Question

A dockworker loading crates on a ship finds that a 34-kg crate, initially at rest on a horizontal surface, requires a 73-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 50 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor

Answer 1

ANSWER:::

1) the coefficients of static =0.219

2)  kinetic friction =0.150

CALCULATION::::

Lets, the crate weighs 34 kg, you can find the normal force:

So, normal force F_normal = ma

F_normal =(34 kg)(9.8 m/s²)

= 333.2 N

The force used to set the box in motion is the applied force against the static friction holding the box in place.

To find the static friction coefficient we can do this

  Applied Force = (Normal Force)*()

Where is static friction coefficient

Then   Applied Force = (Normal Force)*()

  

Static friction coefficient is 0.219

Again we calculate,

The second applied force is less because it has overcome the static friction and is now being acted on by n   the kinetic friction:

the equation is, Applied Force = ()*(Normal Force) .............................(2)

where is Kinetic friction coefficient

from equation (2)

  

Is Kinetic friction coefficient is=0.15