# Homework Help: Two tensions on one rope

1. Sep 28, 2011

### Friendly Hobo

1. The problem statement, all variables and given/known data

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tensions in the rope (a) to the left and (b) to the right of the mountain climber.

The rope coming from the left is 65* below the edge of the cliff. The rope from the right is 80* below the edge of the cliff.

(* denotes degree symbol)
2. Relevant equations

ΣFnet = ma

3. The attempt at a solution

I'm really not sure how to go about this. The question is part of an online homework assignment, and I only have one attempt left for full credit. What I have done (obviously wrong):

Drew a free body diagram of two right triangles with the angles of the left one being 65*, 25*, 90* and the angles of the right one being 10*, 80*, 90*. I thought the answer was gotten by taking weight: 541N/sin(Θ) but have now found out this assumption was wrong. I'm thinking I am supposed to break the tension forces into their components now, but I'm not quite sure what to do with the components themselves. Any help would be appreciated.

2. Sep 28, 2011

### hotvette

This is a statics problem. Do you know the two basic rules of statics problems? Also, you might want to post the diagram.

3. Sep 28, 2011

### Friendly Hobo

Im not sure if I know the two basic rules or not to be honest. I'd love to learn them though so I can hopefully get through these sorts of problems in the future without assistance. I've attached the diagram that is given in the online assignment.

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• ###### ch04p_102.gif
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4. Sep 28, 2011

### Friendly Hobo

I think I may have figured it out, but I want to get someone to check it for me before I submit. After breaking the two tensions forces, which I'll call T1 and T2 from here, I was able to work out a solution.

ΣFx = T1cos(25) - T2cos(10) = ma
a = 0 m/s^2 so

ΣFx = T1cos(25) - T2cos(10) = 0
T1cos(25) = T2cos(10)
T1 = [cos(10)/cos(25)]T2

ΣFy = T1sin(25) - T2sin(10) - w = ma
a = 0 m/s^2
ΣFy = T1sin(25) + T2sin(10) -w = 0

T1sin(25) + T2sin(10) = w

[cos(10)/cos(25)]T2 * sin(25) + T2sin(10) = w
(0.9848/0.9063) * 0.4226T2 + 0.1736T2 = 514N
T2 = 812.171N

T1 = [cos(10)/cos(25)]T2
T1 = (0.9848/0.9063)(812.171N)
T1 = 882.571N

Does this all look correct? It's my only shot at full credit and I don't want math blunders or poor assumptions on my part to mess it up!