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 ?
0 70 N
O 80 N
O 75 N
O 85 N
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The correct option is;

75 N

Explanation:

The given parameters are;

The angle of inclination of the ramp, θ = 42°

The gravitational force of the box = m × g = 112.1 N

Let represent the force of static friction acting on the box

Given that the box is at rest, we have

The component of the gravitational force acting along the inclined plane = The force of static friction acting on the box,

We have that the component of the gravitational force acting along the inclined plane = m × g × sin (θ) =

= m × g × sin (θ)

Substituting the values, gives;

= 112.1 × sin (42°) ≈ 75.0095409728

∴ The force of static friction acting on the box = ≈ 75 N

the answer is b

explanation:

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow