To find the tensions in the rope on both sides of the mountain climber, we can use the equilibrium conditions. Since the climber is at rest, the sum of forces in both the vertical and horizontal directions must equal zero.

Let:

• $T_L$ be the tension in the rope on the left side.
• $T_R$ be the tension in the rope on the right side.
• $\theta_L = 65^\circ$ is the angle of the left side of the rope with the horizontal.
• $\theta_R = 80^\circ$ is the angle of the right side of the rope with the horizontal.
• $W = 550 \, \text{N}$ is the weight of the mountain climber.

Step 1: Vertical force balance

In the vertical direction, the sum of the vertical components of the tensions must balance the weight of the climber:

$T_L \sin(\theta_L) + T_R \sin(\theta_R) = W$

Substituting the known values:

$T_L \sin(65^\circ) + T_R \sin(80^\circ) = 550$

Step 2: Horizontal force balance

In the horizontal direction, the horizontal components of the tensions must cancel each other out:

$T_L \cos(\theta_L) = T_R \cos(\theta_R)$

Substituting the known angles:

$T_L \cos(65^\circ) = T_R \cos(80^\circ)$

Step 3: Solving the system of equations

Now we solve these two equations simultaneously:

1. $T_L \sin(65^\circ) + T_R \sin(80^\circ) = 550$
2. $T_L \cos(65^\circ) = T_R \cos(80^\circ)$

From equation 2, solve for $T_L$:

$T_L = T_R \frac{\cos(80^\circ)}{\cos(65^\circ)}$

Substitute this expression for $T_L$ into equation 1:

$T_R \frac{\cos(80^\circ)}{\cos(65^\circ)} \sin(65^\circ) + T_R \sin(80^\circ) = 550$

Factor out $T_R$:

$T_R \left( \frac{\cos(80^\circ) \sin(65^\circ)}{\cos(65^\circ)} + \sin(80^\circ) \right) = 550$

Now compute the trigonometric values:

• $\sin(65^\circ) \approx 0.9063$
• $\cos(65^\circ) \approx 0.4226$
• $\sin(80^\circ) \approx 0.9848$
• $\cos(80^\circ) \approx 0.1736$

Substitute these into the equation:

$T_R \left( \frac{0.1736 \times 0.9063}{0.4226} + 0.9848 \right) = 550$

Simplify:

$T_R \left( \frac{0.1573}{0.4226} + 0.9848 \right) = 550$ $T_R \left( 0.3722 + 0.9848 \right) = 550$ $T_R \times 1.357 = 550$

Solve for $T_R$:

$T_R = \frac{550}{1.357} \approx 405.5 \, \text{N}$

Step 4: Find $T_L$

Now that we know $T_R$, substitute it back into the expression for $T_L$:

$T_L = 405.5 \times \frac{\cos(80^\circ)}{\cos(65^\circ)} = 405.5 \times \frac{0.1736}{0.4226}$ $T_L \approx 405.5 \times 0.4108 \approx 166.5 \, \text{N}$

Final Answer:

• The tension in the rope to the left of the climber is approximately $T_L \approx 166.5 \, \text{N}$.
• The tension in the rope to the right of the climber is approximately $T_R \approx 405.5 \, \text{N}$.

Do you want more details, or have any questions?

Here are some related questions to explore:

1. How does changing the angle on either side affect the tensions in the rope?
2. What happens to the tension if the climber’s weight increases?
3. What if the climber moves to the exact middle between the cliffs—how would that change the tensions?
4. How would the tensions change if the angle on the right side was less than on the left?
5. How would the solution differ if the angles were exactly the same?

Tip: In equilibrium problems, always remember that the net force in both horizontal and vertical directions must be zero!