/s2`

Frs = 0.76 x 60 x 10 N

Frs = 456 N

So the horizontal force required to move the crate ,F= 456 N

b)

When crate is moving with constant velocity then maximum value of static friction will act on the crate.

The maximum value kinetic friction Frk = μk .N

Frs = 0.41 x 60 x 10 N

Frs = 246 N

So the horizontal force required to slide the crate,F = 246 N