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Sphere With Torque Sliding

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A uniform sphere of radius r and weight W slides
    along the floor due to a constant horizontal force P applied by a
    string. Suppose the coefficient of friction is µ. Find the height of the
    string above the floor h.


    2. Relevant equations
    ƩT = 0 because there cannot be non-zero torque as the sphere must not roll, as it is sliding.
    T = r x F
    Friction = μ*W


    3. The attempt at a solution
    ƩT = 0 = T(P) + T(Friction)
    = r x P + r x F(Friction)
    = rPsinθ + rF(Friction)
    = rPsinθ + rμW
    => -rPsinθ = rμW
    => sinθ = -(μW/P)

    I have it to the definition of the angle of the force applied, but how do I relate the angle to the height above the ground?
     
  2. jcsd
  3. Nov 18, 2013 #2

    ehild

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    What is θ? The force is horizontal, applied at hight h above the floor. Make a figure . What is the torque of the force?


    ehild
     
    Last edited: Nov 18, 2013
  4. Nov 19, 2013 #3
    I guess I'm not sure what you're getting at, since θ is the angle between r and P, and the torque is then r x P = rPsinθ.

    But it got me thinking, if you take the θ between r and P and subtract it from 180 to get the angle that r makes with the horizontal to reach P (because of the parallel horizontal lines), I will then have an angle, say [itex]\phi[/itex] that can be used to describe the y position of the force. As in I will have the angle that relates to the "unit" circle. Then can I just take the sin([itex]\phi[/itex]) and add it to r to get the height? Would that work?
     
  5. Nov 19, 2013 #4

    CWatters

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    I suggest a drawing. In fact always do a drawing. I got a simpler answer with no trig functions.
     

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    Last edited: Nov 19, 2013
  6. Nov 19, 2013 #5
    Here is the picture I have been using. Maybe I'm interpreting the problem wrong?

    Edit: Oh! Just saw your picture. Taking a look at it now...

    Edit2: To reach your solution, did you use the same method, however? Rather, is my next step to find a way to rewrite the sinθ in terms of R and H or should I rethink my process?
     

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    Last edited: Nov 19, 2013
  7. Nov 19, 2013 #6

    ehild

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    Yes, try to write sinθ in terms of H and R. And check the signs of the torques -that of the applied force and that of friction.

    ehild
     
  8. Nov 19, 2013 #7
    I think I got it, hopefully this is right. Since the angle of θ is actually simply the angle that is made with the horizontal, as in a unit circle...

    [itex]sin(θ) = \frac{-uw}{P}[/itex]

    [itex]\frac{r-h}{r} = \frac{-uw}{P}[/itex]

    [itex]r^2 - h = \frac{-uwr}{P}[/itex]

    [itex]h = \frac{uwr}{P} + r^2[/itex]
     
  9. Nov 20, 2013 #8

    ehild

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    No, your derivation is not right. What is uw? h is length, r^2 is length-squared. You can not subtract them.

    The angle between two vectors is the angle enclosed by the direction of the vectors. See picture. You need the angle between the position vector R and the force vector P. Shift the position vector R, so as its tail is at the same point as the tail of P. θ is the angle shown. In the blue right triangle, you can see that sinθ = (R-H)/R.

    The torque of the applied force should be equal in magnitude and of opposite direction as the torque of the friction.
    attachment.php?attachmentid=64108&d=1384928101.jpg
    ehild
     

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    Last edited: Nov 20, 2013
  10. Nov 20, 2013 #9

    CWatters

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    Remembering that torque is just force * tangential distance, if you pick the centre of the sphere to sum them around you can read off the distances from the drawing.

    The tangential distance for the friction force is just R
    The tangential distance for force P is ???

    EDIT: I modified my drawing slightly...
     

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