Parallel and perpendicular force

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  • #1
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A box with a mass of 22kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficient of friction between the box and the ramp are us = 0.78 and uk = 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.

For a, this is what I did:

Fapp = Fs + Fg sin theta
= Usmgcos theta + mg sin theta
= mg (Uscos theta + sin theta)
= (22kg) (9.8m/s^2)(0.78 cos 45 degrees + sin 45 degrees)
= 270N

The largest force that can be applied upward if the box is to stay at rest is 270N.

For b..i'm not totally sure but this is what I got:
Fs = Fg sin theta
= mg sin 45 degrees
= (22) (9.8) (0.707)
= 152N

Fn = Fs * Us
= 152 * 0.78
= 118N

Fapp = Fn - Fg cos theta
= 118 - mg cos 45 degrees
= 118 - (22) (9.8) (0.707)
= 118 - 152
= -34N

Could that be right?
 

Answers and Replies

  • #2
Doc Al
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Your solution for part a is perfectly correct.

For part b, realize that if no extra force were applied, gravity would cause the box to slide down the ramp. (Since [itex]m g \sin \theta > \mu m g \cos \theta[/itex].)

When you add a force (F) perpendicular to the ramp, what changes? Hint: What's the new normal force?
 
  • #3
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Umm..i'm not completly sure but Fn = Fg cos theta?
 
  • #4
Doc Al
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inner08 said:
... Fn = Fg cos theta?
No. That's just the normal force due to the weight of the box. But if you are also pressing down on the box with additional force, what happens to the normal force?
 
  • #5
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If you press down on the box, then the normal force will increase I would assume.
 
  • #6
Doc Al
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Right. So if you press down with a force F, what's the new normal force?
 
  • #7
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Fn = Fg cos theta + F ?
 
  • #8
Doc Al
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That's right.
 
  • #9
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ok..great :).

So i'm assuming that finding this "new" normal force is the answer to question b) since it is the perpendicular force acting on the box. Now i'm still a bit confused since I seem to have two unknown variables. I can find the Fg cos theta using the mass and gravity in the problem as well as the angle of inclination. But how would I go to finding the F (the force thats pressing down on the box)? I appreciate your patience with me...I'm having quite a bit of difficulty with all the notions of forces/motions.
 
  • #10
Doc Al
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Remember my first post in this thread. You are trying to find the added force (F) needed to keep that box from sliding. Consider the forces on the box parallel to the ramp: to keep the box from sliding, those forces must add to zero.
 
  • #11
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ok I think I got this..hopefully.

If, as you said, no extra force is added, gravity would force the box to slide down so mgsin theta > usmgcos theta.

Now if it were to stay at rest with a perpendicular force, then the equation would be mgsin theta = usmgcos theta + F?
 
  • #12
Doc Al
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inner08 said:
Now if it were to stay at rest with a perpendicular force, then the equation would be mgsin theta = usmgcos theta + F?
No. Remember, F acts perpendicular to the ramp. While F does not act parallel to the ramp, it does affect another force that does. (How does friction depend on normal force?)
 
  • #13
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ok so i've been looking/thinking about this for a while now. We are looking for the force that acts perpendicular to the ramp. So, F = Fn - Fg cos theta. Then F = usmgsin theta - mg cos theta?

Weather this is right or wrong, i'm just wondering if there is a trick to doing these because i'm having trouble finding the forces that are indicated in problems (perpendicular or parallel). Any insight would be appreciated!
 
  • #14
Doc Al
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inner08 said:
We are looking for the force that acts perpendicular to the ramp. So, F = Fn - Fg cos theta.
Right!

Then F = usmgsin theta - mg cos theta?
No. For some reason, you are setting Fn = usmgsin theta.

Instead, rewrite your initial equation to show the normal force:
Fn = F + Fg cos theta

Now use this normal force to calculate the maximum static friction force using:
[tex]F_f = \mu F_n[/tex]

That's the friction force, which you can write in terms of our unknown force F. That friction force must balance the component of the box's weight down the ramp.

Weather this is right or wrong, i'm just wondering if there is a trick to doing these because i'm having trouble finding the forces that are indicated in problems (perpendicular or parallel).
My advice is to solve as many problems as you can get your hands on. Don't give up until you understand how to solve each problem. (Getting the answer is not enough--even after you know the answer, things still might not be clear. Keep going over the problem, keep thinking, until it clicks in.)

Other than that I advise you to not look for tricks (although you'll undoubtedly discover a few), but to keep looking for how the basic principles are applied. For example, the basics here are Newton's 2nd law, the idea of forces as vectors with components that can be considered separately, and the relationship of friction with the normal force.
 
  • #15
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My advice is to solve as many problems as you can get your hands on. Don't give up until you understand how to solve each problem. (Getting the answer is not enough--even after you know the answer, things still might not be clear. Keep going over the problem, keep thinking, until it clicks in.)
sorry for digging up this old thread but I am taking a correspondence course with this same question in it...there really isn't many example questions in this course for me to properly grasp this. Any chance someone could point me to a place to get more practice questions like the one above?
 

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