Find tension in three ropes suspending same mass

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SUMMARY

The discussion focuses on calculating the tensions in three ropes suspending a bird feeder, with Rope1 at a 60-degree angle and Rope2 at a 30-degree angle. The tension in Rope3 is established as T3 = mg, where m is the mass of the feeder and g is the acceleration due to gravity. The correct approach to find the tensions in Rope1 and Rope2 involves applying the equilibrium conditions for forces in both the x-direction and y-direction, leading to two equations: ΣF_x = 0 and ΣF_y = 0.

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A bird feeder is hanging off of a branch by 3 ropes. Rope1 and Rope2 are tied to the branch and both conect to rope3. Rope3 is conected to the birdfeeder. Rope 1 is 60 degrees of the branch and rope2 is 30 degrees.
Symbolicly find the tensions of all of the ropes.

We know that the tension of rope 3 would equal mass times gravity. t3=mg

:confused: BUT, i was wondering if this would be a true statement. t1=66.6% of t3 and t2 would be 33.3% of t3?
I got these numbers just from doing a simple ratio assuming that the angle of rope 1 and 2 are proportional to the tension of rope3, but i do not know how to prove if that is right or wrong. Which is why I am posting this, my teacher had a different answer for what he was looking for, but i was curious just to see if that is a true statement about t1 and t2.
 
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You are correct that the tension in rope 3, T3 = mg. But your reasoning about the other tensions is not correct.

The right way to do it is to apply the condition for equilibrium: Forces in the x-direction and y-direction must add to zero. Apply this condition at the point where the three ropes connect and see what you get.
 
Represent the forces relative to where the ropes meet each other.

\Sigma F_x = 0 = R_2 \cos (30) - R_1 \cos (60)

\Sigma F_y = 0 = R_1 \sin (60) + R_2 \sin (30) - mg

2 equations, 2 variables.
 

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