Finding tension of 2 wires at different angles

AI Thread Summary
The discussion focuses on calculating the tensions T1, T2, and T3 in a system where a 325N cement bag is suspended from three wires at different angles. The user has correctly set up the equilibrium equation for the vertical forces, stating that T3 equals the weight of the cement. However, they are struggling to isolate T1 and T2 using the horizontal forces. Another participant suggests combining terms involving T2 and simplifying the trigonometric functions to solve for T2. The conversation emphasizes the importance of correctly applying equilibrium conditions and simplifying expressions to find the tensions.
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A bag of cement of weight 325N hangs from 3 wires as shown in figure (I've attached it). Two of the wires make angle1=60oand angle2=25o with the horizontal. If the system is in equilibrium, determine the tensions of T1, T2, and T3

If you can't see the image, the T1 and T2 are suspended from a horizontal at the angles specified. T1 on the left, T2 on the right. Where they come together T3 is hanging which is attached to the cement with weight 325N. I hope this is helpful.

So I am assuming that T3=325N
I set up the net forces in the y direction as 0=T1sin60+T2sin25-325
That is where I am stuck. I don't understand how to find the tensions from this. I thought about using the forces in the x direction to solve for either T1 or T2 and then substituting that in for one of the other, however that just does not make sense to me. This is my second semester in physics and the first time I took this class this was one of my biggest problems, I just don't know what to do when I get to this point. Thanks for any and all help, I surely appreciate it.
 
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Hi, rollad.
rollad said:
So I am assuming that T3=325N
I set up the net forces in the y direction as 0=T1sin60+T2sin25-325
Looks good.
That is where I am stuck. I don't understand how to find the tensions from this. I thought about using the forces in the x direction to solve for either T1 or T2 and then substituting that in for one of the other, however that just does not make sense to me.
Actually, that makes a lot of sense to me! What bothers you about it?
 
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This is what I get 0=(T2cos25/cos60)sin60+T2sin25-325

I guess I am just confused on how to get everything equally a single T2
 
rollad said:
This is what I get 0=(T2cos25/cos60)sin60+T2sin25-325

I guess I am just confused on how to get everything equally a single T2

OK. That looks good. The rest is just simplifying the expression. The first two terms on the right can be combined into one term involving T2. Recall that ##ax + bx = (a+b)x##. You can boil all those trig functions down to a single number.
 
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