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When the truck is at rest, will the crate also be at rest?

  1. Feb 28, 2017 #1
    1. The problem statement, all variables and given/known data
    I have attached the known information.
    problem-2-png.113891.png
    When the truck is at rest, will the crate also be at rest or will it slide
    downwards?

    2. Relevant equations
    Sum F_y =0
    G_max = (my_s)*N
    G_min = (my_k) * N

    3. The attempt at a solution
    I did a Free body diagram on the crate (see the attached file)

    The first step is finding the normal force.
    Sum F_y =m*a_y =0
    N-mg*cos(10)=1932.19 N

    now i find the friction forces
    G_ max=(my_s)*N= 0.70*1932.19=1352.53 N
    G_min=(my_k)*N=0.50*1932.19=966.10 N

    Now i will find F_II
    F_II = m*g*sin(10) = 340.70 N

    since G_max > F_II there is no motion downwards i,e a_x = 0

    I am not sure if I have done it correct.
    Thank you for your help.
     

    Attached Files:

  2. jcsd
  3. Feb 28, 2017 #2

    haruspex

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    I assume you mean N=mg*cos(10)=1932.19 N
    Only if it is moving. And it is mu, not my.
    You can get the actual Greek character and subscript using the buttons just above the text area, Σ, X2, X2
    Yes.
     
  4. Feb 28, 2017 #3
    Yes i ment N=mg*cos(10)=1932.19 N (isolating for N)

    So I don't have to calculate for G_min? Meaning I can skip this part in the calculations?
     
  5. Feb 28, 2017 #4

    haruspex

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    Yes. You can assume the crate was placed statically. Until static friction is overcome, it is going nowhere.
    I would have guessed there are more parts to the problem.
     
  6. Feb 28, 2017 #5
    Okay. Thank you very much. And yes there are more parts to the problem, i am trying to solve them now.
     
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