Recent content by jarmen

ok...

ok, then how do i find T1 or T2

i tried those answers but they were incorrect.

so T1, 80*, =3253.7N T2, 65*, =1336.9N does this look right, seems like quite a bit of tension

right. so, cos=adj/hyp so hyp=565/cos(65) right?

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

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

no i know that but since the 565 is force downward and the 65* is at one end, isn't the 565 the opposite side?

so, cos(65)=adj/hyp then hyp=565/cos(65) ?

i thought 565 was the opposite?

so for the left it would be... sin(65) (565) right would be sin(80) (565) correct?

565 Newtons

Yes.., the outside angles

close, the 80* is the angle on the outside of DB, 65* is outside DA A--------------c--------------------B 65*\ ...\ --------D

hold on that didnt post correctly ok the ...'s are empty space, needed it to make space. the \\\\\'s are the rope being pulled 65 degrees the ---'s are rope being pulled 80 degrees sorry, best i can come up with