Help with a tension problem.

shastafarian · Mar 28, 2007

Hey, I'm new, obviously, but I need help with a question, and it should be incredibly simple, I know, I'm probobly just overthinking or misunderstanding a concept.

alright, so a guy is hanging from a rope connected on both sides like \./ where the (.) is the guy. He is seen as a particle. The ropes both have an angle of 10 degrees. I need to find the tension in the ropes. the guy weights 822 N.

Please help!

denverdoc · Mar 28, 2007

Tension still eludes me so i wouldn't be surprised to make a blunder here,
but let me give you a little jump start, then show your work, and the cavalry will save the day.

Lets assume for starters that he is dangling from two separate vertical ropes:

what would the tension in each rope be? That seems easy: 822/2

Now just suppose that the two ropes are at 10 degrees off vertical. The tension then must be greater than the first case. Resolve the tension into x and y components--the x components cancel--a tug of war at a standstill. So only the vertical component of the tension carries the weight. This help at all?

shastafarian · Mar 28, 2007

Yes, as far as understanding what is happening, I figured 822/2 as much, but I guess my problem is finding out how to combine the components to realize the tension on it at 10 degrees. oh yeah, and its 10 degrees from horizontal. that is, 10 degrees down from the ceiling.

Thanks!

denverdoc · Mar 28, 2007

thats where trig helps.
In this case the sine of the angle times the tension is that actually keeping the guy from free fall. there are still two ropes.

shastafarian · Mar 28, 2007

Tsin(10)=w ????

shastafarian · Mar 28, 2007

NM, I got it. . . It was (882/2)/cos(80)

Thanks a lot though!

denverdoc · Mar 28, 2007

come back and spread the word,
(BTW cos80=sin10)