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Force exerted on block

  1. May 18, 2017 #1
    1. The problem statement, all variables and given/known data
    Blocks of 1.0, 2.0, and 3.0 kg are lined up on a frictionless table with a 12-N force applied to the leftmost block. What's the magnitude of the force that the rightmost block exerts on the middle one?

    The blocks are lined up in such fashion: [1][2][3] and the 12 N force being applied to the leftmost block is obviously to the rightward direction.

    2. Relevant equations


    3. The attempt at a solution

    The force being applied to the entire system is 12 N and all of the blocks accelerate at the same rate, so letting ##m_i## be the mass of the i'th block, we have ##12 = (m_1 + m_2 + m_3)a \implies a = \frac{12}{m_1 + m_2 + m_3}##. Now, let ##F_{23}## denote the force that the second block exerts on the third block. We have ##F_{23} = \frac{12m_3}{m_1 + m_2 + m_3}##. Plugging in the values for each ##m_i## gives ##a = 6##. I know that ##F_{32}##, which is the force that the third block exerts on the second is then ##-6## N by the third law.

    I have two questions.

    1) The textbook gives an answer of ##6## N to the right. Is this wrong? How could the force that the third block exerts on the second block be to the right?

    2) If I try to find ##F_{32}## directly, without using the third law, I get ##F_{32} = m_2 \cdot (-a) = m_2 \cdot \frac{-12}{m_1 + m_2 + m_3} \neq -6##. Is this because the acceleration, ##a##, that I'm plugging in is the acceleration of the entire system, whereas I would need to plug in the acceleration that block 3 induces on block 2?
     
  2. jcsd
  3. May 18, 2017 #2

    haruspex

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    Yes, it is clearly wrong.
    You are forgetting the other force on the middle block.
     
  4. May 18, 2017 #3
    The only other horizontal force is ##F_{12}## (the force the first block exerts on the second block), but we are told to only find ##F_{32}##.
     
  5. May 18, 2017 #4

    haruspex

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    I understand that, but the law you are trying to apply to the middle block is ΣF=ma. ΣF must include all the horizontal forces acting on the middle block.
     
  6. May 18, 2017 #5

    scottdave

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    I tend to agree with you about the direction. Let's look at a slightly simpler problem, to see if it can help with yours.
    Take two blocks. The left one (call it A) is 3 kg, and the right one (call it B) is 2 kg. Now push (10 Newtons toward the right) on A. Total mass is 5 kg, so (10 N)/(5 kg) = 2 m/s2. Now let's look at B. It is 2 kg, and it is accelerating at 2 m/s2 so it must have 4 Newtons pushing on it (toward the right). So it is pushing back 4 Newtons onto A (to the left). Now A has 10 N right and 4 N left for a net force of 6 N (to the right).
    Let's see if we did it right. A has 6 Newtons net force, and it's mass is 3 kg, so (6 N)/(3 kg) = 2 m/s2, which is what we expect. Can you expand this thought process to your problem of 3 blocks?
     
  7. May 18, 2017 #6
    Thanks for the help guys.
     
  8. May 18, 2017 #7
    I thought I had solved it but I realized I made some mistakes...

    Okay, the only horizontal forces that are on Block 2 are ##F_{12}## and ##F_{32}##. Therefore, ##F_{12} + F_{32} = m_2a = 2m_2## (since a = 2). Therefore, ##F_{32} = 2m_2 - F_{12} = 2m_2 - 2m_2 = 0## since ##F_{12} = 2m_2##. This is clearly wrong. I feel as if I am missing something obvious...
     
  9. May 18, 2017 #8

    haruspex

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    How so?
     
  10. May 18, 2017 #9
    ##F_{12}## is the force that the first block exerts on the second. The object the force is being exerted on is block 2, Hence ##F_{12} = m_2(a) = 2m_2##
     
  11. May 18, 2017 #10

    haruspex

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    That's making the same mistake as before, but from the other side. The acceleration of the middle block results from the net force on it.
     
  12. May 18, 2017 #11
    Okay, I see your point. The acceleration I need for ##F_{12}## is the acceleration the middle block gains from its contact with block 1 alone (right?)

    But I don't see how to calculate that acceleration.
     
  13. May 18, 2017 #12

    haruspex

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    You could start with the 12N and subtract the net force needed to accelerate block 1 to find F12.
     
  14. May 18, 2017 #13
    That solves the question, but why does subtracting the 2N from 12N give me F12?
     
  15. May 19, 2017 #14

    haruspex

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    Because for block 1, 2N=ma=ΣF=12N+F21= 12N-F12. So F12=12N-2N.
     
  16. May 19, 2017 #15
    Thank you for your help and patience. I have to say, I think the first way I solved the problem was a lot easier!
     
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