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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.

Respuesta :

The magnitude of the force of static friction acting on the box is 75 N. Option D is correct.

What is static friction?

It is the frictional force between the two solid objects that are attached to each other.

It can be given by

[tex]F_s = mg \rm \ sin \theta[/tex]

Where,

[tex]F_s[/tex] - static friction

[tex]mg[/tex] - gravitational force =  112. 1 N

[tex]\theta[/tex] - inclined angle  = 42°

Put the values in the formula,

Fs = 112.1   x   sin42°

Fs = 75 N

Therefore, the magnitude of the force of static friction acting on the box is 75 N.

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