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Physics problem

  1. Apr 21, 2005 #1
    I wasn't for sure where to start with this problem. It is the first of 3 so if I understand how to do this one, I will be able to most likely get the others. I'm pretty sure that I need to make a free-body diagram of the forces in this situation but wasn't for sure where to label the friction force.

    A 11 kg sphere is help against a wall by a string being pulled at an angle of 62 degrees. Given: f is the magnitude of the frictional force, and W=Mg.
    If the net torque is equal to zero for the center of the sphere, that leads to which of the following?
    4. F=f
  2. jcsd
  3. Apr 21, 2005 #2


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    None of them look right based on what you have stated, but the statement is not clear. What is F? The force (tension) from the string? Is the angle measured relative to horizontal or vertical?
  4. Apr 22, 2005 #3
    Sorry, I tried to post a picture that helped explain but it didn't work. F is the tension force of the string and the 62 degree angle is relative to the horizontal.
  5. Apr 22, 2005 #4
    The correct answer is 4!!!

    [tex]f\cdot R= F \cdot R[/tex]

    (total torque relative to the center is zero)

    R- radius
    f - friction
    F- tension in the string
  6. Apr 22, 2005 #5


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    That appears to be the only possible answer among the choices given, but it is not the only possible solution to the problem unless you know that the string is on a tangent of the sphere. So either the diagram that we have not seen shows it is tangent, or the question should be asking for a possible solution, not THE solution.
  7. Apr 22, 2005 #6
    You're right OlderDan, we don't have the picture then we don't know if 1 or 4 is correct (for another ones the chances are slim to none whatever the picture would contain). So under the given circumstances 4 CAN BE an "answer". 1 can be an answer only if the angle is measured with respect to the vertical direction and the string is not tangent to the sphere.

    Anyway, we're fooling around with this "without-figure" problem......... :grumpy:
    Last edited: Apr 22, 2005
  8. Apr 23, 2005 #7
    Here, you can learn more :

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