6 m 1m Begin by setting up a vertical y-axis so that y = 0 is at the top of the tank, and y = -1 is at the bottom. We first find the formula for cross-sectional area. The cross-sections are rectangles of length 6m. We need to determine the width of the rectangle can be found using the fact that the front semi-circular face has the formula x + y² = 1 for y < 0. The width of a rectangle is twice the x-coordinate for the rectangle at height y. The x-coordinate is given by x = = /1 – y², so the width is 2/1- y?. Thus A(y) = 6(2) /1 – y². Therefore the force on a cross-section is given by F(y) = 1000(9.8)12/1 – 3²AY. The distance a cross-section travels is D(y) = 0 – y = –y. V Thus V = | 1000(9.8)12/1- y²(-y) dy. W = (a) Suppose instead that the radius of the semi-circular cross-sections was 2 meters instead of 1 meter. How would that change the problem? You can reference some of the above work. Set up the integral that can be used to compute the work required to pump all of the water out of the top of the tank. Make sure to identify cross-sectional force and distance traveled by a cross-section. * ty= 2
6 m 1m Begin by setting up a vertical y-axis so that y = 0 is at the top of the tank, and y = -1 is at the bottom. We first find the formula for cross-sectional area. The cross-sections are rectangles of length 6m. We need to determine the width of the rectangle can be found using the fact that the front semi-circular face has the formula x + y² = 1 for y < 0. The width of a rectangle is twice the x-coordinate for the rectangle at height y. The x-coordinate is given by x = = /1 – y², so the width is 2/1- y?. Thus A(y) = 6(2) /1 – y². Therefore the force on a cross-section is given by F(y) = 1000(9.8)12/1 – 3²AY. The distance a cross-section travels is D(y) = 0 – y = –y. V Thus V = | 1000(9.8)12/1- y²(-y) dy. W = (a) Suppose instead that the radius of the semi-circular cross-sections was 2 meters instead of 1 meter. How would that change the problem? You can reference some of the above work. Set up the integral that can be used to compute the work required to pump all of the water out of the top of the tank. Make sure to identify cross-sectional force and distance traveled by a cross-section. * ty= 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
How would changing the radius of the cross section change the problem
Transcribed Image Text:6 m
1m
Begin by setting up a vertical y-axis so that y = 0 is at the top of the tank, and y = -1 is at the bottom.
We first find the formula for cross-sectional area. The cross-sections are rectangles of length 6m. We need
to determine the width of the rectangle can be found using the fact that the front semi-circular face has the
formula x + y² = 1 for y < 0. The width of a rectangle is twice the x-coordinate for the rectangle at height
y. The x-coordinate is given by x =
= /1 – y², so the width is 2/1– y?. Thus A(y) = 6(2) /1 – y².
Therefore the force on a cross-section is given by F(y) = 1000(9.8)12/1 – y²Ay. The distance a cross-section
travels is D(y) = 0 – y = –y.
V
Thus
W =
V = | 1000(9.8)12/1- y²(-y) dy.
(a) Suppose instead that the radius of the semi-circular cross-sections was 2 meters instead of 1 meter.
How would that change the problem? You can reference some of the above work. Set up the integral
that can be used to compute the work required to pump all of the water out of the top of the tank.
Make sure to identify cross-sectional force and distance traveled by a cross-section.
* ty= 2
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