How Do You Calculate Rope Tension with Different Angles in Physics?

[Friendly Hobo]

Homework Statement

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)

Homework Equations

ΣFnet = ma

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.

 

[hotvette]

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

 

[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.

 

[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!

 

[rwisz]

Hello,

I would approach this problem by first understanding the forces at play. The mountain climber is suspended by a rope, which is attached to two cliffs. The forces acting on the climber are her weight (514 N) pulling her down and the tension in the rope pulling her up. Since she is not moving, the net force on her must be zero according to Newton's First Law. This means that the tension in the rope must be equal to her weight.

Next, we need to find the individual tensions in the left and right sides of the rope. To do this, we can use trigonometry. Since the left side of the rope makes a 65* angle with the cliff, we can use the sine function to find the tension in that side of the rope: T_left = 514 N/sin(65*) = 587 N. Similarly, for the right side of the rope, we can use the sine function again to find the tension: T_right = 514 N/sin(10*) = 2966 N.

So to summarize, the tension in the left side of the rope is 587 N and the tension in the right side of the rope is 2966 N. This makes sense because the climber is closer to the left cliff, so the tension on that side of the rope is less.

I hope this helps! Let me know if you have any further questions.