Question

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 555 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 tension in the rope to the left of the mountain climber.

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The tension in the rope on the left of the mountain climber is $T_a = 1106 \ N$

Explanation:

From the question we are told that

The weight of the mountain climber is m = 555 N

Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as  

       $T_a* cos 65 -555 + T_b * cos(85) = 0$

Here  $T_a, T_b$ are the tension of the rope on the left and on the right hand side

 So

    $0.423T_a + 0.0871T_b = 555$

=>   $0.0871T_b = 555 - 0.423T_a$

=>   $T_b = \frac{555 - 0.423T_a}{0.0871}$

Generally from the diagram , the total amount of force acting on the rope along the horizontal  axis at equilibrium is mathematically represented as

      $T_a* sin 65 - T_b * sin(85) = 0$

=>     $0.9063T_a - 0.9962T_b = 0$

=>     $0.9063T_a = 0.9962T_b$

=>     $0.9063T_a = 0.9962[\frac{555 - 0.423T_a}{0.0871}]$

=>     $0.9063T_a = [\frac{552.891 - 0.421T_a}{0.0871}]$

=>    $0.0789T_a = [552.891 - 0.421T_a$

=>    $0.4999T_a = 552.891$

=>      $T_a = 1106 \ N$