Determine the tension T CB in cable CB for equilibrium if the hook supports a 46

Question

Determine the tension TCB in cable CB for equilibrium if the hook supports a 46 lb crate and s = 5 ft. Units in pounds are required as part of your answer!

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<< For this question, Free-Body Diagram (FBD) must be submitted through "FBD Dropbox" located below the test link.>>

Determine the tension TcB in cable CB for equilibrium if the hook supports a 46 Ib crate and s -5 ft. Units in pounds are required as part of your answer! 8 ft 2 ft 6 ft 2t 2 ft

Explanation / Answer

>> Writing all the coordinates:

A = (2,0,0)

B = (-2,0,0)

C = (0,5,0)

D = (0,11,8)

>> Considering CA

Vector CA, CA = 2 i - 5 j

>> Magnitude = [22 + 52]1/2 = 5.385

=> Unit Vector along CA = (2 i - 5 j)/5.385 = 0.371 i - 0.928 j

So, Tension in Cable CA = Tca(0.371 i - 0.928 j)

>> Considering CB

Vector CB, CB = - 2 i - 5 j

>> Magnitude = [22 + 52]1/2 = 5.385

=> Unit Vector along CB = (- 2 i - 5 j)/5.385 = - 0.371 i - 0.928 j

So, Tension in Cable CB = Tcb(- 0.371 i - 0.928 j)

>> Considering CD

Vector CD, CD = 6 j + 8 k

>> Magnitude = [62 + 82]1/2 = 10

=> Unit Vector along CD = (6 j + 8 k)/10 = 0.6 j + 0.8 k

So, Tension in Cable CD = Tcd(0.6 j + 0.8 k)

>> As, Weight acting on crate, W = - 46*32.2 k = - 1481.2 k

>> Now, For Equilibrium, Net forces = 0

=> Tca + Tcb + Tcd + W = 0

=> Tca(0.371 i - 0.928 j) + Tcb(- 0.371 i - 0.928 j) + Tcd(0.6 j + 0.8 k) - 1481.2 k = 0

>> Comparing Coefficients,

=> 0.371*Tca - 0.371*Tcb = 0

=> - 0.928*Tca - 0.928*Tcb + 0.6*Tcd = 0

=> 0.8*Tcd = 1481.2

>> Solving above,

=> Tca = 598.545 lb-f

=> Tcb = 598.545 lb-f

=> Tcd = 1851.5 lb-f

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