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Figuring out the forces

  1. Aug 25, 2013 #1
    Hi,

    Please take a look at this figure:
    Screenshot by Lightshot

    There is a force that is applied so that block B will not fall under the influence of gravity.

    I have set up the problem as:
    F= (ma+mb)a

    μR=mbg

    where μ is the coefficient of friction and R is the reaction force.

    Now my question is I can't figure out why mba is the reaction force.

    What exactly is maa and mbb? Shouldn't maa be the force of block a on b? Then what about mbb? What is this force? The force direction should be pointing to the right side (since the applied force is to the right)? So it is as if mbb is acting on thin air?

    I can't wrap my head around what is maa and mbb, please help.

    Thanks.
     
  2. jcsd
  3. Aug 25, 2013 #2

    Andrew Mason

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    The 3rd law pair to the force of A on B is the equal and opposite force of B on A. But you do not really need to use the third law. This is a second law problem.

    The force of A on B, FA→B = mba, which is normal to the surfaces between A and B. The maximum static friction force between B and A is the force normal to the surface between B and A multiplied by the co-efficient of static friction (μs).

    Ffmax = μsFA→B = μsmba

    In order for B not to fall, how must this maximum static friction force be in relation to mbg ?

    AM
     
  4. Aug 26, 2013 #3
    Thanks Andrew. I know the solution to this problem, but I need to know what is maa and mba. Why is the force of A on B equals to mba instead of maa? Then what is maa?
     
  5. Aug 26, 2013 #4

    Andrew Mason

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    Draw a freebody diagram and apply the second law.

    F = ma+ba = maa + mba

    so: maa = F - mba

    Examining the forces on A: there is F pushing A forward and the force from B on A (= -mba) which is in the opposite direction to F. Together they must sum to the net force on A which is necessarily maa (second law).

    For B, there is only the force of A on B. This must be equal to mba (second law).

    AM
     
    Last edited: Aug 26, 2013
  6. Aug 31, 2013 #5
    Thanks, I think it's clearer now.

    So, F only acts on A but not B right? This is because F is in direct contact with A but not B, correct? But since B is in direct contact with A, it will feel the force of A on B.
     
  7. Aug 31, 2013 #6

    Andrew Mason

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    You can divide the bodies any way you like. The forces that apply to all parts of A and B all originate with F: eg. without F, A could apply no force to B.

    If you want to make a distinction between A and B, then you can say that F acts on A and A acts on B.

    You could divide A into two parts and say that the left half of A applies a force to the right half of A so that the force applied by the left half of A causes the right half to accelerate.

    Or you could make no distinction between bodies or parts of bodies treating A+B as a single mass and simply say that F applies a force to a mass A+B.

    AM
     
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