A mountain climber, in the process of crossing. Left side 65.0 degrees and right side 80 degreesbetween two cliffs by a rope, pauses to rest. She weighs 515 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.

To solve this problem, we can use the principles of equilibrium. Since the climber is stationary, both the horizontal and vertical forces must balance. The forces at play are the tensions in the rope and the climber's weight.

Let’s break down the problem step by step.

### Step 1: Define the forces

1. The weight of the climber is \( W = 515 \, \text{N} \) acting vertically downward.
2. The tension in the left side of the rope is \( T_L \), and the angle it makes with the horizontal is \( \theta_L = 65.0^\circ \).
3. The tension in the right side of the rope is \( T_R \), and the angle it makes with the horizontal is \( \theta_R = 80.0^\circ \).

### Step 2: Resolve the forces into components

We will resolve the tension forces into their horizontal and vertical components:

- For the left tension \( T_L \):
  - Horizontal component: \( T_L \cos \theta_L \)
  - Vertical component: \( T_L \sin \theta_L \)

- For the right tension \( T_R \):
  - Horizontal component: \( T_R \cos \theta_R \)
  - Vertical component: \( T_R \sin \theta_R \)

### Step 3: Set up equilibrium equations

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

1. **Vertical equilibrium**: The sum of the vertical forces must equal the climber’s weight:
   \[
   T_L \sin \theta_L + T_R \sin \theta_R = W
   \]
   Substituting \( W = 515 \, \text{N} \):
   \[
   T_L \sin 65^\circ + T_R \sin 80^\circ = 515
   \]

2. **Horizontal equilibrium**: The horizontal components of the tensions must cancel out:
   \[
   T_L \cos \theta_L = T_R \cos \theta_R
   \]
   This equation relates \( T_L \) and \( T_R \).

### Step 4: Solve the system of equations

1. From the horizontal equilibrium equation:
   \[
   T_L \cos 65^\circ = T_R \cos 80^\circ
   \]
   Solving for \( T_L \) in terms of \( T_R \):
   \[
   T_L = \frac{\cos 80^\circ}{\cos 65^\circ} T_R
   \]

2. Substitute this expression for \( T_L \) into the vertical equilibrium equation:
   \[
   \left( \frac{\cos 80^\circ}{\cos 65^\circ} T_R \right) \sin 65^\circ + T_R \sin 80^\circ = 515
   \]

Now we can solve this equation for \( T_R \). After finding \( T_R \), we can substitute back to find \( T_L \).

Let me calculate these values.The tensions in the rope are as follows:

- The tension in the left side of the rope, \( T_L \), is approximately **155.91 N**.
- The tension in the right side of the rope, \( T_R \), is approximately **379.46 N**.

This difference in tension arises because the climber is closer to the left cliff, causing the rope on the right side to bear more weight.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. How would the tensions change if the climber moved closer to the right cliff?
2. What would the tensions be if the angles on both sides were the same?
3. How would the problem change if the climber weighed 600 N instead of 515 N?
4. Can we calculate the exact distance between the climber and the two cliffs using this information?
5. How would the forces change if the climber started to ascend?

**Tip**: When dealing with equilibrium problems, always break forces into components and apply the conditions for both horizontal and vertical force balance.

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 515 N. She is closer to the left cliff than 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. The angles are 65 degrees to the left and 80 degrees to the right.

Solve a statics problem involving a mountain climber resting between two cliffs. Learn how to calculate the tensions in the rope using trigonometry and equilibrium of forces. The climber weighs 515 N, with angles of 65 and 80 degrees on the left and right sides. Ideal for Grades 11-12 students studying statics and trigonometry.

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