How Do You Solve a Vector Tension Problem with Three Ropes?

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To solve the vector tension problem with three ropes supporting weights of 20 N and 10 N, two equations are established based on the forces in the vertical direction. The first equation combines the tensions T1, T2, and T3 with the total weight, while the second focuses on the horizontal components. The discussion emphasizes the importance of separating forces into x and y components for clarity. Participants suggest using force diagrams at each junction to visualize the tensions better. Ultimately, the goal is to derive a third equation to solve for the unknown tensions effectively.
elfmotat
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Homework Statement

2 weights of 20 N and 10 N are hung from 3 ropes. Find the tension of T1, T2, and T3:
vector.jpg


The attempt at a solution

If I designate one of the points as the origin, I get:

T_1sin45+T_2sin0+T_3sin30=30 N

and

T_2cos0+T_3cos30-T_1sin45=0

I'm having trouble finding a third equation... if those two are even right in the first place.
 
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Try and stick to separating the forces into x and y components, at least mentally.

If you think about the forces that only act in the vertical direction for each weight separately you should be able to pretty much read off the values for T1 and T3, just use one of those to get you on to the shared tension.

You have four equations at your disposal, both axes for both weights.
 
Kabbotta said:
Try and stick to separating the forces into x and y components, at least mentally.

If you think about the forces that only act in the vertical direction for each weight separately you should be able to pretty much read off the values for T1 and T3, just use one of those to get you on to the shared tension.

You have four equations at your disposal, both axes for both weights.

Sorry... I don't understand what you're saying. I don't know how to separate it into components with a system with more than one weight.
 
Sorry I thought that might be the problem,

At each junction of wires you can draw a little force diagram like this.

You can create four equations by focusing on only one axis at a time.

First one, this is only the force that acts parallel to the y-axis w/ the left knot at it's center.

\sum F_{y} = T_{1}Sin(45) -20 N. = 0
 

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