+149 I drew fbd, and I thought that the vertical component of the red rope is the only thing pulling up (against gravity) because the horizontal component is, well, horizontal and the other line is also horizontal. B/c is equal to gravity, value is 1.25*9.81 which was 12.263. I then used trig to find the horizontal component of the force, and got 16.796N. Now i thought that this had to be equal to the force of the blue one, idk why but i just thought that they were equal cuz like it was in the middle lol, and i set it equal then used hookes law (16.796 = 1320*x,) then i got x = 0.012 cm, and i tried 0.01 cm and it said it was wrong. Can somone help me solve this problem???
Thanks in advance
If R is the tension in the red rope and B is the tension in the blue rope, then resolving horizontally,
\(R\cos\theta = B\)
and resolving vertically
\(R\sin\theta = 1.25g.\)
Dividing the second equation by the first,
\(\tan\theta=1.25g/B, \text{ so } B=1.25g/\tan\theta.\)
As an alternative, you could take moments about the top lefthand corner.
Call the top left hand corner P and the the point at which the two ropes meet Q, and let the length PQ = x.
Then taking moments about P,
\(1.25g.(x\cos\theta)= B(x\sin\theta) \\ \text{so} \\ B = 1.25g/ \tan\theta.\)
