# Hanging chandelier

1. Feb 5, 2006

### nick727kcin

**** nvm**** i accidentally gave up. im actually emailing my prof right now to see if i can undo it

why must they make this so difficult?

here is the problem:

t1= mg/[cos(90-theta1)+ sin(90- theta1)]

this is definately right. the only problem is that i have to include theta 2 in my formula. this has stumped me

Last edited: Feb 5, 2006
2. Feb 5, 2006

### grief

is it a right triangle, because if it's a right triangle, you know that 90 -theta1 is just theta2

3. Feb 5, 2006

### finchie_88

I got this when I did it.
$$T_1 = \frac{mg - T_2sin\theta_2}{sin\theta_1}$$
Surely $$cos(90 - \theta_1)$$ is the same as $$sin(\theta_1)$$?

Edit: Dam it, beaten by a minute

4. Feb 5, 2006

### nick727kcin

o. well i thought it was right

5. Feb 5, 2006

### nick727kcin

i do not believe that it is a right triangle, but it could be i guess

this is what led me to believe that was right:

http://www.astro.sunysb.edu/iyers/home/Midterm1_Sol1.pdf

look at problem number 4 on page 3

6. Feb 5, 2006

### nick727kcin

i get what you guys are staying, but that wouldnt include theta2 without using T2

7. Feb 5, 2006

### finchie_88

The expression you gave in your original post was re-arranged wrong, but the correct one is the same as the one I posted earlier. To get rid of theata2, resolve vertically and substitute in.

8. Feb 5, 2006

### nick727kcin

but doesnt it say not to include t2 in it?

9. Feb 5, 2006

### nick727kcin

i did it your way and it said:

10. Feb 5, 2006

### grief

in the link that you gave, theta2 was 45. The answer includes a tangent of theta2, but since tangent of 45 is just 1, it does not appear in the answer there.

11. Feb 5, 2006

### nick727kcin

guys forget it, i accidentally gave up on this question