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Math Help - Tension & Magnitude

  1. #1
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    Tension & Magnitude

    A tetherball weighing 1 lb is pulled outward from the pole by a horizontal force u until the rope makes an angle of theta degrees with the pole

    1. Determine the resulting tension in the rope and magnitude of u when theta = 30 degrees
    2. Write the tension T in the rope and the magnitude of u as functions of theta
    3. Determine the domains of the functions
    4. Compare T and ||u|| as theta increases

    I know that ||u|| = sqrt a^2+b^2 but I just don't know what the values of a and b are.

    Thanks in advance!!!

    Jennifer
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  2. #2
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    Quote Originally Posted by jmoran View Post
    A tetherball weighing 1 lb is pulled outward from the pole by a horizontal force u until the rope makes an angle of theta degrees with the pole

    1. Determine the resulting tension in the rope and magnitude of u when theta = 30 degrees
    2. Write the tension T in the rope and the magnitude of u as functions of theta
    3. Determine the domains of the functions
    4. Compare T and ||u|| as theta increases

    I know that ||u|| = sqrt a^2+b^2 but I just don't know what the values of a and b are.

    Thanks in advance!!!

    Jennifer
    Draw a sketch.

    You deal with a right triangle with the hypotenuse \vec t and the legs \vec w and \vec u.

    You know: \cos(\theta) = \dfrac{|\vec w|}{|\vec t|}

    Since you know the magnitude of w and the angle \theta you can determine the magnitude of t

    \tan(\theta) = \dfrac{|\vec u|}{|\vec w|}

    For a specific value of \theta you can calculate the magnitude of u.
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  3. #3
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    I don't know if I understand.

    I am assuming the magnitude of w is = 1??

    If that is the case then in order to get the value of t (which is the tension) when angle is 30 degrees, I would use

    t = 1/cos(30)

    So now I have the t = 1.15.

    Next -

    Is this the formula to determine the magnitude of u??



    If so, again assuming the magnitude of w is = 1 and the angle is 30 degrees, would it be

    i * tan(30) = .58

    so |u| = .58 ??

    I think I am missing something but I'm not sure what....

    Thanks,

    Jennifer
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  4. #4
    Member pflo's Avatar
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    The fact that w's magnitude is 1 is given in the problem (the ball is 1 pound). Look again at the diagram posted by earboth. The only assumption is that we are disregarding the weight of the rope itself.

    Other than that, your work is correct. It is all right-triangle ratios (SOH, CAH, TOA).
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  5. #5
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    So if the question was

    1. Determine the resulting tension in the rope and magnitude of u when theta = 30 degrees

    The answer would be

    tension = 1/cos(30) = 1.15
    |u| = 1 * tan(30) = .58

    Is that right?

    Jennifer
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  6. #6
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    Quote Originally Posted by jmoran View Post
    So if the question was

    1. Determine the resulting tension in the rope and magnitude of u when theta = 30 degrees

    The answer would be

    tension = 1/cos(30) = 1.15
    |u| = 1 * tan(30) = .58

    Is that right?

    Jennifer
    Yes

    (Keep in mind that the forces are measured in pounds (very unusual) and if you have to use Newtons you must transform these values)
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  7. #7
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    OMG !!!

    I actually understand.....

    Thank you thank you thank you!!!

    Jennifer
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  8. #8
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    Re: Tension & Magnitude

    ?
    Last edited by Wulfman; January 31st 2013 at 02:19 PM.
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  9. #9
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    Re: Tension & Magnitude

    Quote Originally Posted by Wulfman View Post
    ?
    If we are to help you then we need more information about what you need help on than "?"

    -Dan
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  10. #10
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    Re: Tension & Magnitude

    At angle \theta, tension T, in pounds, the horizontal force on the ball is [tex]Tsin(\theta)[/itex] and that must be u. The vertical force is T cos(\theta) and that must be the balls weight, 1 lb. Solve the equations T sin(\theta)= u and T cos(\theta)= 1 for T and \theta.

    Notice that if you square each equation and add you get T^2 sin^2(\theta)+ T^2 cos^2(\theta)= T^2(sin^2(\theta)+ cos^2(\theta))= T^2= u^2+ 1. If you divide the first equation by the second, you get \frac{T sin(\theta)}{T cos(\theta)}= tan(\theta)= frac{u}{1}= u.
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