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Homework Help: Forces acting on 2 boxes connected by taut thread

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Two boxes $A$ and $B$ are connect4d by a lightweight cord and are resting on a frictionless surface. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force $F_p$ of 40 N is applied to the 10 kg box. Find the acceleration of each box and the tension in the cord between them.

    2. Relevant equations
    Newton's 2nd law: $\sum F_{x} = ma_x$

    Is the force $F_p$ the same on both the boxes? If so, why?
    To calculate the answers, I need to assume that $\sum F_{x}$ on box A and box B are the same, $F_T$, the force on the thread, is acting in one direction opposing $F_{p}$ on box $A$ and $F_T$ is acting on box $B$ in the same direction as box A.

    I can solve it but am unable to find a rationale for $F_x$ being the same on both boxes where $F_x$ is the sum of all forces acting on each box.
  2. jcsd
  3. May 4, 2008 #2


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    Gold Member

    F_x is NOT the same on both boxes, that is why you can't find a rational for saying that. The net force on each block will be different.

    There is a quantity that is the same for both boxes. What is it?

    HINT: They are tied together, thus, what quantities describing their motion must be the same?

    Also, if you want to use LaTeX on this forum you have to surround all code with the following tags, not $:

    [ tex ] [ /tex ]

    (without the spaces, of course)
  4. May 4, 2008 #3
    Since they are tied together, their acceleration must be the same. In the solution, however, it says that [tex]F_x[/tex] is the same for both boxes.
  5. May 4, 2008 #4


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    Yes, there accelerations must be the same. Work from there and use newton's second law on each block.

    You should end up with two equations and two unknowns, a, and the force of tension in the middle rope.

    Also, the force may indeed be the same on each block, but this does not have to be the case. The accelerations however, will always have to be equal. Thus, I suggest working from there.
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