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Homework Help: Two boxs

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data

    In Figure 5-50a, a constant horizontal force F->a is applied to block A, which pushes against block B with a 23.0 N force directed horizontally to the right. In Figure 5-48b, the same force is applied to block B; now block A pushes on block B with a 11.0 N force directed horizontally to the left. The blocks have a combined mass of 12.0 kg. What are the magnitudes of (a) their acceleration in Figure 5-50a and (b) force F->a?


    2. Relevant equations

    Fnet=m*a

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c05/fig05_50.gif


    3. The attempt at a solution

    Ok so for block scenario one I have to find the Fnet to solve for m*a.
    So I have Fnet= sum of forces=m*a

    for case two

    Fnet= sum of forces =m*a


    So I have

    12*a= Fapp+23-F_AB
    12*a=Fapp-11+F_AB


    I am stuck right here any tips?
     
  2. jcsd
  3. Feb 16, 2009 #2

    LowlyPion

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    Write down what you know.

    Ma + Mb = 12

    23N = Mb*a

    11N = Ma*a

    Since a is the same, then

    Mb/Ma = 23/11

    Mb = 12 - Ma = 23/11*Ma

    Ma = 12*11/34 ...
     
  4. Feb 16, 2009 #3
    Thank very much
     
  5. Feb 16, 2009 #4
    Wait how would I found F->


    Do I use: 23=F-> + X and 11=F-> -X ?


    Then F->= 17 N?
     
  6. Feb 16, 2009 #5

    LowlyPion

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    What is your total mass and the acceleration of the system?
     
  7. Feb 16, 2009 #6
    So the F->applied would be 12*a which I found?
     
  8. Feb 16, 2009 #7

    LowlyPion

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    Isn't a the acceleration of both blocks?

    And total mass is Ma + Mb.

    So ...
     
  9. Feb 16, 2009 #8
    Thanks
     
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