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  1. #1
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    questions in physics help

    questions in physics help-kmb88039.jpg

    1 ) A traffic light weighing 100 N hangs from a vertical cable to two other cables that are fastened to a support.
    The upper cabels make angle of 37 and 53 with the horizontal . Find the tension in each of the three cables

    -------------------------------------------
    Here I now first must find x component and y component

    How I find the
    x component and y component in every Q I do mistake

    and why you say -T cos 37 why ( -) here ??

    plese Help me
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  2. #2
    Senior Member BAdhi's Avatar
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    here is my solution

    for the equilibrium of the traffic light,

    T_3 = 100N

    considering the equilibrium of the joining point of cables,

    T_2cos37 = T_1cos53

    T_1 = \frac{T_2cos37}{cos53}

    on Y direction,

    T_1sin37+T_2sin53=T_3

    you have three equations

    I cannot find where you go a minus sign
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  3. #3
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    How you get the point of cables ?
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  4. #4
    Senior Member BAdhi's Avatar
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    since all the points on the system is at a static equillibrium, you can apply \sum F = 0

    so applying that to the point where three cables meet you can get an equation with three tensions.
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  5. #5
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    ok the system at equillibrium

    then

    F = 0

    Fx = T2 cos 53 + T1 cos 37 = 0
    Fy = T2 sin 35+ T1 sin 37 = 0

    After that what I do ?
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  6. #6
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    e^(i*pi)'s Avatar
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    Since you have two equations and two unknowns solve simultaneously
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  7. #7
    Senior Member BAdhi's Avatar
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    Fy is incorrect reffer above post
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  8. #8
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    Fx = T2 cos 53 + T1 cos 37 = 0
    Fy = T2 sin 53+ T1 sin 37 - 100 = 0

    sorry but now what will I do
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  9. #9
    Senior Member BAdhi's Avatar
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    infact Fx=> T2cos53 -T1cos 37 =0

    take T2 (or T1) from 1st equation and substitute it into the second equation.
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  10. #10
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    Ok Now I take T2

    T2 = T1cos37/cos53 = 1.327T1

    Then I substitute this value ( 1.327 T1 ) second equation
    T1 sin37 +(1.327T1)sin53 - 100 = 0
    T1[sin37 +(1.327)sin53 ] = 100
    T1 = 100/sin37 +(1.327)sin53 = 167
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  11. #11
    Senior Member BAdhi's Avatar
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    correct...
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  12. #12
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    ...

    but the answer is wrong not as ny book
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by r-soy View Post
    Ok Now I take T2

    T2 = T1cos37/cos53 = 1.327T1

    Then I substitute this value ( 1.327 T1 ) second equation
    T1 sin37 +(1.327T1)sin53 - 100 = 0
    T1[sin37 +(1.327)sin53 ] = 100
    T1 = 100/sin37 +(1.327)sin53 = 167
    T1 = 100 / [sin(37) +(1.327) sin(53)]

    Redo your arithmetic, and make sure the calculator is in the correct mode and you put the brackets in the right place..

    CB
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  14. #14
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    Yes now ok
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