Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
3. On a very rainy day, Nick drags a box of old books to the bookstore to sell. The box moves through the puddles at U = 10 ft/min. For all calculations use an approximation that the box follows the figure. Calculate the force required to keep the box moving at U. Hints: obtain drag coefficient by calculating the Reynolds number, consider the projected area as the bottom plus the two long sides. • Determine the power (in watt) required to overcome the skin friction drag B=2ft L=3 ft Water line D = 2 in

3. On a very rainy day, Nick drags a box of old books to the bookstore to sell. The box moves through the puddles at U = 10 ft/min. For all calculations use an approximation that the box follows the figure. Calculate the force required to keep the box moving at U. Hints: obtain drag coefficient by calculating the Reynolds number, consider the projected area as the bottom plus the two long sides. • Determine the power (in watt) required to overcome the skin friction drag B=2ft L=3 ft Water line D = 2 in

icon
Related questions
Question
## Problem Statement ### Scenario: On a very rainy day, Nick drags a box of old books to the bookstore to sell. The box moves through the puddles at a velocity of \( U = 10 \, \text{ft/min} \). For all calculations, use an approximation that the box follows the given dimensions. ### Task: 1. **Calculate the force required to keep the box moving at \( U \).** - **Hints:** Obtain the drag coefficient by calculating the Reynolds number. Consider the projected area as the bottom plus the two long sides. 2. **Determine the power (in watts) required to overcome the skin friction drag.** ### Diagram: The diagram shows a rectangular box with the following dimensions: - Length (\( L \)) = 3 feet - Breadth (\( B \)) = 2 feet - Depth (\( D \)) = 2 inches (converted to feet: \( \frac{2}{12} = \frac{1}{6} \, \text{ft} \)) The box is submerged partially in water, up to a "water line" shown on the sides, with velocity \( U \) indicated as the direction of motion. ### Considerations for Calculations: - The projected area to take into account includes the bottom area and two side areas submerged in water. - Use appropriate formulas to calculate the Reynolds number to determine the drag coefficient. - Use physics principles to determine the force and power needed.
Transcribed Image Text:## Problem Statement ### Scenario: On a very rainy day, Nick drags a box of old books to the bookstore to sell. The box moves through the puddles at a velocity of \( U = 10 \, \text{ft/min} \). For all calculations, use an approximation that the box follows the given dimensions. ### Task: 1. **Calculate the force required to keep the box moving at \( U \).** - **Hints:** Obtain the drag coefficient by calculating the Reynolds number. Consider the projected area as the bottom plus the two long sides. 2. **Determine the power (in watts) required to overcome the skin friction drag.** ### Diagram: The diagram shows a rectangular box with the following dimensions: - Length (\( L \)) = 3 feet - Breadth (\( B \)) = 2 feet - Depth (\( D \)) = 2 inches (converted to feet: \( \frac{2}{12} = \frac{1}{6} \, \text{ft} \)) The box is submerged partially in water, up to a "water line" shown on the sides, with velocity \( U \) indicated as the direction of motion. ### Considerations for Calculations: - The projected area to take into account includes the bottom area and two side areas submerged in water. - Use appropriate formulas to calculate the Reynolds number to determine the drag coefficient. - Use physics principles to determine the force and power needed.
Expert Solution
Step 1: Drag force

Given,

the velocity of the box, 

U equals 10 space f t divided by m i n U equals 10 cross times 0.00508 space m divided by s U equals 0.0508 m divided by s

length of the box, 

L equals 3 space f t L equals 3 cross times 0.3048 m L equals 0.9144 m

the breadth of the box, 

B equals 2 space f t B equals 2 cross times 0.3048 m B equals 0.6096 m

depth of the box, 

D equals 2 f t D equals 2 cross times 0.3048 m D equals 0.6096 m

let the normal drag coefficient be C subscript D equals 0.2

The required force, 

F subscript D equals C subscript D A rho subscript w U squared over 2 F subscript D equals 0.2 cross times open parentheses L cross times B close parentheses cross times 1000 cross times fraction numerator 0.0508 squared over denominator 2 end fraction F subscript D equals 0.2581 cross times open parentheses 0.9144 cross times 0.6096 close parentheses F subscript D equals 0.144 space N


steps

Step by step

Solved in 3 steps with 14 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer