Part A If the tension developed in either cable AB or AC can not exceeded 1000 l

Question

Part A

If the tension developed in either cable AB or AC can not exceeded 1000 lb, determine the maximum tension that can be developed in cable AD when it is tightened by the turnbuckle.

Also, what is the force developed along the antenna tower at point A?

FAD = Part A If the tension developed in either cable AB or AC can not exceeded 1000 lb, determine the maximum tension that can be developed in cable AD when it is tightened by the turnbuckle. FAD = Also, what is the force developed along the antenna tower at point A?

Explanation / Answer

AB = sqrt(10^2 + 15^2 + 30^2) = 35 ft

AC = sqrt(10^2 + 15^2 + 30^2) = 35 ft

AD = sqrt(12.5^2 + 30^2) = 32.5 ft
Tab_x = Tab*(10/35)

Tab_y = Tab*(-15/35)

Tab_z = Tab*(-30/35)

Tac_x = Tac*(-15/35)

Tac_y = Tac*(-10/35)

Tac_z = Tac*(-30/35)

Tad_x = 0

Tad_y = Tad*12.5/32.5

Tad_z = Tad*(-30/32.5)

For equilibrium, Tab_x + Tac_x = 0

Tab*10/35 - Tac*15/35 = 0

Tab = 1.5*Tac

Hence, Tab > Tac. Hence, as the load increases AB will break before AC.

Putting Tab = 1000 lb we get Tac = 1000/1.5 = 667 lb

For equilibrium of y-direction, Tab_y + Tac_y + Tad_y = 0

1000*(-15/35) + 667*(-10/35) + Tad*12.5/32.5 = 0

Tad = 1609.8 lb

Force developed along the antenna = Tab_z + Tac_z + Tad_z

= 1000*(-30/35) + 667*(-30/35) + 1609.8*(-30/32.5)

= -2914.8 lb

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