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Math Help - Boom vector problem

  1. #1
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    Boom vector problem

    Hi can anybody help me in this question that i have???

    A boom OB carries a load F of manetude 500N and is supported by cables BC as shown in the diagram where the dimensions of the system are given. Determine the tensions in the cables so that equilibrium is maintained and the resultant force at piont B as shown in drawing attached.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by sarnekat View Post
    Hi can anybody help me in this question that i have???

    A boom OB carries a load F of manetude 500N and is supported by cables BC as shown in the diagram where the dimensions of the system are given. Determine the tensions in the cables so that equilibrium is maintained and the resultant force at piont B as shown in drawing attached.

    In addition to the tension T in cable BC, you want to know also the tension S in cable BD, I think.
    In your sketch, I understand that the foot of the boom, point O, is at (0,0,0) of the (x,y,z) axes.
    Also, the points B,C,D are all in the same z = 75cm plane....and this plane is horizontal.
    And, point B is at (100cm, 0 , 75cm).

    In the vertical (x,z) plane....for the compression U in the boom:
    U is in the hypotenuse of the right triangle whose legs are 75cm vertical and 100cm horizontal.

    At point B,
    Break the U into its vertcal and horizontal components Uz and Ux respectively.
    For equilibrium in the z-direction,
    Uz = F = 500N.

    By proportion, Uz / 75cm = Ux / 100cm
    So, Ux = (100 / 75)Uz = (100 / 75)(500N) = 666.66667 N -------------------------------------**
    This Ux will be supported by the combined tensions S and T in the two cables.

    ----------------------------------
    In the horizontal z = 75cm plane:

    For the tension T in the BC cable,
    T is in the hypotenuse of the right triangle whose legs are 50cm and 100cm....so, the hypotenuse is sqrt(50^2 +100^2) =
    111.8034 cm.
    Break the T into its Tx and Ty components,
    By proportion,
    T / 111.8034 = Tx / 100 .......Tx = (100 / 111.8034)T = (0.89443)T
    T / 111.8034 = Ty / 50 ......Ty = (50 / 111.8034)T = (0.44721)T

    For the tension S in the BD cable,
    S is in the hypotenuse of the right triangle whose legs are 75cm and 100cm....so, the hypotenuse is sqrt(75^2 +100^2) = 125cm.
    Break the S into its Sx and Sy components,
    By proportion,
    S / 125 = Sx / 100 .......Sx = (100 / 125)S = (0.8)S
    S / 125 = Sy / 75 ......Sy = (75 / 125)S = (0.6)S

    For equilibrium in the y-direction at point B,
    Ty = Sy
    (0.44721)T = (0.6)S
    T = (0.6 / 0.44721)S = (1.34165)S ------------------**

    For eqilibrium in the x-direction,
    Ux = Tx +Sx
    666.66667 N = (0.89443)T +((0.8)S
    Substituting (1.34165)S for T,
    666.66667 = (0.89443)(1.34165 S) +(0.8)S
    S = 666.6667 / 2.0
    S = 333.33333 newtons ---------------------------------answer.

    And, T = (1.34165)(333.33333) = 447.21667 newtons --------answer.
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  3. #3
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    Jul 2008
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    Answers to boom questions

    I showed this to my lecturer and the answers do not match the answers that he gave was FBC= 538.8N, FBD = 388N the resultant force is= 1033N and the force at point OB = -666.5 N and the angles are alpha = 180 degrees Beta = 90 degrees and gamma = 90 degrees i understand the way that shown but is there another way of calculating using unit vectors i,j and k??
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