Discovering the Maximum Force to Keep a Box at Rest on an Inclined Ramp

[F.B]

This question is driving me nuts. I really need help with it, actually i just need help setting up my equations.

Mu = the weird friction sympol

A box of a mass m=22 kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficient of friction between the box and the ramp are "Mu s" = 0.78 and "Mu k" = 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.

My equation are:

Fn=Fgy - Fapp
Fnet = 0
also
Fnet = Fgx - Ffs

After that i end up getting two equations for Fapp but when i solve for it i get a negative answer. I also tried flipping Fapp so Fn=Fapp - Fgy but that also won't work, i get the wrong answer.

So can anyone please help get my two equations, there's something wrong with mine.

 

[Päällikkö]

Please clarify your expressions, what is Fn, what are Fapp and Ffs?

Which way does the friction act?

 

[quicknote]

First of all, it's important to understand the forces acting on the box on the inclined ramp. There are three main forces at play: the weight of the box (mg) acting downward, the normal force (Fn) acting perpendicular to the ramp, and the friction force (Ff) acting parallel to the ramp. The force parallel to the ramp is the one we are interested in, as it is the force that will keep the box at rest.

To determine the maximum force that can be applied upward, we need to consider the maximum value of friction that can be exerted on the box. This is where the coefficient of friction comes into play. The coefficient of static friction (μs) is the measure of the friction force between two surfaces when they are not moving relative to each other. The coefficient of kinetic friction (μk) is the measure of the friction force between two surfaces when they are in motion relative to each other.

In this case, since the box is at rest, we will use the coefficient of static friction. The formula for maximum friction force is Ff = μsFn. So in order to keep the box at rest, the force applied upward must be equal to or less than the maximum friction force.

Now, to set up your equations, you can use the following:

Fnet = 0 (since the box is at rest)
Fgx = mg (weight of the box)
Fn = Fgy (since the box is at rest, the normal force is equal to the weight component perpendicular to the ramp)

Substituting these values into the equation Fnet = Fgx - Ffs, we get:

0 = mg - μsFn

Since we want to find the maximum force that can be applied upward, we need to solve for Fn. So we rearrange the equation to get:

Fn = mg/μs

Substituting the values of m, g, and μs given in the problem, we get:

Fn = (22 kg)(9.8 m/s^2)/0.78 = 279.49 N

So the maximum force that can be applied upward, parallel to the ramp, without the box moving is 279.49 N.

I hope this helps to clarify the problem and set up the equations correctly. Remember, it's important to understand the forces at play and use the appropriate values for the coefficient of friction depending on the situation (static or kinetic). Good luck!