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Math Help - calulating tension in a string

  1. #1
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    calulating tension in a string

    [b]1. The problem

    Calculate the tension in a 3m string attached to a 3kg bob that is moving in horizontal circles of a 0.6m radius

    [b]2. Relevant equations

    The only formula i have been able to find is tension = mass1 w2 with w2 being the circumfance but this does not that in to account gravity or the length of the string

    [b]3. The attempt at a solution

    i know the circumfance is .6 x 2 x pie which is 3.7692 but should this figure be timesed by gravity? ive spent 2 hours looking for other formulas but have had no luck. Excuse the less than great english. Im not so much after an answer but a way of getting there

    Thanks
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  2. #2
    MHF Contributor Unknown008's Avatar
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    There is no need to find the circumference.

    Make a simple sketch.

    Code:
            /|
           / |
          /  |
         /   |
      3 /    |
       /     |
      /      |
     /_______|
    O   0.6
    You'll find that the angle the string does with the horizontal to be:

    \theta = cos^{-1}(\dfrac{0.6}{3})

    Then, resolving vertically, the vertical component of the tension is given by:

    Tsin\theta

    Since the bob is in equilibrium vertically, the resultant force = 0.

    The downwards force acting on the bob is given by:

    F = mg = 3 \times 9.8 = 29.4\ N

    Hence, Tsin\theta = 29.4

    Solve for T.
    Last edited by Unknown008; October 3rd 2010 at 10:19 AM. Reason: Corrected typo on the cos ratio, see Earboth's post
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by Unknown008 View Post
    There is no need to find the circumference.

    ...
    You'll find that the angle the string does with the horizontal to be:

    \theta = cos^{-1}(\dfrac{3}{0.6})

    ...
    I don't want to pick at you but in my opinion the equation should read:

    \theta = cos^{-1}\left(\dfrac{0.6}{3} \right)
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Woops, I didn't spot that typo. Thanks!
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  5. #5
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    thanks for your help guys but im still strugglng. to get the engle do i need to times the result of .6 dived by 3 by cos. If so i cant work out how? and you say tsin0 - 29.4n. does that mean that t = 29.4 dived by the sin of the angle?

    Sorry to be a pain but im still unsure how to get the angle and how to work out the tension. i understand basic trig

    Thanks
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Maybe it's time for a more realistic sketch...



    Does that help to understand?

    To find the angle, take the inverse of the trigonometric ratio:

    cos\theta = \dfrac{0.6}{3}

    \theta = cos^{1-}\left(\dfrac{0.6}{3}\right)

    In your calculator, you type in 0.6/3, then you type in 2nd function, cos, ANS. This should give you 78.5 degrees
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  7. #7
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    I got that bit thanks but i just didnt get how i got the tension using the force and angle? so if t sin0 = 29.4
    Sorry to be pain
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  8. #8
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    It helps to take a little time thinking about what forces are involved - all of them. Read up on Newton's third law!
    You know that the bob is moving in a circle. The only thing holding it up is the string, T, and the only thing that pulls it down is gravity, F=mg. Now identify how these forces act on the bob. Gravity only has a single component, pointing downward. However, since the string is attached to the bob at an angle from the vertical, it has both a vertical and a horizontal component. Thus the force T must be greater than mg, since only part of T acts to counteract mg. But since you do know that mg is needed to keep it up, you also know the vertical component of T. And since you know the angle at which T is acting, you know that

    Tsin(theta) = mg
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  9. #9
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    So if Tsin = mg does that mean that T = mg/sin ? and what unit are we measuring tension in?
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Yes, and Tension is in Newtons.
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