How Do You Calculate Tensions T1 and T2 in a Two-Cable System?

[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?

 

[Astronuc]

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.

 

[Demianu27]

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

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.

 

[Astronuc]

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°

 

[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.