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Static equilibrium with a suspended sphere on a wall

  1. Oct 26, 2007 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations
    [tex]\Sigma[/tex]Fi = 0
    [tex]\Sigma[/tex][tex]\tau[/tex] = 0

    Nw = force of the wall
    Ff = Frictional force
    [tex]\mu[/tex] = frictional coefficient
    T = Tension

    3. The attempt at a solution
    [tex]\Sigma[/tex]Fx = Nw - Tsin(30) = 0

    [tex]\Sigma[/tex]Fy = -mg + Tcos(30) + Ff = 0

    [tex]\Sigma[/tex][tex]\tau[/tex] = (R/2)mg - (3/2)R*Ff = 0

    That's what i have, but i have a feeling that i'm missing something. Any help would be great. Especially if it's Doc Al!

  2. jcsd
  3. Oct 26, 2007 #2

    Doc Al

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    Staff: Mentor

    Right idea, but check your sines and cosines.


    The only thing you're missing is the relationship between the force of friction and [itex]\mu[/itex]. What is it?
  4. Oct 26, 2007 #3

    I think the trig is correct. Unless i'm smoking something, but the opposite of that 30 degree angle is the x component of T, right? And the adjacent is the y component? Unless you want me to use Tcos60 for the xcomponent and Tsin60 for the y?

    Anyway --->

    Ff = [tex]\mu[/tex]Nw --->

    Nw = Tsin30 --> from the x-components, so -->

    Ff = [tex]\mu[/tex]*Tsin30
  5. Oct 26, 2007 #4

    Doc Al

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    Staff: Mentor

    Yep, you're right. (D'oh!)

    Right. So solve those equations together and you'll find the value for [itex]\mu[/itex].
  6. Oct 26, 2007 #5
    worked out on paper, i got 0.866 which is the right answer!

    You know what confuses me the most is that you can solve for an actual number in a question with all variables. What is the best way to tell that variables will actually cancel out and you'll end up with computable trig functions and a constant? My brain is still trying to get used to it.
  7. Oct 26, 2007 #6

    Doc Al

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    Staff: Mentor

    If the number of unknowns matches the number of independent equations--that's a good sign.

    But the real secret is to have the confidence to set up the equations and attempt to solve them without knowing in advance that it will work out. That's how you learn. Some students (not you!) are afraid to make a move unless they see how to get the final answer in one step; with that attitude they'll never improve their problem-solving skills.

    After you've solved a zillion problems, you'll have an intuition about what's solvable and what's not. But don't wait for that!
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