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Calculate max acceleration such that A does not slide

  1. May 6, 2016 #1
    1. The problem statement, all variables and given/known data
    image.jpg

    2. Relevant equations
    Friction ≤ μR

    3. The attempt at a solution:
    Ok so I need help for (ii) and this is what I did:
    I thought that in order for A not to slide, the net force on B ≤ Friction between the boxes
    450a ≤ 0.2(200g)
    a≤0.89

    however, when I tried this,
    200a≤0.2(200g)
    a≤2

    What's the correct physics behind this?
     
  2. jcsd
  3. May 6, 2016 #2

    BvU

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    Friction ≤ μR might be correct. You don't say what R is ?
    What ##\mu## did you find in part (i) (just checking :smile:) ?

    450 a ≤ 0.2(200g) would that be 200 g like in 200 times the acceleration from gravity ? Again, you don't say what 200 g stands for...
     
  4. May 6, 2016 #3
    Hi. R is the normal reaction force, sorry for not stating it
    I got μ=0.7 for part (i)
    200g is the normal reaction force of A due to its weight , a is the acceleration of the the whole body
     
  5. May 6, 2016 #4

    BvU

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    So if you draw a free body diagram of A, you get a maximum friction force of mA g Newton. The force to accelerate block A should therefore not be bigger than that. And that force is mA times the acceleration. The mass of B does not come into the equations, only the maximum force it can exert on A.
     
  6. May 6, 2016 #5
    Ah I see. So the force that actually accelerates A is the net force, P-(friction between B and floor), right?
     
  7. May 6, 2016 #6

    BvU

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    So the force that actually accelerates A is friction force between A and B. that is the only horizontal force that works on A.
    No. If you want to see why not, draw a free body diagram of box B !
     
  8. May 6, 2016 #7
    But how does frictional force accelerates A? I thought there must be some forward force?
     
  9. May 6, 2016 #8
    The static friction between the boxes will accelerate box ##A##.
    There is a forward force acting on ##A##. Imagine this, instead of saying that box ##B## is moving forward. You could reverse this and say that box ##A## is moving backwards. It is just saying that box ##A## is moving backward relative to box ##B##.
    And What is the friction direction if an object is trying to move backward? The direction is Forward. The static friction wants box ##A## to keep up with box ##B##.

    Another way to imagine this (which I like the most), You could put two fingers on your left hand. Now try to move your left hand in any direction. You will feel a tendency or a force acting on your fingers in the same direction.

    Now you may ask why is it static friction not kinetic? Well, you want it to move with the same acceleration, don't you? so as long as the acceleration of box ##A## equals the acceleration of box. We can say that at any moment box ##A## is still relative to box ##B##. You could visualize this two by watching two cars racing each other. The drivers see each other as if they were standing still but they see their surrounding move.
     
    Last edited: May 6, 2016
  10. May 6, 2016 #9

    BvU

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    There is. Block B is dragging block A along ! Newton 3: action = - reaction. So on the free body diagram of block A there is one forward force, the opposite of a backward force on block B. That's why I invited you to draw these free body diagrams. Once you understand those, you are well equipped to handle this and similar excerises!
     
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