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

Answer:

f = 8.7 N                        

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 :

$f=\mu_s\times N$

N is the normal force

$f=\mu_s\times mg\ cos\theta$  

$f=0.1\times 100\times \ cos(30)$  

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.