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Magnitude of force between blocks

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    The last question on this page(#44): http://dacc.edu/~ksturgeon/106c5webpage.pdf
    I haven't had any problems with any homework till now so I'm hoping for a little guidance here. I guess my main issue is I don't understand what it means by magnitude of force. Because it seems to me that the Force of the small block is already given at 6N. I haven't had any other questions quite like this and I can only find solutions online and not method.

    2. Relevant equations

    f=ma

    3. The attempt at a solution
    a=4 for the 3kg block and 3 for the 1kg block
     
    Last edited by a moderator: Feb 26, 2012
  2. jcsd
  3. Feb 26, 2012 #2

    Astronuc

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    Staff: Mentor

    One has to determine the net force imposed by the small block on the large block.

    First one has determine the acceleration of the blocks, and the force imposed by the small block as a result of that acceleration. Then one has to determine the additional force from the imposed force.
     
  4. Feb 26, 2012 #3
    So, can I combine the blocks to have 1 block at 5kg with 1 force pulling at 12newtons and another in the opposite direction at 6 newtons? So a 5kg block with 6 newtons in one direction total. 6=5a so a is 1.2m/s^2. Then in addition to the 6 newton force from the small box I have an additional F=2*1.2 = 2.4 newtons. Therefore a total of 8.4 newtons of force.

    That is the answer I found online so I assume its right. Is this the best method for this or is there another way? It seems wrong to me to add the net acceleration to the small box since it is actually in the opposite direction of the way its 6 newtons is pushing it. I would have though it should be subtracted instead. Or, perhaps subtracting net acceleration from the original 3m/s^2 to give 1.8m/s^2 a for the box.

    Or, should I think of the 2.4 newtons as the force from the large box on the small one and add it as part of the magnitude of the equal and opposite reaction force?
     
  5. Feb 26, 2012 #4

    Astronuc

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    Staff: Mentor

    8.4 N is correct.

    The way to think about this is look at the net unbalance force on the mass.

    --- 2F ---> M(3kg), m(2kg) <----- F.

    The net unbalanced force is 2F - F or F toward the right.

    Now F is acting on the total mass M+m. So a = F/(M+m).


    Now the small mass resists the acceleration with a force = ma. So the small block imposes a force on the larger block at their interface (surfaces) of f = (2 kg)(1.2 m/s2). There is also the force F pushing against the smaller block into the larger block, so the total force is f + F or (2 kg)(1.2 m/s2) + 6 N = 8.4 N.
     
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