Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Work Calculation for Lifting a Chain**
**Problem:**
A 20-foot chain with a density of 4 lb/ft hangs off the roof of a building. How much work is required to pull the chain to the roof?
**Explanation:**
Consider the given problem: we have a uniform chain hanging from a building, and we need to calculate the work required to pull the entire chain to the roof.
### Given Data:
- Length of the chain: 20 feet
- Density of the chain: 4 pounds per foot (lb/ft)
### Steps to Solve:
1. **Determine the Force:**
The force required is based on the weight of the chain per foot. Since the chain's density is uniform:
\[
\text{Weight per unit length} = 4 \text{ lb/ft}
\]
2. **Set Up the Work Integral:**
Work (\(W\)) is defined as the integral of force (\(F\)) times distance (\(d\)):
\[
W = \int_{0}^{L} F(x) \, dx
\]
Where \(L\) is the length of the chain (20 feet) and \(x\) represents an incremental length of the chain being lifted.
3. **Integrate the Force over the Distance:**
Since the weight is linearly distributed, let \(x\) be the distance from the bottom of the chain:
\[
F(x) = 4 \, \text{lb/ft} \times (20 - x) \, \text{feet}
\]
The integral becomes:
\[
W = \int_{0}^{20} 4 \times (20 - x) \, dx
\]
4. **Calculate the Integral:**
Break down the calculation:
\[
W = 4 \int_{0}^{20} (20 - x) \, dx
\]
Perform the integral:
\[
W = 4 \left[ 20x - \frac{x^2}{2} \right]_{0}^{20}
\]
Evaluate the integral at the bounds:
\[
W = 4 \left[ (20 \times 20) - \frac{20^2}{2} \right]
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffddfa066-ae14-4710-b68e-f1412b1153a3%2Fe40f5265-f1f9-4d00-8b1e-db162b1252bb%2F6pbvhjf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Work Calculation for Lifting a Chain**
**Problem:**
A 20-foot chain with a density of 4 lb/ft hangs off the roof of a building. How much work is required to pull the chain to the roof?
**Explanation:**
Consider the given problem: we have a uniform chain hanging from a building, and we need to calculate the work required to pull the entire chain to the roof.
### Given Data:
- Length of the chain: 20 feet
- Density of the chain: 4 pounds per foot (lb/ft)
### Steps to Solve:
1. **Determine the Force:**
The force required is based on the weight of the chain per foot. Since the chain's density is uniform:
\[
\text{Weight per unit length} = 4 \text{ lb/ft}
\]
2. **Set Up the Work Integral:**
Work (\(W\)) is defined as the integral of force (\(F\)) times distance (\(d\)):
\[
W = \int_{0}^{L} F(x) \, dx
\]
Where \(L\) is the length of the chain (20 feet) and \(x\) represents an incremental length of the chain being lifted.
3. **Integrate the Force over the Distance:**
Since the weight is linearly distributed, let \(x\) be the distance from the bottom of the chain:
\[
F(x) = 4 \, \text{lb/ft} \times (20 - x) \, \text{feet}
\]
The integral becomes:
\[
W = \int_{0}^{20} 4 \times (20 - x) \, dx
\]
4. **Calculate the Integral:**
Break down the calculation:
\[
W = 4 \int_{0}^{20} (20 - x) \, dx
\]
Perform the integral:
\[
W = 4 \left[ 20x - \frac{x^2}{2} \right]_{0}^{20}
\]
Evaluate the integral at the bounds:
\[
W = 4 \left[ (20 \times 20) - \frac{20^2}{2} \right]
\]
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