Recent content by Sdawg1969

ok, if both lengths of rope were equal T1=Fg/2 T2=Fg/2 they are not but this gives me an idea...

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

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

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

so let T1 = hyp1 then cos(80)=565N/T1 let T2 = hyp2 then cos(65)=565N/T2

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

check out my pic

SohCahToa Sin=Opp/Hyp Cos=Adj/Hyp Tan=Opp/Adj sohcahtoa

not quite- cos\Theta=adjacent/hyp, 565N is the adjacent, so...

right, so you have an angle, and side (Newtons). the rest should be simple

you don't need to figure out the vertical component of force, you should know that. What is the vertical component of force in Newtons?

got it, ok --------------- |65\.../80|

so are the angles ----------------- \ 65*...|...80*/

does the problem look like this? A --------- c ----------------- B ---------- D AB would be the rope, D would be the climber, 80* would be the angle between Dc and DB, and 65* would be the angle between Dc and DA?

no, it does not. but depending on where the angles are, you are going to use sin or cos to figure out your 2 tensions