A metal chain ladder, 20 meters long, has a linear density of 3 Newtons per meter, and hangs down from the roof along the side of a building 30 meters tall. How much work is needed to pull the chain ladder to the top of the building? How much work is done in pulling up the first half of the chain ladder?

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Chapter63: Volumes Of Pyramids And Cones
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My answers were 600 and 150 respectively. If they are wrong, please show me how to do it correctly. If they are correct, please only write “correct.” Thank you!
### Problem: Work Required to Pull Up a Chain Ladder

#### Given:
- A metal chain ladder, 20 meters long
- Linear density: 3 Newtons per meter
- Hanging from the roof of a building that is 30 meters tall

#### Questions:
1. How much work is needed to pull the chain ladder to the top of the building?
2. How much work is done in pulling up the first half of the chain ladder?

---

### Solution

To solve this problem, we need to understand the concept of work and how it applies to lifting a distributed load like a chain ladder.

#### 1. Calculating Total Work

##### Work when lifting the entire chain ladder:

1. **Determine the weight per unit length of the chain ladder:**
   \[
   \lambda = 3 \, \text{N/m}
   \]

2. **Total length of the chain ladder:**
   \[
   L = 20 \, \text{m}
   \]

3. **Divide the chain ladder into infinitesimal segments:**
   Each tiny segment \( dy \) at a height \( y \) above the ground.

4. **Weight of an infinitesimal segment:**
   \[
   dW = \lambda \cdot dy = 3 \, \text{N/m} \cdot dy
   \]

5. **Integrated work to pull each infinitesimal segment to the top of the building:**
   \[
   dW = \lambda \cdot y \cdot dy = 3y \, dy
   \]
   
   The limits are from \(0\) to \(20\) since the height ranges from ground level to the top of the chain ladder.

6. **Integrate to find the total work:**
   \[
   W = \int_{0}^{20} 3y \, dy
   \]

7. **Calculate the integral:**
   \[
   W = \left[ \frac{3y^2}{2} \right]_{0}^{20} = \frac{3}{2} \left[ 20^2 - 0^2 \right] = \frac{3}{2} \cdot 400 = 600 \, \text{J}
   \]

#### 2. Calculating Work for the First Half

##### Work when lifting the first half of the
Transcribed Image Text:### Problem: Work Required to Pull Up a Chain Ladder #### Given: - A metal chain ladder, 20 meters long - Linear density: 3 Newtons per meter - Hanging from the roof of a building that is 30 meters tall #### Questions: 1. How much work is needed to pull the chain ladder to the top of the building? 2. How much work is done in pulling up the first half of the chain ladder? --- ### Solution To solve this problem, we need to understand the concept of work and how it applies to lifting a distributed load like a chain ladder. #### 1. Calculating Total Work ##### Work when lifting the entire chain ladder: 1. **Determine the weight per unit length of the chain ladder:** \[ \lambda = 3 \, \text{N/m} \] 2. **Total length of the chain ladder:** \[ L = 20 \, \text{m} \] 3. **Divide the chain ladder into infinitesimal segments:** Each tiny segment \( dy \) at a height \( y \) above the ground. 4. **Weight of an infinitesimal segment:** \[ dW = \lambda \cdot dy = 3 \, \text{N/m} \cdot dy \] 5. **Integrated work to pull each infinitesimal segment to the top of the building:** \[ dW = \lambda \cdot y \cdot dy = 3y \, dy \] The limits are from \(0\) to \(20\) since the height ranges from ground level to the top of the chain ladder. 6. **Integrate to find the total work:** \[ W = \int_{0}^{20} 3y \, dy \] 7. **Calculate the integral:** \[ W = \left[ \frac{3y^2}{2} \right]_{0}^{20} = \frac{3}{2} \left[ 20^2 - 0^2 \right] = \frac{3}{2} \cdot 400 = 600 \, \text{J} \] #### 2. Calculating Work for the First Half ##### Work when lifting the first half of the
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