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5. A tank is full of water. Find the work required to pump the water out of the spout. Water weighs 62.5 lb/ft3³ 12 ft → T 6 ft ↓ 10 ft ft →

5. A tank is full of water. Find the work required to pump the water out of the spout. Water weighs 62.5 lb/ft3³ 12 ft → T 6 ft ↓ 10 ft ft →

Calculus: Early Transcendentals
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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**Topic: Work Required to Pump Water from a Tank** **Problem Statement:** 5. A tank is full of water. Find the work required to pump the water out of the spout. Water weighs 62.5 lb/ft³. **Diagram Explanation:** The diagram illustrates a rectangular tank with dimensions: - Length: 12 feet - Width: 10 feet - Height: 6 feet Each dimension is labeled clearly, with arrows indicating the respective measurements. **Objective:** Determine the work required to pump all the water out of this tank through a spout located at the top. **Given Data:** - Water Weight: 62.5 lb/ft³ - Tank Dimensions: 12 ft (length) x 10 ft (width) x 6 ft (height) **Steps to Find the Solution:** 1. **Calculate Volume of Water in the Tank:** Volume = Length x Width x Height Volume = 12 ft x 10 ft x 6 ft = 720 ft³ 2. **Calculate Weight of Water in the Tank:** Weight = Volume x Weight per unit volume Weight = 720 ft³ x 62.5 lb/ft³ = 45,000 lb 3. **Determine the Distance Each Layer of Water Must Be Pumped:** Since water is pumped out of the spout at the top, the distance each layer must be pumped will vary from 0 feet (top layer) to 6 feet (bottom layer). 4. **Use Calculus to Integrate Work Done Over the Height of the Tank:** Work required (W) is given by integrating the weight of each infinitesimal slice of water (dW) times its distance (y) from the spout: W = ∫ (from 0 to 6) (Volume of slice x Weight per unit volume x Distance to spout) dy Specifically: - Volume of slice = area of base x thickness (dy) = 12 ft x 10 ft x dy - Weight of slice = Volume of slice x Weight per unit volume = 12 x 10 x 62.5 dy = 7500 dy - Distance to spout varies from 0 to 6 feet Therefore: W = ∫ (from 0 to 6) 7500y dy
Transcribed Image Text:**Topic: Work Required to Pump Water from a Tank** **Problem Statement:** 5. A tank is full of water. Find the work required to pump the water out of the spout. Water weighs 62.5 lb/ft³. **Diagram Explanation:** The diagram illustrates a rectangular tank with dimensions: - Length: 12 feet - Width: 10 feet - Height: 6 feet Each dimension is labeled clearly, with arrows indicating the respective measurements. **Objective:** Determine the work required to pump all the water out of this tank through a spout located at the top. **Given Data:** - Water Weight: 62.5 lb/ft³ - Tank Dimensions: 12 ft (length) x 10 ft (width) x 6 ft (height) **Steps to Find the Solution:** 1. **Calculate Volume of Water in the Tank:** Volume = Length x Width x Height Volume = 12 ft x 10 ft x 6 ft = 720 ft³ 2. **Calculate Weight of Water in the Tank:** Weight = Volume x Weight per unit volume Weight = 720 ft³ x 62.5 lb/ft³ = 45,000 lb 3. **Determine the Distance Each Layer of Water Must Be Pumped:** Since water is pumped out of the spout at the top, the distance each layer must be pumped will vary from 0 feet (top layer) to 6 feet (bottom layer). 4. **Use Calculus to Integrate Work Done Over the Height of the Tank:** Work required (W) is given by integrating the weight of each infinitesimal slice of water (dW) times its distance (y) from the spout: W = ∫ (from 0 to 6) (Volume of slice x Weight per unit volume x Distance to spout) dy Specifically: - Volume of slice = area of base x thickness (dy) = 12 ft x 10 ft x dy - Weight of slice = Volume of slice x Weight per unit volume = 12 x 10 x 62.5 dy = 7500 dy - Distance to spout varies from 0 to 6 feet Therefore: W = ∫ (from 0 to 6) 7500y dy
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