To solve this problem, we need to analyze the forces acting on the mountain climber at equilibrium. Since the climber is at rest, the net force on her must be zero. We'll use Newton's Second Law and break down the forces into horizontal and vertical components.

Given:

• The climber’s weight: $W = 550 \, \text{N}$
• Angle with the left side of the rope: $\theta_L = 65.0^\circ$
• Angle with the right side of the rope: $\theta_R = 80.0^\circ$

Let:

• $T_L$ be the tension in the left side of the rope.
• $T_R$ be the tension in the right side of the rope.

Step 1: Set up the force equations

At equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.

1. Vertical equilibrium (sum of vertical forces): $T_L \sin \theta_L + T_R \sin \theta_R = W$ $T_L \sin 65^\circ + T_R \sin 80^\circ = 550$

2. Horizontal equilibrium (sum of horizontal forces): The horizontal components of the tension forces must cancel each other: $T_L \cos \theta_L = T_R \cos \theta_R$ $T_L \cos 65^\circ = T_R \cos 80^\circ$

Step 2: Solve the system of equations

We now have two equations:

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

From the second equation: $T_L = \frac{T_R \cos 80^\circ}{\cos 65^\circ}$

Step 3: Substitute into the vertical equation

Substitute the expression for $T_L$ into the first equation: $\frac{T_R \cos 80^\circ}{\cos 65^\circ} \sin 65^\circ + T_R \sin 80^\circ = 550$ Simplify this expression and solve for $T_R$: $T_R \left( \frac{\cos 80^\circ \sin 65^\circ}{\cos 65^\circ} + \sin 80^\circ \right) = 550$

Numerically:

• $\sin 65^\circ \approx 0.906$
• $\cos 65^\circ \approx 0.4226$
• $\sin 80^\circ \approx 0.985$
• $\cos 80^\circ \approx 0.174$

Now plug these values into the equation: $T_R \left( \frac{0.174 \times 0.906}{0.4226} + 0.985 \right) = 550$ $T_R \left( 0.373 + 0.985 \right) = 550$ $T_R \times 1.358 = 550$ $T_R = \frac{550}{1.358} \approx 405 \, \text{N}$

Step 4: Solve for $T_L$

Now that we have $T_R$, use the horizontal equilibrium equation to find $T_L$: $T_L = \frac{T_R \cos 80^\circ}{\cos 65^\circ} = \frac{405 \times 0.174}{0.4226} \approx 167 \, \text{N}$

Final Answer:

• The tension in the left side of the rope: $T_L \approx 167 \, \text{N}$
• The tension in the right side of the rope: $T_R \approx 405 \, \text{N}$

Would you like more details or have any questions?

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Here are 5 related questions for further exploration:

1. What would happen to the tensions if the climber were exactly in the middle?
2. How would the tension change if the angle on the left side decreased to 45 degrees?
3. What if the climber's weight increased to 700 N? How would the tensions change?
4. Can you calculate the horizontal force components for both sides?
5. How would a slight movement of the climber to the right affect the balance of forces?

Tip: Always check the sum of forces in both vertical and horizontal directions to ensure equilibrium.