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Three blocks

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Blocks of mass 5, 10, and 25 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 14 N is applied to the left-most block.


    2. Relevant equations

    [a] What is the magnitude of the force that the middle block exerts on the rightmost one?
    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?

    3. The attempt at a solution

    this is what i tried doing but i think im totally off
    Fm3m2 = m2a + 14 - m1a
     

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    Last edited: May 9, 2010
  2. jcsd
  3. May 9, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi farrah003! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    They all have the same acceleration, so use F = ma on all-three-together to find a.

    Then use F =ma (with that a) on individual blocks. :wink:
     
  4. May 9, 2010 #3
    i tried f = ma
    this is what i did
    m = 5+10+25 = 40
    14 = 40(a)
    a = .35

    but it said thats not the right answer
     
  5. May 9, 2010 #4
    sorryy my mistake i got it for part a which was 8.75 but when i did it again for part b it said wrong
     
  6. May 9, 2010 #5

    tiny-tim

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    I'm confused :confused:

    show your equation for part b. :smile:
     
  7. May 9, 2010 #6
    F = (5)(.35) = .175
     
  8. May 9, 2010 #7

    tiny-tim

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    That F is the total force.

    Apart from the right-most block (which you solved for in part a), each block has two forces on it.
     
  9. May 9, 2010 #8
    so then what formula do i use ?
     
  10. May 9, 2010 #9

    tiny-tim

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    Ftotal = ma
     
  11. May 9, 2010 #10
    so then i do 14=5a
    then the a is 2.8 n if i multiply that im gonna get 14 again ... im confused !!:confused:
     
  12. May 9, 2010 #11

    tiny-tim

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    No, a is the same for all three blocks.

    The left block has two forces on it: 14 from the left, and the reaction force from the right.

    The vector sum of those two forces is 5a.
     
  13. May 9, 2010 #12
    lol i still dunno what numbers to use in what formula
     
  14. May 10, 2010 #13

    tiny-tim

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    (just got up :zzz: …)

    The left block has two forces, 14 and R (the reaction force from the middle block), so F = ma means 14 - R = ma.
     
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