# 2D Particle Equilibrium

1. Jan 21, 2009

### jonnyboy

1. The problem statement, all variables and given/known data
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.

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

3. 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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Last edited: Jan 22, 2009
2. Jan 21, 2009

### rl.bhat

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.