- 9

- 0

**[SOLVED] Magnitude of force of box on ramp.**

Hello PF, this is my first post on PH. I am hoping that someone will be able to walk me through the question below. I’d really appreciate your help. Thanks.

So, the question is as follow:

A box with a mass of 22 kg is at rest on a ramp inclined at 45º to the horizontal. The coefficients of friction between the box and the ramp are µs = 0.78 and µk = 0.65.

a. Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.

b. 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.

So, it is given that:

m = 22 kg

∠θ = 45º

µs = 0.78

µk = 0.65

I haven’t yet attempted the second part of the question, but this is what I have so far for the first part:

Since the box has an applied force (FA) going upwards,

∴ friction is directed down on the ramp.

Since the box is at rest, the net force must be at equilibrium

∴ Fnet = 0, and

Ff = FA (where Ff is the force of friction) <<< Is this statement correct?

So, I know that I need to find FN:

FN = Fg • cos45º

Since, Fg = mg

∴ FN = mg • cos45º

And then I get confused…

I started with finding Ff when the friction is static:

Ff = µsFN

Ff = 118.91N

Then, it occurred to me that since the object is being pushed upwards, then it would come across kinetic friction, and not static. So…

Ff = µkFN

Ff = 99.09N

Or would it come across both types of friction? And if so, how would I calculate it. Also… Ff = FA, therefore largest force applied upwards is FA? Would that be the answer? It seems that I am missing something…