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Homework Help: Fnet = ma problem

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Blocks A and B are sitting side by side (touching) on a horizontal surface. A constant horizontal force Fa is applied to block A, which pushes against block B with a 32.0 N force directed horizontally to the right. The same force Fa is applied to block B; now block A pushes on block B with a 16.0 N force directed horizontally to the left. The blocks have a combined mass of 30.0 kg.

    (a) What is the magnitude of the acceleration of the blocks in figure (a)?
    (b) What is the magnitude of the force Fa?

    2. Relevant equations

    Fnet = ma

    3. The attempt at a solution

    Part 1: Fa pushes on block A:
    Block A has Fa pushing to the right, and block B pushing to the left so

    Fnet = massblockA *acceleration = Fa - 32N

    Block B pushes back on block A (only horizontal force on it)

    Fnet = massblockB *acceleraton = 32N

    mass blockA + massblockB = 30kg

    Using these three equations I solved for Fa and got that Fa = 30kg*acceleration...
    I thought that I could use the next set of data (Fa pushing on block B) to come up with another equation, then I could solve for Fa and acceleration, but the equations are exactly the same, so pretty useless.
    Any help would be great, thanks!
    (ps~ I sometimes have trouble with FBDs, so if something is wrong with how I laid out the forces, it would be great to know!)
  2. jcsd
  3. May 6, 2008 #2

    Doc Al

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    Staff: Mentor

    They aren't exactly the same. Write an equation for the force on block A using the second set of data.
  4. May 6, 2008 #3
    Ok, Now I'm really not so sure:
    For Fa pusing on block A:

    massA + massB = 30kg
    massblockA *acceleration = Fa - 32N
    massblockB *acceleraton = 32N


    (Fa - 32)/ acceleration = massA
    32/acceleration = mass B

    (Fa - 32) + 32 = 30 * acceleration
    So Fa = 30 * acceleration

    For Fa pusing on B:

    massA = 16N / acceleration
    masB = (Fa -16) / acceleration
    16N + (Fa - 16) = 30 * acceleration
    or, Fa = 30 * acceleration

    .... am I having a math problem then?
  5. May 6, 2008 #4

    Doc Al

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    Staff: Mentor

    Combine the two equations I highlighted.
  6. May 6, 2008 #5
    Oh.... how did you know to use those equations? (I have a final coming up, and any tips would be great)
    Thanks for your help!
  7. May 6, 2008 #6

    Doc Al

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    Staff: Mentor

    To me, those two equations are the most direct mathematical statements of the given data:

    That tells me that: 32.0 = massB*a

    From Newton's 3rd law, I immediately know the force on block A is 16.0 N.
    That tells me that: 16.0 = massA*a

    Add those equations to get:
    32 + 16 = (massA + massB)*a

    Then apply:
    To get the acceleration.

    The best thing to do is to write clear FBD diagrams of both blocks for both sets of data. Write all the equations and look for any easy ways to combine them.
  8. May 6, 2008 #7
    Thanks for your help :)
  9. Sep 29, 2011 #8
    Sorry if I bring it up, but I have the same problem, and I'm trying to figure out how to solve it. I read all the post above, and one thing I don't understand is, the acceleration in those 2 figures should be different, they can't be the same, and if they're not the same, how can we set up the equations ?
  10. Sep 29, 2011 #9
    unloved .... did you draw FBD ?

    EDIT:first assume that the acceleration is different in two cases, when you set up equations , you will see that they turn out to be same
    Last edited: Sep 29, 2011
  11. Sep 29, 2011 #10

    Doc Al

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    Staff: Mentor

    Why do you think the accelerations of the two blocks are different? They are in constant contact.
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