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Homework Help: Friction, Newton's Laws

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data
    The two blocks shown in the diagram are at rest on a horizontal surface when a force P is applied to block B. Blaocks A and B have masses 20kg and 35kg respectively. The coefficient of friction between the two blocks is 0.35 and between the horizontal surface and Block B is 0.3.
    Determine the maximum force, P, before A slips on B.


    2. Relevant equations
    F = ma , Friction max = uR


    3. The attempt at a solution
    Okay this thing has been killing me for months.
    Friction between B and the ground is (20 + 35)g(0.3) = 16.5g and between B and A is
    (0.3)(35)g = 10.5g. This produces anti-friction which propells A forward. Friction between A and B is (20)(0.3)g = 6g.
    So, P should equal 16.5g +10.5g + (10.5g - 6g) = 31.5g N.
    I'm supposed to be getting 35.75g N.

    Where am I going wrong?
     

    Attached Files:

  2. jcsd
  3. May 16, 2010 #2

    Doc Al

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    :bugeye: Anti-friction??? The only force propelling A forward is the friction from B.

    Hint: When the static friction between A and B is at its maximum value, what is the acceleration?
     
  4. May 16, 2010 #3
    Sorry, that's what I mean.

    The acceleration of A or B?
    The acceleration of A relative to B will be 0 if slipping doesn't occur.
     
  5. May 16, 2010 #4

    Doc Al

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    Consider A first. What's the acceleration of A when slipping is just about to occur?
    Exactly! :wink:
     
  6. May 16, 2010 #5
    Well if the force pushing A forward is (10.5)g, the acceleration should be (10.5)g/20, =
    (0.525)g m/s^2.
     
  7. May 16, 2010 #6

    Doc Al

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    What's the maximum static friction force acting on A?
     
  8. May 16, 2010 #7
    The normal reaction by the Coefficient of Friction. (20g)(0.3) = 6g N.
     
  9. May 16, 2010 #8

    Doc Al

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    Good! Except that here μ = 0.35.
     
  10. May 16, 2010 #9
    Okay, I did it again, and got the right answer.
    So let me see if I've understood.

    I've included a force diagram.

    The frictional forces acting on B are [(55g)(0.3) + (35)(0.35)] = 28.75g N

    The force pushing A forward is (20g)(0.35) = 7g N.

    To make B move, P must be 28.75g, and to ensure that there is no relative acceleration between A and B it must be (28.75g + 7g) = 35.75g N.
     

    Attached Files:

  11. May 16, 2010 #10

    Doc Al

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    The friction force of A on B is equal and opposite to the friction force of B on A (which you have calculated correctly).
    Good. So what must be the acceleration of A?
     
  12. May 16, 2010 #11
    If the friction force on B is the coefficient of friction by the normal reaction, should it not be
    (0.35)(20g)?

    F = ma, ma = 7g, so a = 7g/20 = 0.35g m/s^2.
     
  13. May 16, 2010 #12

    Doc Al

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    Exactly.

    Good! So what must be the acceleration of both A and B at that point? What force P is required to produce such an acceleration?
     
  14. May 16, 2010 #13
    If the acceleration of A is 0.35g, the acceleration of B must also be 0.35g, so that there is no relative acceleration between them and slipping will not occur.

    F = ma, so P = (35)(0.35g).
    Evidently I've gone wrong somewhere. Haha.
     
  15. May 16, 2010 #14

    Doc Al

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    Good!

    P is not the only force acting on B.

    Hint: Try treating both blocks as a single object.
     
  16. May 16, 2010 #15
    P minus the two frictional forces acting on B has to produce an acceleration of 0.35g.

    So [P - 16.5g-7g = 35(0.35g)]
    P = 35.75g N

    And if both blocks were a single object, the 7g N forces would not affect the system, and the mass of the system would be 55kg.

    So [P -16.5g = 55(0.35g)]
    P = 35.75g N
     
  17. May 16, 2010 #16

    Doc Al

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    There you go. :approve:
     
  18. May 16, 2010 #17
    Thank you very much for your help. :smile:
     
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