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 ?
0 70 N
O 80 N
O 75 N
O 85 N
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Answer 1

Answer:

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 $F_f$ 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, $F_f$

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

$F_f$ = m × g × sin (θ)

Substituting the values, gives;

$F_f$ = 112.1 × sin (42°) ≈ 75.0095409728

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

Answer 2

Answer:

C: 75N

Explanation: