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Homework Help: Rotational problem

  1. Oct 27, 2008 #1
    1. The problem statement, all variables and given/known data
    My friend wanted me to explain this problem, it seems pretty simple but I can't get the back of the book answer.

    a wheel 60 cm in radius is connected to an axle that is 1.5 cm in radius. Someone pulls with a force of 50 N on a thin cord wrapped around the axle. How much force do you have to exert on the wheel's rim to hold it still?

    3. The attempt at a solution

    First I tried a simple torque balancing but that doesnt work: 60cm * x N = 1.5 cm * 50 N -> x = 1.25 N.

    Then I thought, maybe the wheel has friction with the ground, because torque AND force must be balanced in static equilibrium, so:

    x (the force) * 60 cm - (50N - x) (frictional force that prevents slipping) * 60 cm = 50 N * 1.5 cm. But that makes an answer near 25ish N, obviously because these torques are large compared to the one around the axle.

    So the correct answer is supposed to be 6.1 N but I can't get it, can anyone explain it to me? I really like this friend and I would appreciate it :)
  2. jcsd
  3. Oct 27, 2008 #2
    My interpretation of the question is the exact same as your first attempt, and likewise i get the same answer. Is anything else given to you in the problem? Width of the cord, angle a force is applied and diagrams you can provide?
  4. Oct 27, 2008 #3
    No, just this text ;) I think the first one is wrong because the forces are not balanced, so it should move forward unless there was something pushing back... but yeah this problem seems pretty odd.
  5. Oct 27, 2008 #4
    Well, assuming it is rotating on an axle, i would assume the normal for between the axel and the wheel would counter any transverse motion... making your solution of 1.25N correct again. I don’t know what to tell you other than maybe we're interpreting the question wrong or there is a mistake in the back of the book.
  6. Oct 27, 2008 #5
    Can anyone else please help? I really need an answer, I REALLY like this friend ;)
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