Tension of climber with known angles and weight.?

Question

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. she weighs 510 N. As the drawing shows, she is close to the left cliff than to the right cliff, with the results 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.

|.\.............../.|
|..\............/...|
|...\......../......|
|....\..../.........|
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(ignore the dots i'm trying to give a rough estimate of the picture)
the left top angle is 65 degrees
the right top angle is 80 degrees

I have been trying this question for a while now. and i guessed that -T1sin65 + T2sin80 = 0
+T1cos65 - T2cos80 = 0

and i further went on to finally get T1 = 2048.3 N
and T2 = 1885.2 N

but i know these are wrong. am I doing it partially corrected? How do I do this problem, its been since high school since I did it. Any help is appreciated. thanks.

Answer 1

Assuming your climber is not moving, the vertical forces acting on her have to add up to zero. Otherwise, she'd be accelerating (a(V)) and moving up or down.

Thus W - (T1cos(65) + T2cos(80)) = 0 = ma(V) = f(V) the net vertical force.

Similarly, since she's not moving sideways:

T1sin(65) - T2sin(80) = 0 = ma(H) = f(H) in the horizontal.

You now have two unknowns (T1 and T2) and two equations. So you can solve for T1 and T2.

You were astute to know you were doing something wrong. Clearly the tensions cannot be greater than the weight of the climber when there are no other forces at work.

BY THE WAY, your climber needs to go on a diet. She weighs over 1,600 pounds (= (510/9.8) X 32.2) on Earth's surface. What has she been eating?

Answer 2

Nor do I and, i think, my TC colleagues recognize the position to start up. What you've given us is inadequate files to charm to out the considered necessary stress diagrams to remedy the mission. case in factor, the climber, is she going up a vertical cliff, an prone hill, what? I easily have authentic worry visualizing her on the top of a 40 deg re the horizontal rope at the same time as mountaineering a vertical cliff. usually, the rope will be way steeper than 40 deg, like 80 deg or better, case in factor. If she's going up an incline, then a number of her weight is supported by skill of the hill. that could want to evidently decrease the stress on the rope. the position is the rope connected to the climber re her center of gravity? As a corollary, what variety of torque is advanced, if any? So "the position to start up" is with resubmitting this mission note for note from its source; so no longer some thing is omitted.

Answer 3

Assuming that the angles are with the upper horizontal

T1cos65-T2cos80 = 0

T1sin65 + T2sin80 = 510 (The climber's weight)

Now solve.