Question

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 535 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 to the left and to the right of the mountain climber.

Answer 1

Answer:

$T_{2}=\frac{535}{sin(\theta _{2})+cos(\theta _{2})tan(\theta _{1})}$

$T_{1}=\frac{535}{sin(\theta _{1})+cos(\theta _{1})tan(\theta _{2})}$

Explanation:

Consider the general position of the climber as shown in the figure

we have for equilibrium along x and y   directions we have

$T_{1}cos(\theta _{1})=T_{2}cos(\theta _{2})\\\\T_{1}sin(\theta _{1})+T_{2}sin(\theta _{2})=W$

upon solving we get

$T_{1}sin(\theta _{1})+\frac{T_{1}cos(\theta_{1})}{cos(\theta _{2})}sin(\theta _{2})=W\\\\T_{1}=\frac{W}{sin(\theta _{1})+cos(\theta _{1})tan(\theta _{2})}$

Similarly

$T_{2}=\frac{W}{sin(\theta _{2})+cos(\theta _{2})tan(\theta _{1})}$