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Math Help - Assistance Needed in Calculating Cable Tension

  1. #1
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    Talking Assistance Needed in Calculating Cable Tension

    I'm having so much trouble with this and it's due in 3 days.

    Here is what it looks like;




    Question:
    For W=175 pounds, computer tension force in the cables (both sides) for L=11 and X=(1,2,3,4,5). Which X position creates the greatest tension? Where is the best place to hang the weight?

    I'm not asking for a solution, I just wonder if anyone can give me an equation I can use.
    Last edited by Lance2497; September 25th 2008 at 05:58 PM.
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  2. #2
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    the system is in equilibrium ... \sum F = 0

    let \theta = angle T_1 makes with the horizontal
    \phi = angle T_2 makes with the horizontal

    T_1 \cos{\theta} = T_2 \cos{\phi}

    T_1 \sin{\theta} + T_2 \sin{\phi} = 175

    you'll have to determine the values of the trig ratios from your given dimensions.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    the system is in equilibrium ... \sum F = 0

    let \theta = angle T_1 makes with the horizontal
    \phi = angle T_2 makes with the horizontal

    T_1 \cos{\theta} = T_2 \cos{\phi}

    T_1 \sin{\theta} + T_2 \sin{\phi} = 175

    you'll have to determine the values of the trig ratios from your given dimensions.


    How did you get 175 as the angle measure?

    I apologize for the error, but T1 and T2 refer to the suspended cable pieces, not angles.
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  4. #4
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    I'm even more sorry that you have no idea what you are solving for ...

    T_1 and T_2 refer to tensions (forces) in the cable, or didn't you know that?

    175 lbs is a force, not an angle.
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  5. #5
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    I appreciate your clarity on my confusion, as well as you snarky remarks. Truly makes everything go much smoother when one in need of assistance is given grief for his lack of understanding.

    Thank you
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  6. #6
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    Quote Originally Posted by skeeter View Post
    the system is in equilibrium ... \sum F = 0

    let \theta = angle T_1 makes with the horizontal
    \phi = angle T_2 makes with the horizontal

    T_1 \cos{\theta} = T_2 \cos{\phi}

    T_1 \sin{\theta} + T_2 \sin{\phi} = 175

    you'll have to determine the values of the trig ratios from your given dimensions.


    i dont understain why theta 1 and theta 2 are equal??? they dont look to be equal...looking at the horizontal line they dont look the same...could you please explain me that??? (how are the angles equal
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  7. #7
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    I did not say the angles were equal.
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  8. #8
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    Quote Originally Posted by skeeter View Post
    I did not say the angles were equal.
    can you explain me what did you do there?
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  9. #9
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    He didn't use the angles, he used the fact that the triangle formed by the horizontal and vertical forces has the same angles and is similar to the triangle formed by the roof, vertical, and cable. The horizontal force on each side, divided by the Tension in the rope on that side, is equal to the horizontal distance (x on the left, 10-x on the right) divided by the length of the hypotenuse ( \sqrt{x^2+ H^2} on the left, \sqrt{(10-x)^2+ H^2} on the right). Those must be equal in order that the weight must not move from side to side. That was the first of the two equations skeeter gave.

    The vertical force on each side, divided by the tension in the rope on that side, is equal to the vertical distance, H, divided by the length of the hypotenuse ( \sqrt{x^2+ H^2} on the left, \sqrt{(10-x)^2+ H^2} on the right). The total of those two must be the weight itself. That is the second equation skeeter gave.
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    He didn't use the angles, he used the fact that the triangle formed by the horizontal and vertical forces has the same angles and is similar to the triangle formed by the roof, vertical, and cable. The horizontal force on each side, divided by the Tension in the rope on that side, is equal to the horizontal distance (x on the left, 10-x on the right) divided by the length of the hypotenuse ( \sqrt{x^2+ H^2} on the left, \sqrt{(10-x)^2+ H^2} on the right). Those must be equal in order that the weight must not move from side to side. That was the first of the two equations skeeter gave.

    The vertical force on each side, divided by the tension in the rope on that side, is equal to the vertical distance, H, divided by the length of the hypotenuse ( \sqrt{x^2+ H^2} on the left, \sqrt{(10-x)^2+ H^2} on the right). The total of those two must be the weight itself. That is the second equation skeeter gave.

    i agree that
    T_1 \sin{\theta} = T_2 \sin{\phi}

    but why is
    T_1 \sin{\theta} + T_2 \sin{\phi} = 175 is like saying that H+H=175 and i think that just 1 H is=175
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  11. #11
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    Quote Originally Posted by juanfe_zodiac View Post
    i agree that
    T_1 \sin{\theta} = T_2 \sin{\phi}

    the above equation is incorrect

    but why is
    T_1 \sin{\theta} + T_2 \sin{\phi} = 175 is like saying that H+H=175 and i think that just 1 H is=175
    the horizontal components of T_1 and T_2 are equal ...

    T_1 \cos{\theta} = T_2 \cos{\phi}
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