Magnitude of force of box on ramp.

[BitXBit]

[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…

 

[Tentothe]

Are friction and the applied force the only two forces acting along the ramp's surface? How about the other component of the gravitational force?

 

[BitXBit]

Thank you for your reply.

Um... do you mean the component of the force of gravity acting on the box on the ramp? (FgR)

If so...
FgR = Fg • sin45º
FgR = mg • sin 45º
FgR = 152.24N

I was thinking about that, but I wasn't sure where I would go next...

 

[alphysicist]

Hi BitXBit,

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?

Once you include the gravitational component in this equation, what result do you get?

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

The box is not moving upwards, so the friction is static friction.

 

[BitXBit]

Hmmm... I'm sure if this is correct, but do you mean...

Fnet = FgR + FA - Ff = 0?

 

[alphysicist]

Looks good; does it look right to you? The applied force is up the ramp; the friction is helping to keep the box from moving upwards and so must be down the ramp; and the gravity component is down the ramp; these have to cancel and so give your equation.

You can combine that with the perpendicular direction equation to find the answer.

 

[BitXBit]

Hi alphysicist,

I'm not sure if it looks right, actually. I was thinking maybe I have it wrong, and it should be:
Fnet = FA - FgR - Ff = 0
(Since both the gravity component and friction are headed down the ramp, and the applied force is going up the ramp.)

Also, I did the calculations for:
Fnet = FgR + FA - Ff = 0
Fnet = 152.45N + FA - 118.91N
Fnet = 33.54N + FA
∴ FA = –33.54N, but doesn't this mean that the applied force is going DOWN the ramp? Or does it not matter, because its the magnitude?

I was also wondering what you meant by the perpendicular direction equation. I tried to find it in my book, and I tried to google it, but I could not figure it out.

 

[alphysicist]

Hi alphysicist,

I'm not sure if it looks right, actually. I was thinking maybe I have it wrong, and it should be:
Fnet = FA - FgR - Ff = 0
(Since both the gravity component and friction are headed down the ramp, and the applied force is going up the ramp.)

This is right; I thought Fgr was negative in the other post but I see it's not. Sorry!

This is the equation of the components parallel to the ramp; the other equation you have is the components perpendicular to the ramp, which relates the normal and gravitational force.

That way you can plug the normal force into the frictional like you did before.

 

[BitXBit]

Thank you so much for your patience alphysicist! :D

Okay, so from my previous calculations, I have:
FN = mg • cos45º

So, Ff = µsFN
Ff = 118.91N

And FgR = Fg • sin45º
FgR = mg • sin 45º
FgR = 152.24N

So...
Since FA - FgR - Ff = 0
∴ FA = Fgr + Ff
FA = 152.45N + 118.91N
FA = 271.36N
FA = 2.7 x 10^2 (Correct to two significant digit)

Therefore, the largest force that can be applied upwards is 2.7 x 10^2N! (:D?)

And so, the second part of the question is exactly the same, except that I need to concentrate on the components perpendicular to the ramp. :O! I think I finally understand this! Thank you so much for walking me through this! :D! Now to try the second part!

 

[alphysicist]

Sounds good! (At least, that's the answer I got.)

 

[KOKA]

The force of gravity parallel to the ramp is pushing the box down, and the force of friction is acting up the ramp in order to prevent its motion. In this state, Fnet is not equal to 0, and the box is being pulled downwards by gravity. The equation in this situation is:
Fnet = FgR - Ff
You've concluded that FgR is equal to mgsin45 and that Ff is equal to umgcos45.
Therefore, Fnet = mgsin45 - usmgcos45
= (22)(9.8)(sin45) - (0.78)(22)(9.8)(cos45)
= 33.5 N
= 34 N
34 N is the force that must be applied up the ramp to counteract the force of gravity parallel to the ramp that is continuing to push the box down.

 

[alphysicist]

KOKA,

The force of gravity parallel to the ramp is pushing the box down, and the force of friction is acting up the ramp in order to prevent its motion.

This is not correct. The force of friction is acting down the ramp to prevent the motion of the box.

If you change that then you will get the correct answer that BitXBit found in post #9.

In this state, Fnet is not equal to 0, and the box is being pulled downwards by gravity. The equation in this situation is:
Fnet = FgR - Ff
You've concluded that FgR is equal to mgsin45 and that Ff is equal to umgcos45.
Therefore, Fnet = mgsin45 - usmgcos45
= (22)(9.8)(sin45) - (0.78)(22)(9.8)(cos45)
= 33.5 N
= 34 N
34 N is the force that must be applied up the ramp to counteract the force of gravity parallel to the ramp that is continuing to push the box down.

 

[KOKA]

KOKA,
This is not correct. The force of friction is acting down the ramp to prevent the motion of the box.

If you change that then you will get the correct answer that BitXBit found in post #9.

I'm just a senior at high school so I'm no pro at this, but doesn't friction act in the opposite direction of motion? The box is moving down the ramp, so friction has got to be in the opposite direction. If you look at it, 270N is a lot of force to be applying to a box.

Edit:
Looking back, if you calculate the amount of force needed to lift the box upward (mg), it's only 215.6 N. How can the amount of applied force needed to maintain the box's position on the ramp be greater than the force needed to lift the box vertically?

 

[alphysicist]

I'm just a senior at high school so I'm no pro at this, but doesn't friction act in the opposite direction of motion?

There's no motion in this case; but they are asking for the maximum force that can be applied before the box slides up the ramp.

The box is moving down the ramp, so friction has got to be in the opposite direction. If you look at it, 270N is a lot of force to be applying to a BOX.

Edit:
Looking back, if you calculate the amount of force needed to lift the box upward (mg), it's only 215.6 N. How can the amount of applied force needed to maintain the box's position on the ramp be greater than the force needed to lift the box vertically?

Since they want the maximum force you can apply and still not have the box move, the applied force has to overcome both the component of the weight along the incline, plus the maximum static frictional force.

 

[KOKA]

There's no motion in this case; but they are asking for the maximum force that can be applied before the box slides up the ramp.
Since they want the maximum force you can apply and still not have the box move, the applied force has to overcome both the component of the weight along the incline, plus the maximum static frictional force.

Yeah! I see now! I kept looking at the question as if the box was in motion already, and we needed to apply a force for it to stay at rest. Instead, the box is already at rest and we just need to know the amount of force that can be applied to challenge the maximum static friction and force of gravity parallel to the ramp, but not overcome it causing the box to move. When I mentioned it was a box, I just pictured a plain cardboard box and neglected to take in the actual mass of the box. Can't believe I did that, I'm such a silly girl, huh?Thanks a lot for helping me clear this up alphysicist, you've really helped me out.

 

[emadtheman]

So for part b.) Fnet would equal to zero?
if so would the equation go about like this?:

Fnet = Fg_y cos45 - U_sFn cos 45 + F_app ?

Fg_y = force of gravity in the y direction
U_s = coefficient of static friction
Fn = normal force
U_sFn = force of static friction (Fs)or would i have to instead subtract F_app rather than adding it?

Also, for part a.) I checked some other websites and all of them seem to favour the answer gotten from koka meaning that the applied froce = 43N rather than 271.4N
271.4 makes much more sense to me yet I'm still in doubt

 

[alphysicist]

Hi emadtheman,

So for part b.) Fnet would equal to zero?

That's right; since the box is to remain at rest the net force acting on it must be zero.

if so would the equation go about like this?:

Fnet = Fg_y cos45 - U_sFn cos 45 + F_app ?

That does not look right to me. How did you derive this equation? (Also, by 'y-direction' do you mean the vertical direction, or something else?)

Fg_y = force of gravity in the y direction
U_s = coefficient of static friction
Fn = normal force
U_sFn = force of static friction (Fs)


or would i have to instead subtract F_app rather than adding it?

Also, for part a.) I checked some other websites and all of them seem to favour the answer gotten from koka meaning that the applied froce = 43N rather than 271.4N
271.4 makes much more sense to me yet I'm still in doubt

You don't trust me?

But seriously, I don't know how to respond to the statement that some other websites don't agree. I explained to Koka why she was wrong. What do you not agree with?