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Homework Help: Car moving with two blocks attached to it.

  1. Jan 12, 2013 #1
    1. The problem statement, all variables and given/known data
    What force must be applied in the car of the image in order to all blocks remain stationary in relation to the car? Supposed that there's no friction in all surfaces, pulleys and wheels.

    2. Relevant equations

    Fr = ma

    3. The attempt at a solution
    First thing: That's a doubt I have, since there is no friction, the force F applied on the cart won't actually affect the block m1 in anyway, right? I mean, if there was no m2, the block m1 would stand in the air while M rushed in a straight line, right?

    Ok, so I made all the equations and free body diagrams:

    For block m1:
    [itex]T = m_1a[/itex]

    For block m2(F_32 is the force that the car M exerts on block with mass m2, by the way):
    [itex]\\T - P_2 = 0\\
    F_{32} = m_2a[/itex]

    And finally, for the car(Again, F_23 is the reaction to F_32):
    [itex]F - F_{23}= Ma [/itex]

    Substituting everything, I find:
    [itex]F = \frac{m_2g(m2+M)}{m1}[/itex]

    The answer, however, is:
    [itex]F = \frac{m_2g(m1+m2+M)}{m1}[/itex]

    Right, so, the thing is, I've learned that there are mainly two ways of working this kind of problem:
    The first one is to look at the whole system, and only analyze the external forces. The downside is that I won't know the internal forces.
    The second one is to look at each free body diagram and calculate the relevant forces. The downside being that it takes more time to do in this way.

    That being said, both ways always give the same answers, they're just different paths.

    However, in this case, I've noticed that if I consider the system as a whole, I will get the right answer (Since mtotal = m1 + m2 + M), when considering the free body diagrams though, the block 1 won't act on car M, so he doesn't get in.

    Where am I wrong?

    Thanks in advance!
    Last edited: Jan 12, 2013
  2. jcsd
  3. Jan 12, 2013 #2


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    Staff: Mentor

    That's true, so the idea is to make sure that the tension in the rope just keeps m2 from falling. Then that tension will also keep m1 stationary with respect to the cart.

    What tension in the rope will keep block m2 from falling? (What tension balances the vertical components of its FBD?)

    What acceleration of m1 will produce this tension? (think inertial force)
  4. Jan 13, 2013 #3
    Right! But in fact, I did that. P2 = T means it will remain stationary (I actually wrote that as P2 - T = 0, but it's the same thing of course). I also found the acceleration correctly.

    My biggest problem was, as I said, the last part when analyzing the big cart alone.
  5. Jan 13, 2013 #4


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    Homework Helper

    The string presses the pulley: it is attached to the cart, so the cart experiences force from the tension in he string.

  6. Jan 13, 2013 #5
    Wow, that is... amazing, and does make sense. That is truly amazing, thanks a lot!
  7. Jan 13, 2013 #6


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    Homework Helper

    You are welcome :)

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