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, [tex]\mu_s=0.1[/tex]
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 :
[tex]f=\mu_s\times N[/tex]
N is the normal force
[tex]f=\mu_s\times mg\ cos\theta[/tex]
[tex]f=0.1\times 100\times \ cos(30)[/tex]
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.
