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The traffic light shown in Figure is supported by a system of cables. Points A a

ID: 1825643 • Letter: T

Question

The traffic light shown in Figure is supported by a system of cables. Points A and B are in a vertical plane which is parallel to x-z plane. Point C is in y-z plane. The traffic light has a mass of 75 kg. Determine the following: (a). Position vectors for OA, OB and OC (b). Write equilibrium equations in the x, y and z directions (c). Find the tensions in cables OA, OB and OC (show all your workings that you do to find these) Image is at http://www.imagesup.net/?di=3135702364213 Question has been answered here: http://www.chegg.com/homework-help/questions-and-answers/the-access-door-is-held-in-the-200-open-position-by-the-cable-ab-as-shown-in-figure-5-the--q3496608 , but there is no explination or workings about how the answer for question a) was found? Also the image posted is very hard to read....

Explanation / Answer

cordinate of A = (4,-8,5)
B =(6,-8,5)
C =(0,8,5)
O =(0,0,0)

(a) OA =(4-0)i -(8-0)j +(5-0)k = 4i -8j +5k
OB = 6i -8j +5k
OC = 8j +5k

(b) let tension in cable A = Ta
                =Ta* unit vector in OA direction
                  =Ta (4i -8j +5k)/105


tension in cable B = Tb*(6i -8j +5k)/105
tension in cable C =Tc*(8j +5k)/89

x componet of tension = 4Ta/105 +6Tb/105

y component of tension = -8Ta/105 -8Tb/105 +5Tc/89

z component of tension =5Ta/105 +5Tb/105 +5Tc/89

weight of trafic light = -75g k

total force in x direction = x componet of tension = 4Ta/105 +6Tb/105

total force in y direction =y componet of tension = -8Ta/105 -8Tb/105 +5Tc/89

total force in z direction = z component of tension + weight of trafic light

                                      =5Ta/105 +5Tb/105 +5Tc/89 -75g

(c) total force in x direction is zero because there is no motion in x direction

hence 4Ta/105 +6Tb/105 =0

             Ta =-3Tb/2

total force in x direction is zero because there is no motion in y direction

hence    -8Ta/105 -8Tb/105 +5Tc/89= 0

           or   -12Tb/105 -8Tb/105 +5Tc/89=0

           or    Tc = 4Tb89/105

total force in x direction is zero because there is no motion in z direction

hence 5Ta/105 +5Tb/105 +5Tc/89 -75g =0

           -7.5Tb/105 +5Tb/105 +20Tb/105-75g = 0

              17.5Tb = 75g105

                    Tb = 75g105/17.5 =

                   Ta =-2Tb/3 = -50g105/17.5

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