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Homework Help: Tension in2 wires

  1. Mar 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Two wires are tied to the 200 g sphere shown in figure. The sphere revolves in a horiIzontal circle at a constant speed of 6.80 m/s.

    http://img684.imageshack.us/img684/6537/knightfigure0761.jpg [Broken]

    2. Relevant equations
    Ca=m*v^2/r
    f=ma

    3. The attempt at a solution
    I was able to find the radius using triangles, and with that
    I know the centrifical acceleration is .200*6.8^2/.866=10.679
    But i have no idea where to go from here
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 3, 2010 #2

    Delphi51

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    Use trig to find what tension provides a horizontal component of half the centripetal force m*v^2/r
     
  4. Mar 3, 2010 #3
    Does that meen the tension of the 2 strings will be equal? that is whats throwing me off
     
  5. Mar 3, 2010 #4

    Delphi51

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    Yes. The tensions must be equal because it is perfectly symmetrical.
     
  6. Mar 4, 2010 #5

    rl.bhat

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    What about the weight of the sphere? Will it not contribute to the tension.
     
  7. Mar 4, 2010 #6
    I may be wrong, but I don't see how it's possible for the tensions to be equal. In addition to the radial acceleration, there is also the vertical weight of the sphere. So, the upper string will be supporting the weight of the sphere, but the bottom string won't.

    Imagine the system is just starting to spin. The sphere will hang, making the upper string tight, but the lower string will be slack. As the system spins, the ball will be forced outward. If the lower string weren't there, the top string might almost reach a horizontal angle if the rotation is fast enough. But the bottom string will kick in and keep the sphere from rising above a certain point. No matter how fast it spins, the top string will never be slack, and the way I figure it, the top string will always have greater tension than the bottom string.

    Is that incorrect?
     
  8. Mar 4, 2010 #7

    PhanthomJay

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    I think you may be correct, however, in order to find out, the OP will have to assume that they are not equal and sum forces in the x direction and y directions and use Newton's laws, and then be sure that ther lower rope does not stay slack. Juggalomike, what's the acceleration in the x direction, and in the y direction?
     
  9. Mar 4, 2010 #8
    i believe acceleration in the x direction is 10.679 and the y direction is M*G
     
  10. Mar 4, 2010 #9

    PhanthomJay

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    that's the net centripetal force, in Newtons
    Is there any acceleration in the y direction? And you are confusing acceleration with force, which are related by Newton's 2nd law: F_net = ma.
     
  11. Mar 7, 2010 #10

    rl.bhat

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    Let T1 and T2 are the tensions in upper and lower strings. Let 2θ be the angle between two strings. Then
    T1*cosθ + T2*cosθ = m*v^2/R........(1)
    T1*sinθ = T2*sinθ + mg........(2)
    Solve these two equations to find T1 and T2.
     
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