Question

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.

Answer 1

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$