How Do You Calculate the Weight of Signal C in a Suspended Cable System?

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SUMMARY

The discussion focuses on calculating the weight of signal C in a suspended cable system, given that signal B weighs 300N. The correct weight of signal C is determined to be 97.7N. The user attempted to solve the problem using free body diagrams and equations of equilibrium, specifically analyzing the tensions in the cables connecting the signals. The calculations for tension T_bc were initially found to be 251N, leading to discrepancies in the final weight calculation for signal C.

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  • Understanding of free body diagrams in physics
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Homework Statement


Two traffic signals are temporarily suspended from a cable as shown. Knowing that the signal at B weighs 300N. Determine the weight at signal C.

Homework Equations


Red arrows show lengths of cables and their components.
Answer is : W_c = 97.7N

The Attempt at a Solution


I started by drawing two free body diagrams. One for particle B and the other for particle C. They both share the cable BC, hence, they share the same tension, T_bc, right? So, I tried to solve for T_bc with the free body diagram of particle B resulting in the sum of the forces along x: -T_ab(3.6/3.9) + T_bc(3.4/3.42) =0
(Sigma = addition of forces) SigmaF_y: T_ab(1.5/3.9) + T_bc(.4/3.42) - 300 = 0

So, solving for T_bc I get 251 N. I then follow the same steps as above for particle C.
I get Sigma F_x: T_bc(3.4/3.42) + T_cd(2.4/2.5) = 0
Sigma F_y: T_bc(.4/3.42) + T_cd(.7/2.5) - W_c = 0
so far? when I plug in what I got for T_bc to the second eqns. I get a different answer for w_c than what I'm supposed to get. please help
 

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the free body diagram of particle B resulting in the sum of the forces along x: -T_ab(3.6/3.9) + T_bc(3.4/3.42) =0
simaF_y: T_ab(1.5/3.9) + T_bc(.4/3.42) - 300 = 0
So, solving for T_bc I get 251 N. I then follow the same steps as above for particle C.

Check this calculation. I am getting differenct answer.
 

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