# Tension of rope problem

SohCahToa

Sin=Opp/Hyp

sohcahtoa

then hyp=565/cos(65) ?

no i know that but since the 565 is force downward and the 65* is at one end, isnt the 565 the opposite side?

check out my pic

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i cant... But i just realized what u are saying.. Ok i get it

so the triangles are on the outside of the ropes not inside... ok i understand

565 is the force upward! it is the Normal force (Fn). the sumation of all forces on her is zero (she is not falling ie. accelerating down, or up) so Fn=Fg and as you said Fg=565N=Fn

so hyp=565/cos(65) right?

so let T1 = hyp1

then cos(80)=565N/T1

let T2 = hyp2

then cos(65)=565N/T2

right
I wish the pictures would work, as that would have made this so much easier, but hopefully that helped you

so T1, 80*, =3253.7N

T2, 65*, =1336.9N

does this look right, seems like quite a bit of tension

i tried those answers but they were incorrect.

no- I think we did something wrong. T1+T2=565N so you could say

T1=565-T2

ok, then how do i find T1 or T2

I messed you up more then I helped, sorry man. I need my notes/book/calculator and they are all in my car.. sorry to take you down the wrong track like that

ok, if both lengths of rope were equal

T1=Fg/2

T2=Fg/2

they are not but this gives me an idea...

ok...

Doc Al
Mentor
There are three forces acting on the person:
- Tension (T1) from left rope (acting at the angle of the rope)
- Tension (T2) from right rope (acting at the angle of the rope)
- Weight (W) acting down (which is given)

Since she's in equilibrium:
- The sum of the vertical force components must equal 0
- The sum of the horizontal force components must equal 0

That will give you two equations, which you can solve to find the two unknowns: T1 & T2.