1. Nov 11, 2007

### princessfrost

Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff. Suppose that an 82.5 kg climber, who is 1.54 m tall and has a center of gravity 1.3 m from his feet, rappels down a vertical cliff with his body raised 28.9 degree above the horizontal. He holds the rope 1.46 m from his feet, and it makes a 24.2 degree angle with the cliff face.
http://session.masteringphysics.com/problemAsset/1039093/3/YF-11-33.jpg

a) What tension does his rope need to support?

I found the tension to be equal to - mg(1.3) sin(28.9) + T(1.3) sin(180 - 24.2-28.9)=0 and then i solve for T. T= (mg(1.3)sin(28.9))/(1.3sin(180-24.2-28.9)). I plugged in the numbers and found the answer to be 488.61 N but it's wrong. Can someone please help me? Thank you!

2. Nov 11, 2007

### Staff: Mentor

You are using the wrong angles. When using $\tau = rF\sin\theta$, the angle is the angle between the position vector (r) and the force.

3. Nov 11, 2007

### princessfrost

ok i got it. thank you sooo much for your help! i really appreciated it!

:)