# Tension on strings holding a weight [closed]

It's a 100% standard problem in any basic physics course, and I cannot solve it.

Here's the scenario:

What is the tension in the strings (T1 and T2)?

My thoughts were (cos/sin rewritten as cos(60°)=½ or cos(45°)=½√2, for example):

From the x components:

cos(45°) T2 + cos(60°) T1 = 0
T1 = −√2 T2

From the y components:

sin(60°) T1 + sin(45°) T2 = m g
½√3 T1 + ½√2 T2 = m g

But using that as the basis to algebraically get the Ts leads to the wrong solution.

PS: Yes that's a homework (this edX course), but I already have the accepted solution (there is a "Show answer" button when one has exhausted ones attempts), and after this I'll drop the course anyway because it just takes too much time and I'm taking it just for fun.

## closed as off-topic by Pranav Hosangadi, John Rennie, Bernhard, Danu, JamalSJan 11 '15 at 13:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Pranav Hosangadi, John Rennie, Bernhard, Danu, JamalS
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• Try $T_1 = \sqrt{2}T_2$ instead of $T_1 = -\sqrt{2}T_2$ – John Rennie Jan 11 '15 at 10:18
• – Kyle Kanos Jan 14 '15 at 21:02

It is important to chose a frame of reference which you stick with. In the reference frame with positive X axis towards the right, and positive Y axis straight up, angles are measured from the positive X axis. The first angle is $\theta_1 = 60°$, and the second is $\theta_2 =180°-45°$. So, From the x components: $T_2 \cos{ (180-45)}+T_1 \cos{60} = 0$ from which $T_1 = \sqrt{2}T_2$

It doesn't make a difference in the Y components as $\sin{\theta} = \sin{(180° - \theta)}.$