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Basic 3 Block Question

  1. Sep 5, 2010 #1
    Three blocks, each of mass 8 kg are on a frictionless table. A hand pushes on the left most box (A) such that the three boxes accelerate in the positive horizontal direction as shown at a rate of a = 0.7 m/s2.

    Here are the ones I got right:
    1. What is the magnitude of the force on block A from the hand?
    Answer: 16.9-Correct
    2. What is the net horizontal force on block A ?
    Answer:5.6-Correct

    And now the ones I cant figure out:
    3. What is the horizontal force on block A due to block B?
    and
    4. What is the net horizontal force on block B?

    For number 3, I tried subtracting (0.7*3) from total force of 16.9 but that was incorrect. Not sure how to approach this one.
     
  2. jcsd
  3. Sep 5, 2010 #2

    PhanthomJay

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    For #3,why 0.7*3? Is that a typo error? If the hand force on A is 16.8 N to the right, and the net force on A is 5.6 N to the right, then the force of B on A must be _____ to the ____.

    For #4, F_net = ma. What is m and what is a? What is the direction of F_net?

    Always draw free body diagrams.
     
  4. Sep 7, 2010 #3
    Ok number 3 is -11.2

    Number 4-wouldnt that answer be the same as the net force on block A? Why or why not? In this case fnet=ma would be f=8(0.7)=5.6, same as number 2. Whats wrong with this reasoning? Or would I have to also add to that the force coming from block A as well? The direction would be positive.
     
  5. Sep 7, 2010 #4

    PhanthomJay

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    that is correct, where the minus sign indicates that the force of B on A acts to the left.
    Nothing, it is correct. But be sure to indicate its direction is to the right.
    There are two horizontal forces acting on B .One is from Block A, 11.2 N acting to the right (per newton 3). the other is from Block C, which you can calculate if you were asked, by knowing that the NET force on B is 5.6 N to the right.
     
  6. Sep 7, 2010 #5
    Wait so the answer to 4 is not just 5.6? So the net horizontal force on B would have to be the sum of the forces acting on it from the left and right. Since A is giving B a force of 11.2 N positive direction and C is also giving B a force in the negative direction, the answer could not be 5.6?
     
  7. Sep 7, 2010 #6

    PhanthomJay

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    the answer to #4 IS 5.6 N to the right, using F_net=ma.
    that is correct
    Why not? 11.2 N to the right from A , and 5.6N to the left from C , so the net force is 11.2-5.6 = 5.6 N to the right. Convice yourself that the force of C on B is 5.6 N to the left, by looking at a free body diagram of C.
     
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