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Elden [556K]
3 years ago
6

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 frict

ional force acting on the box? Select one: a. 5.0 N
b. 8.7 N
c. 10N

Physics
1 answer:
Semmy [17]3 years ago
8 0

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.

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