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Net force to keep box at rest

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data

    (a) A box is placed on an frictionless inclined plane with an angle of 18° from the horizontal. The box has a mass of 35 kg, what is the magnitude of the acceleration of the box along the inclined plane?
    (b) What magnitude of force would you need to apply to the box in order to keep it at rest? Assume there is no friction and the force applied is parallel to the ground.


    2. Relevant equations

    F=ma
    ma=mgsin

    3. The attempt at a solution

    I got the answer to the first part. I am just confused about the second part. I get that Fnet=0 N in order for the entire system to not move. However, I was thinking that, in order for that to happen, whatever the force would be to begin with, that would have to be the opposing force, as well (i.e. if F=mgsin then for Fnet=0, F=mgsin-mgsin=0, therefore F=mgsin in order for Fnet=0).....but that feels very confusing to me....I know that I have to go against gravity....but...I don't really know how to set up the equation for this problem...
     
  2. jcsd
  3. Oct 2, 2007 #2

    Doc Al

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    Staff: Mentor

    Sounds to me like you are on the right track but just don't realize it. Gravity exerts a force down the incline equal to [itex]mg\sin\theta[/itex]. To cancel that force, you need to exert the same force up the incline. Done! (Just compute the actual value of the force.)
     
  4. Oct 2, 2007 #3
    okay...that was what I was thinking...but would it actually be the value of mgsin? (because that was what I originally thought and it turns out that I was wrong...) would I have to use the acceleration that I found in the first part? maybe do something like mgsin-masin?....or would that just not work....
     
  5. Oct 2, 2007 #4

    Doc Al

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    Ah... I misread the problem. The force is applied parallel to the ground, not parallel to the incline. (Sorry about that.) That's why [itex]mg\sin\theta[/itex] is wrong.

    You need a horizontal force whose component parallel to the incline equals [itex]mg\sin\theta[/itex].

    If you have a horizontal force F, what's its component parallel to the incline?
     
  6. Oct 2, 2007 #5
    ...I think I understand what you are saying, but I'm not sure about setting up the equations....
    it can't be just mgcos because, even though that is a horizontal force it doesn't equal to mgsin...right?
     
  7. Oct 2, 2007 #6

    Doc Al

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    Start by answering this question: If you have a horizontal force F, what's its component parallel to the incline (given the incline angle)?
     
  8. Oct 2, 2007 #7
    ...would it be mgsin?
     
  9. Oct 2, 2007 #8

    Doc Al

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    Forget about the details of this problem, just answer the question in general for a force F.
     
  10. Oct 2, 2007 #9
    would it be Fsin?
     
  11. Oct 2, 2007 #10

    Doc Al

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    No. (If F were a vertical force like gravity, then that would be true.)
     
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