Question

A 100 N box sits on a 30 degree incline. If the static coefficient of friction is 0.1, what is the magnitude of the static frictional force acting on the box? Select one: a. 5.0 N

b. 8.7 N
c. 10N

Answer 1

f = 8.7 N

Step-by-step explanation:

It is given that,

Weight of the box, W = 100 N

It is inclined at an angle of 30 degrees.

The coefficient of friction,
`\mu_s=0.1`

We need to find the magnitude of the static frictional force acting on the box. Let f is the frictional force. It is given by :

`f=\mu_s* N`

N is the normal force

`f=\mu_s* mg\ cos\theta`

`f=0.1* 100* \ cos(30)`

f = 8.66 N

or

f = 8.7 N

So, the the static frictional force acting on the box is 8.7 N. Hence, this is the required solution.