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Three Blocks

  1. Oct 11, 2004 #1
    Ok, I don't know what I'm really doing in this problem. It does not seem as hard as I am making it seem, but I guess I need help to be guided through it. Here's the problem:
    Blocks of mass 4, 8, and 12 kg are lined up from left to right in that order on a frictionless surface so each block is touching the next one. A rightward-pointing force of magnitude 18 N is applied to the left-most block.
    A) What is the magnitude of the force that the middle block exerts on the rightmost one?
    B) What is the magnitude of the force that the leftmost block exerts on the middle one?
    C) Suppose now that the left-right order of the blocks is reversed. Now find the magnitude of the force that the leftmost block exerts on the middle one?

    Can anyone help me??
     
  2. jcsd
  3. Oct 11, 2004 #2
    Conservation of forces.

    See,

    On the entire system the force is 18 N, considering all three objects to form a single system, which leads to the conlusion that all the internal forces should balance each other. In the absence of friction all the forces are basically internal reactions of one block on the other.

    Use the following diagram and use balance of forces...
    and use the systems acceleration as a whole to see what the individual reactions would be.... and ofcourse use the system acceleration on each object individually...

    First Object :
    ---->O<---- N1_2 (Normal Reaction of 1 on 2)
    18 N

    Second Object:
    ------> O <-------
    N1_2 ......... N2_3

    Third Object :
    ------> O <-------
    N2_3 ....... Zero.. no force from the right side....

    So use the free body force diagrams and remember each block accelerates at the system acceleration... so Net Force = Mass * Acceleration. Use it and calculate the reactions...

    Good luck,
    Gopi
     
  4. Oct 11, 2004 #3
    It might be instructive to first think about how all the blocks are moving together. What is the overall acceleration of these blocks? (recall F=ma). Now, for part A) the middle block is pushing on the right block. We already know what the right block's acceleration is right? What force is necessary for a block of 12kg to accelerate like that?

    For part B) think about treating the middle and right blocks as one "mass".
     
  5. Oct 12, 2004 #4
    Ok, I'm still a bit confused. Even with the help that is provided with the questions, it states what is the acceleration of the rightmost block, is it the same as the others? How can you determine the acceleration of the system of three blocks? I believe that to find the acceleration, it is F/m , but I maybe wrong. This would however give me .75 for acceleration on the third block or for the system.
     
  6. Oct 12, 2004 #5

    Pyrrhus

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    Homework Helper

  7. Oct 12, 2004 #6
    Does his explanation fit for mine of three blocks of different masses. The original post said m1=m2=m3 . Also, by using his explanation which i can follow clearly, I would obtain 15 N. What does this number stand for, I don't think it can be acceleration if it is in Newtons.
     
  8. Oct 12, 2004 #7

    Pyrrhus

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    Homework Helper

    the masses are different in yours, in that example they are the same, but the similarities lie in the contact forces, and in Newton's 3rd Law application to solve it.
     
  9. Oct 12, 2004 #8
    Ok, this is what I obtain, and the online program keeps saying it is wrong. Am I doing this right?

    Fnet = 18 - Fm1m2 = m1a
    Fnet = Fm2m1 - Fm2m3 = m2a
    Fnet = Fm3m2 - 0 = Fm3m2 = m3a

    From these and Newton's third law, I substitute and work my way to solving for the acceleration.

    m3a = Fm3m2
    m2a = Fm2m1 - Fm2m3
    m1a = 18 - Fm1m2

    solving for a :

    m3a = m2a - Fm2m1
    m3a = m2a + 18 - m1a
    m2a - m2a + m1a = 18
    a ( m3 - m2 + m1 ) = 18
    a = 18 / (m3 - m2 + m1) = 2.25

    Therefore:
    Fm3m2 = m2a + 18 - m1a = 8 (2.25) + 18 - 4 (2.25) = 27 N

    Is this how you would solve for part a?
     
  10. Oct 12, 2004 #9

    Doc Al

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    Staff: Mentor

    OK, as long as you realize that Fm2m3 = Fm3m2. (By the way, this notation is horrible! :yuck: )

    OK. Now just add them up to solve for a.

    You made a mistake somewhere in here. Instead, just add the three equations above (where I say to) to find a.
     
  11. Oct 12, 2004 #10
    ok, I found my mistake. In taking Newtons Third law, I took the negative of one of the forces to equal another(which is assumed to be the opposite force already), which led to Fm3m2 = 18 - m1a - m2a instead of
    Fm3m2 = m2a + 18 - m1a . Thanks everyone for your help!!
     
    Last edited: Oct 12, 2004
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