# Homework Help: Tension HW problem.

1. Feb 11, 2006

### Demianu27

alright, i'm having trouble finding a formula to answer this problem.

a 58N weight is being suspended by 2 cables. the cable on the left side is horizontal, while the other cable is at a 60' angle.

What is the tension of the right side cable? (T1)

What is the tension of the left side cable, the horizontal one? (T2)

I started off by trying to find the sum of all the forces, but I'm already stuck.
I have;

{Fx = T2 - T1cos60 = 0
and
{Fy = T1sin60 - 58N = 0

is this right so far, or do i have the cos and the sin switched?

2. Feb 11, 2006

### Astronuc

Staff Emeritus
That depends on whether the 60° is with respect to horizontal or vertical.

If the angle is with respect to the horizontal, then your equations are correct.

3. Feb 12, 2006

### Demianu27

http://img123.imageshack.us/img123/2411/118dj.th.png [Broken]

that's the problem.

i've done this;

{Fx = T2 - T1cos60 = 0
and
{Fy = T1sin60 - 58N = 0

then to find T2 i thought i could do this;

T1sin60 = 58N
to
T1=(sin^-1)60 + 58N

but the answer is apparently wrong... so i must not have done it right.

Last edited by a moderator: May 2, 2017
4. Feb 12, 2006

### Astronuc

Staff Emeritus
T1sin60 = 58N is correct.

T1=(sin^-1)60 + 58N is not correct.

T1 * sin 60° = 58N => T1 = 58N / (sin 60°)

Try T2 = T1 * cos 60°

5. Feb 13, 2006

### Demianu27

ahhh i totally forgot i could just divide both sides by it

insted i was doing something else i learned recently.... damnit so many formulas!!! lol

thank you a lot.