Questions: A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 585 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. TL=mathbfi TR=mathbfi

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 585 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.

TL=mathbfi
TR=mathbfi

Transcript text: A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 585 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. \[ \begin{array}{l} T_{L}=\mathbf{i} \\ T_{R}=\mathbf{i} \end{array} \] $\square$ $\square$ eTextbook and Media

Solution

Solution Steps

Step 1: Identify the forces acting on the mountain climber

The forces acting on the mountain climber are:

The gravitational force (weight) $ W = 585 \, \text{N} $ acting downward.
The tension in the left rope $ T_L $ acting at an angle of $ 65^\circ $ to the horizontal.
The tension in the right rope $ T_R $ acting at an angle of $ 80^\circ $ to the horizontal.

Step 2: Resolve the tensions into horizontal and vertical components

For the left rope:

Horizontal component: $ T_L \cos(65^\circ) $
Vertical component: $ T_L \sin(65^\circ) $

For the right rope:

Horizontal component: $ T_R \cos(80^\circ) $
Vertical component: $ T_R \sin(80^\circ) $

Step 3: Apply equilibrium conditions

Since the climber is at rest, the sum of the forces in both the horizontal and vertical directions must be zero.

Horizontal equilibrium:

\[ T_L \cos(65^\circ) = T_R \cos(80^\circ) \]

Vertical equilibrium:

\[ T_L \sin(65^\circ) + T_R \sin(80^\circ) = 585 \, \text{N} \]

Step 4: Solve the system of equations

First, solve the horizontal equilibrium equation for one of the tensions: \[ T_L \cos(65^\circ) = T_R \cos(80^\circ) \] \[ T_L = T_R \frac{\cos(80^\circ)}{\cos(65^\circ)} \]

Next, substitute this expression into the vertical equilibrium equation: \[ T_R \frac{\cos(80^\circ)}{\cos(65^\circ)} \sin(65^\circ) + T_R \sin(80^\circ) = 585 \, \text{N} \]

Simplify and solve for $ T_R $: \[ T_R \left( \frac{\cos(80^\circ) \sin(65^\circ)}{\cos(65^\circ)} + \sin(80^\circ) \right) = 585 \, \text{N} \] \[ T_R \left( \frac{\cos(80^\circ) \sin(65^\circ)}{\cos(65^\circ)} + \sin(80^\circ) \right) = 585 \, \text{N} \]

Calculate the numerical values: \[ \cos(65^\circ) \approx 0.4226 \] \[ \sin(65^\circ) \approx 0.9063 \] \[ \cos(80^\circ) \approx 0.1736 \] \[ \sin(80^\circ) \approx 0.9848 \]

\[ T_R \left( \frac{0.1736 \times 0.9063}{0.4226} + 0.9848 \right) = 585 \, \text{N} \] \[ T_R \left( \frac{0.1573}{0.4226} + 0.9848 \right) = 585 \, \text{N} \] \[ T_R \left( 0.3722 + 0.9848 \right) = 585 \, \text{N} \] \[ T_R \left( 1.357 \right) = 585 \, \text{N} \] \[ T_R = \frac{585 \, \text{N}}{1.357} \] \[ T_R \approx 431.3 \, \text{N} \]

Finally, solve for $ T_L $: \[ T_L = T_R \frac{\cos(80^\circ)}{\cos(65^\circ)} \] \[ T_L = 431.3 \, \text{N} \times \frac{0.1736}{0.4226} \] \[ T_L \approx 177.3 \, \text{N} \]