Question

A box at rest is in a state of equilibrium half way up on a ramp. The ramp has an incline of 42° . What is the force of static friction acting on the box if box has a gravitational force of 112.1 N ?

70 N
80 N
75 N
85 N

Answer 1

75 N

Step-by-step explanation:

Given that,

The angle of incline of the ramp is 42°.

The gravitational force of the box, W = 112.1 N

We need to find the force of static friction on the box.

We have that the component of the gravitational force acting along the inclined plane i.e.

`F=W\sin\theta\\\\F=112.1* \sin(42)\\\\F=75\ N`

So, the force of static friction is equal to 75 N.