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Urgent, mechanics question

  1. Nov 16, 2004 #1
    two astronauts initially at rest in free space, pull on either end of a rope. Astronaut Alex played football in high school and is stronger than astronaut Bob, whose hobby was chess. The maximum force with which Alex can pull, Fa, is larger than the maximum force with which Bob can pull, Fb. Their masses are Ma and Mbi, and the mass of the rope, Mr, is negligible. Find their accelerations and forces if each pulls on the rope as hard as he can.

    Thanks in advance for the help
     
  2. jcsd
  3. Nov 16, 2004 #2

    DaveC426913

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    This is easily solvable in two steps:

    1. Pay attention in class
    2. Read the textbook
     
  4. Nov 16, 2004 #3
    Trust me, I've paid a lot of attention in class and read what text we have. If I use impulse, I can prove that Fa = Fb. This generates an answer, however, the fact that Fa = Fb contradicts the problem. Another approach would be to find the difference in the two forces Fa - Fb. Set that to the force and find acceleration by diving that by Ma + Mb (the total mass)
     
  5. Nov 17, 2004 #4

    cepheid

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    I don't see how impulse enters into it at all. Fa > Fb, that is given in the problem.

    Your last approach (find the NET force and use it to find the total acceleration of the system) sounds about right.
     
  6. Nov 18, 2004 #5

    Doc Al

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    the tension in the rope

    It's a bit of a trick question. You can only pull a rope with a force equal to the tension in the rope, and a massless rope has a single tension throughout. Since the weaker Bob can only supply a force of Fb, that's the most the tension can be. So both astronauts are pulled with a force of Fb.

    Note that a massless rope merely transfers force between the two astronauts. From Newton's 3rd law you know that whatever force Bob exerts on Alex, Alex exerts equal and opposite on Bob.
     
  7. Nov 19, 2004 #6
    Turns out, the answer is, Fa + Fb (for the force). but why? :confused:
     
  8. Nov 19, 2004 #7

    russ_watters

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    We did this in a lab once using a small girl and a big guy on skateboards. When both pulled, the guy pulled so hard the girl wasn't really pulling at all - her arms stayed extended. So the force was just the larger of the two.

    I don't see how it could be Fa+Fb.
     
  9. Nov 20, 2004 #8

    cepheid

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    Ok, so isn't the problem being misleading when it says that each astronaut pulls on the rope as hard as he can, when in fact each astronaut pulls on the rope only as hard as the weaker guy can?
     
  10. Nov 20, 2004 #9

    Doc Al

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    I know what you mean russ, but in this context the girl was "pulling" (that is, exerting a force on the rope) exactly as much as the guy was. If she wasn't, then she couldn't hang on to the rope. (Ignoring the mass of the rope, that is.)
    Agreed. What if two guys of equal strength pulled on the rope ends with force F? Would the force be 2F? Nonsense.
     
  11. Nov 20, 2004 #10

    Doc Al

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    That's why I said that this is a bit of a trick question. It turns out that under these conditions, the most that they can pull happens to be equal to the grip strength of the weaker astronaut. :smile:

    Actually, the problem statement is tricky at many levels. What does "maximum force" that each can exert even mean? Under what conditions was it measured? Standing on ground pulling a horizontal rope attached to a wall? (And thus relying on friction with the ground.) A more relevant measure would be: have each astronaut stand on a frictionless pad and then pull on a rope attached to the wall as hard as they could. (Lot's of luck.) The maximum tension generated would in some sense measure their relative strength under conditions relevant to the actual problem.

    So, on further thought, this problem is very misleading. I wouldn't use it.
     
    Last edited: Nov 20, 2004
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