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Static Equilibrium of massless rope

  1. Dec 5, 2005 #1
    A uniform sphere of mass m and radius r is held in place by a massless rope attached to a frictionless wall a distance L above the Centre of the sphere. Find the tension in the rope and the force on the sphere from the wall.

    I tried to get a picture, but it is apparently too big to host.

    Do I start with a FBD? If so what should it look like?, How can I find T?
     
  2. jcsd
  3. Dec 5, 2005 #2

    Fermat

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    Homework Helper

    Is this what it looks like ?

    Since the sphere is in static equilibrium, then all three forces must pass through the COM of the sphere.
    Also, since the sphere is in static equilibriulm then the sum of the vertical forces is zero and the sum of the horizontal forces is zero.

    ΣFV = 0
    ΣFH = 0

    [​IMG]
     

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  4. Dec 5, 2005 #3

    Astronuc

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    Staff: Mentor

    Well the way to look at this is that the rope is fixed to the surface of the sphere, and one would have to draw a diagram with the rope at some angle to the wall.

    The rope has length Lr, which is related to the height L and radius r.

    For L >> r, the tension would simply be given by mg, the weight of the sphere, but as L -> r the tension becomes a function of the angle between the wall and the rope.
     
  5. Dec 5, 2005 #4
    Yes..Thats what it looks like..

    How can I find the tension...There are no angles given...
    The answer in the back of the book says (mg/L)root(L^2 + r^2)
     
  6. Dec 5, 2005 #5

    Astronuc

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    Draw a triangle and think about what L/root(L^2 + r^2) is, i.e. is one has a triangle of legs r and L, and hypotenuse sqrt(L^2 + r^2), how does one express the sin and cos of the angles?
     
  7. Dec 5, 2005 #6
    so, [tex]sin \theta = \frac{L}{T}[/tex] where [tex]T = root(L^2 + r^2)[/tex]

    Where does [tex] \frac{L}{mg}[/tex] come into place?
     
  8. Dec 5, 2005 #7

    Astronuc

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    Staff: Mentor

    The tension T in the rope can be resolved into vertical and horizontal components. The vertical component of the tension must be equal to mg, the weight of the ball.

    Therefore T cos [itex]\theta[/itex] = mg, where [itex]\theta[/itex] = [itex]\frac{L}{\sqrt{L^2+r^2}}[/itex], so

    T = mg/cos [itex]\theta[/itex] = [itex]\frac{mg\sqrt{L^2+r^2}}{L}[/itex].

    Then the horizontal force is simply equal to T sin [itex]\theta[/itex] and provides the force on the wall. I leave it to the student to finish.
     
    Last edited: Dec 5, 2005
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