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Math Help - Tension in a string

  1. #1
    Senior Member topher0805's Avatar
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    Tension in a string



    I need to find the tension in the horizontal string.

    Mass of the bar:

    m_1 = 4.7 kg

    Hanging mass:

    m_2 = 3.4 kg

    Length of the bar:

    l_1 = 2.3 m

    Angle between bar and horizontal:

    \theta = 42^{\circ}

    Length between hanging mass and string-bar connection:

    l_2 = 0.8 m

    Thanks in advance for the help!
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  2. #2
    Senior Member topher0805's Avatar
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    So far all I've got is that the tension in the string is going to be equal to the torque on the bar, which should just be the force on the bar in the perpendicular direction.

    So, I can find the perpendicular components of the force due to gravity on the bar and the force due to the hanging mass.

    This is where I'm a bit confused. At this point, I would use the formula:

    |{\tau}| = rf

    But I'm not exactly sure what to plug in and where.
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  3. #3
    Senior Member topher0805's Avatar
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    Ok I am positive that I've got it this time, but my answer sheet has a different number. Can anyone check over my work to see if I made any mistakes?

    \tau = rf\sin{\theta}

    Torque from bar itself:

    \tau_1 = 1.15\cdot9.8\cdot4.7\cdot\sin{48}

    \tau_1 = 39.36  N\cdot m

    Torque from hanging mass:

    \tau_2 = 1.5\cdot9.8\cdot3.4\cdot\sin{48}

    \tau_2 = 37.14  N\cdot m

    Add them together to get the total torque:

    \tau_T = \tau_1 + \tau_2

    \tau_T = 76.51 N\cdot m

    Divide by the length of the bar to get the force at the end point:

    <br />
F = \frac{\tau_T}{r}

    F = 33.26 N

    That's what I got for my final answer, but the answer sheet is telling me that it is 49.7 N.

    Any suggestions?
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  4. #4
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    skeeter's Avatar
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    \sum \tau = 0

    torques CCW = torques CW

    (3.4g)(1.5)\sin(48) + (4.7g)(1.15)\sin(48) = T(2.3)\sin(42)

    \frac{(3.4g)(1.5)\sin(48) + (4.7g)(1.15)\sin(48)}{(2.3)\sin(42)} = T

    T = 49.7 \, N
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