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

AI Thread Summary
To calculate the weight of signal C in a suspended cable system, the weight of signal B is known to be 300N. The tension in the cable connecting signals B and C is denoted as T_bc, which is derived from free body diagrams for both signals. The calculations yield T_bc as 251N, but when applying this value to determine the weight of signal C, the result differs from the expected 97.7N. The user seeks assistance to verify the calculations and resolve the discrepancy in the weight of signal C. Accurate tension and force balance equations are crucial for solving this problem correctly.
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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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