Tension in swing chains supporting boy, girl, and swing

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Schaus
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Homework Statement


A 355.0N boy and a 300.0.0N girl sit on a 50.5N porch swing that is 1.5m long. The swing is supported by a chain on each end. What is the tension in each chain when the boy sits .35m from one end and the girl sits .45m from the other? ( ans: boys side = 390 N, girls side = 320 N)

Homework Equations


τcw=τccw

The Attempt at a Solution


I don't have much practice with torque questions especially not ones that are hanging.
Στ= 0 = (T2)(1.5m) - (355N)(0.35m) - (50.5N)(0.75m) - (300N)(1.05m)
Στ = 0 = (T2)(1.5m) - 124.25Nm - 37.875Nm - 135Nm
(T2)(1.5m) = 477.125Nm → divide both sides by 1.5m
318N = T2
I'm trying to figure out how to get the tension on T1 now.
 
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It looks like you have positioned the girl sitting somewhat on the boy.
 
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Awesome! Didn't notice that, thanks! I can't figure out the tension on the boys side though.
 
Well, if you figured out one tension, then you know all of the forces except one - the other tension. And since the swing is in equilibrium (not accelerating), you know that the sum of the forces have to equal 0, true?
 
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Yes. The T1=T2. 320N = 320N? If I added his extra 55N it would only equal 375N
 
I haven't worked this problem yet so I don't know the answer. But, you can't just say that T1 = T2. Neither can you add his weight to one tension to get the other tension. The sum of ALL of the forces have to equal 0. Or, how I prefer to think about it: The upward forces have to equal the downward forces. But you have to make sure that you include all of them.
 
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Ah I see! Thanks a lot! I've gotten my answer again!
 
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Schaus said:
Yes. The T1=T2. 320N = 320N?

Huh? They can't be equal.

Have you drawn a force diagram? What are the 5 (4 known and 1 unknown) vertical forces acting on the swing? What must they add up to?

EDIT: Sorry cross posted
 
Yes I drew a force diagram and I realize that statement is wrong. I added up all my weights going down and minused the tension I found in T2 to get my last tension.