Box resting on an inclined plane

[JamesW]

Homework Statement

A box with a mass of 22kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficient of static friction is 0.78

determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp if the box is to remain at rest?

Homework Equations

F=ma

The Attempt at a Solution

F f (static friction) + Fgx = 0

F n = Fgy + F a

Fa = F n - Fgy

I have viewed a thread and to calculate F n (normal force) they divided Fgx/mu static. I was wondering why this was?

 

[JamesW]

WHY IS Fn = Fgx/ mu static ?

 

[JamesW]

because Fs=(mu static)(Fn)

 

[lightgrav]

which is valid if friction _exactly_ cancels "g sin theta" ...

 

[asteriatic]

I would like to clarify that the equations and approach used in the attempt at a solution are correct. The normal force (F n) is equal to the weight of the box (Fgy) plus any additional forces acting on the box (Fa). In this case, the only additional force is the force applied perpendicular to the ramp, which is what we are trying to find.

The reason for dividing by the coefficient of static friction (mu static) is to account for the friction force that is present on the ramp. The normal force is the perpendicular force exerted by the ramp on the box to counteract the weight of the box and any other forces acting on it. The coefficient of static friction represents the maximum amount of friction that can be present before the box starts to slide down the ramp. By dividing by this coefficient, we can ensure that the normal force is enough to counteract the weight and the friction force, thus keeping the box at rest.

In summary, dividing by the coefficient of static friction is necessary to accurately calculate the normal force and determine the smallest force that can be applied to the box to keep it at rest on the inclined plane.