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Practice Test- Box sliding down a slope

  1. Mar 3, 2014 #1
    1. A 30kg box is placed on a 15° slope and a person pushes on the box up the slope to keep it from sliding. If the coefficient of static friction is 0.15, what is the minimum force that the person must exert to keep the box in place?

    Answer Choices:
    A. 273N
    B. 241N
    C. 33.5 N
    D. 64.7 N
    E. 294 N




    2. Fs≤μs*N
    F=μ*N
    N=mg <-----not sure about that because I know normal force changes depending on the situation but I think this applies here.

    Also my teacher has us use g= 9.8 NOT 9.81



    3. So first I calculated the normal force: N=mg -----> (30)(9.8) = 294 N
    Then I put my normal force into the equation F=μ*N ------> (.15)(294)= 44.1 N
    This was not one of the answer choices. However I recalled this equation : Fs≤μs*N.
    Since I was solving for the force that was a minimum I assumed that out of my answer choices C. 33.5 N was correct because 33.5 ≤ 44.1 which was my calculated force. This turned out to be the correct answer on the answer sheet. However I feel that this was more of a guess that happened to be right rather than the correct way of doing this.


    Any help is appreciated!
     
    Last edited by a moderator: Mar 3, 2014
  2. jcsd
  3. Mar 3, 2014 #2

    Doc Al

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    Staff: Mentor

    First step is to fix this value for normal force. The normal force would equal mg if the surface were horizontal, but it's not. What's the component of the weight perpendicular to the surface?

    Once you have the correct normal force you can figure out the maximum value of static friction.

    Next is to ask this question: If the person didn't push the box up the incline, what would happen and why?
     
  4. Mar 3, 2014 #3
    Box pushed up an incline

    So the component would be (30)cos15= 28.97. Cosine because I defined the incline as the x direction.
     
  5. Mar 3, 2014 #4

    Doc Al

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    Staff: Mentor

    Almost. You want mgcosθ, not just mcosθ.
     
  6. Mar 3, 2014 #5
    Okay so my normal force would be : (30)(9.8)cos (15°) = 283.982
    Then from this I would plug my normal force into F=μN?
     
  7. Mar 3, 2014 #6

    Doc Al

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    Staff: Mentor

    Good.

    That would give you the maximum value of static friction.
     
  8. Mar 3, 2014 #7
    Okay so F= .15 * 283.98= 42.59 N of maximum static friction. Obviously I could just pick the correct answer at this point, but I'm sure there is a way to calculate the minimum force. My first reaction would be to calculate the F= μM because that would account for just the objects weight against the friction force and then I could subtract that from my maximum force to get minimum force. However that would give me 4.5 N which isn't enough to get me to the 33.5 N answer
     
  9. Mar 4, 2014 #8

    Doc Al

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    Staff: Mentor

    Good.

    Of course there is.

    μM has no physical meaning. It's not even a force! (Wrong units, for one thing.)

    Answer the question I asked earlier: If there were no person pushing the box up the incline, what would happen and why? (What force acts to push the box down the incline?)
     
  10. Mar 4, 2014 #9
    If there was no person pushing the box up the incline then the box would slide down because there are no forces to act on it going up the ramp. The forces that push the box down would be gravity and static friction.
     
  11. Mar 4, 2014 #10

    Doc Al

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    Staff: Mentor

    Gravity acts to push the box down. Compare the component of the weight acting down the incline to the max static friction which is trying to prevent the box from sliding down.
     
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