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**Homework:10.4 HW - Vectors**

**Question 39,** **10.4.91**

Part 1 of 2

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Part 1

A tightrope walker located at a certain point deflects the rope as indicated in the figure. If the weight of the tightrope walker is 

170170 

​pounds, how much tension is in each part of the​ rope?

170 pounds170 pounds

4.5 degrees4.5°

3.6 degrees3.6°

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Part 1

The tension in the left end of the rope is about 

enter your response here 

pounds.

​(Do not round until the final answer. Then round to one decimal place as​ needed.)

The tension in the left end of the rope is about(Do not round until the final answer. Then round to one decimal place as needed.)

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To determine the tension in each side of the rope, we can use vector components. Let’s solve for the tensions in each part, assuming they are represented by \( T_1 \) (left side) and \( T_2 \) (right side) and that the angles with the horizontal are \( \theta_1 = 4.5^\circ \) for the left and \( \theta_2 = 3.6^\circ \) for the right.

Since the walker is stationary, the vertical and horizontal components of the tension forces must balance.

### Step 1: Setting Up the Equations
1. **Vertical Forces**: The sum of the vertical components of \( T_1 \) and \( T_2 \) must equal the weight of the walker (170 pounds):
   \[
   T_1 \sin(4.5^\circ) + T_2 \sin(3.6^\circ) = 170
   \]

2. **Horizontal Forces**: The horizontal components of \( T_1 \) and \( T_2 \) must balance each other out since the walker is not moving sideways:
   \[
   T_1 \cos(4.5^\circ) = T_2 \cos(3.6^\circ)
   \]

### Step 2: Solving for \( T_1 \) and \( T_2 \)
1. From the horizontal forces equation, solve for \( T_2 \) in terms of \( T_1 \):
   \[
   T_2 = T_1 \frac{\cos(4.5^\circ)}{\cos(3.6^\circ)}
   \]

2. Substitute \( T_2 \) into the vertical forces equation:
   \[
   T_1 \sin(4.5^\circ) + T_1 \frac{\cos(4.5^\circ)}{\cos(3.6^\circ)} \sin(3.6^\circ) = 170
   \]

3. Now, solve this equation for \( T_1 \), then use the result to find \( T_2 \).

Would you like me to carry out the calculations for the final values of \( T_1 \) and \( T_2 \)? 

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Here are related questions for further practice:
1. What are the horizontal and vertical components of each tension force individually?
2. How would the tension change if the angle on one side of the rope increased?
3. What would happen to the tensions if the weight of the tightrope walker changed?
4. Can you calculate the tensions if the angle values were reversed?
5. How would the solution differ if there were friction on the rope?

**Tip:** Always check if vector components are balanced in both vertical and horizontal directions when solving equilibrium problems.

A tightrope walker located at a certain point deflects the rope as indicated in the figure. If the weight of the tightrope walker is 170 pounds, how much tension is in each part of the rope? The angles of deflection are 4.5 degrees and 3.6 degrees, respectively.

Solve a tension problem involving a tightrope walker weighing 170 pounds. Use vector components and equilibrium to determine the tension in each part of the rope, given angles of 4.5 degrees and 3.6 degrees. This problem covers key concepts in trigonometry and force equilibrium, suitable for Grades 10-12.

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