Question

A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773 what is the normal force on the box

Answer 1

424.3 N

Step-by-step explanation:

In order to understande the normal force on the box, we have to consider only the forces acting along the direction perpendicular to the surface of the ramp.

In fact, there are only to forces acting along this direction:

- The normal force, N, which pushes upward perpendicular to the ramp

- The component of the weight of the box perpendicular to the ramp, which pushes downward (perpendicular to the ramp), and which is equal to

`mg cos \theta`

where m = 50 kg is the mass of the box, g = 9.8 m/s^2 is the acceleration due to gravity, and
`\theta=30^(\circ)` is the angle of the ramp.

Since the box is in equilibrium in the direction perpendicular to the ramp, these two forces must be equal, so we find:

`N=mg cos \theta =(50 kg)(9.8 m/s^2)(cos 30^(\circ))=424.3 N`