Answer The total hydrostatic force on the gate is 11263467 N . Explanation Given: The tank contains water. The diameter of the gate is 4 m . The height of the gate below water surface is 12 − 2 = 10 m . Concept Used: The hydrostatic pressure of a water of density ρ at a depth d from the surface of the water is P = ρ g d . The force on a surface due to pressure P exerted is F = P A , where A is the area of the surface being considered. The radius of the gate is 4 2 = 2 m The depth of the center of the gate from the surface of water is 10 m So, if θ is the angle with the horizontal made by a radius passing through a point p located on the circular edge of the gate, then The depth of the point p is 10 − r sin θ = 10 − 2 sin θ . The width of a horizontal strip of the gate passing through the point is 2 r cos θ . The length of an infinitesimal section of perimeter around the point p making an angle d θ with the center of the disc is r d θ . The height (thickness) of the horizontal strip of disc that corresponds to this small section of perimeter is r d θ ⋅ cos θ Therefore, the area of this infinitesimally thin strip is d A θ = w i d t h × h e i g h t = 2 r cos θ × r d θ cos θ = 2 r 2 cos 2 θ d θ and the hydrostatic pressure at the level of point p is P θ = ρ g h = ρ g ( 10 − 2 sin θ ) Therefore, the force on this thin strip is d F = P θ d A θ = ρ g ( 10 − 2 sin θ ) ⋅ 2 r 2 cos 2 θ d θ = 784000 ( 10 − 2 sin θ ) ⋅ cos 2 θ d θ Now the total force on the gate is the integral of d F from θ = 0 to θ = π . So F = ∫ 0 π d F = ∫ 0 π 784000 ( 10 − 2 sin θ ) ⋅ cos 2 θ d θ = 784000 ( 2 cos 3 θ 3 + 10 ( θ 2 + s i n 2 θ 4 ) ) ] 0 π = 784000 ( 2 ( − 2 3 ) + 10 ( π 2 + 0 ) ) = 784000 × 14.3 6 ¯ = 11263467 Therefore, the total hydrostatic force on the gate is 1 1 2 6 3 4 6 7 N .

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 6.6, Problem 40E

To determine

To Calculate: The hydrostatic force on a semicircular gate at the bottom of a dam with water

Expert Solution & Answer

Answer to Problem 40E

The total hydrostatic force on the gate is $11263467 N$ .

Explanation of Solution

Given: The tank contains water.

The diameter of the gate is $4 m$ .
The height of the gate below water surface is $12−2=10 m$ .

Concept Used:

The hydrostatic pressure of a water of density $ρ$ at a depth $d$ from the surface of the water is $P=ρgd$ .
The force on a surface due to pressure $P$ exerted is $F=PA$ , where $A$ is the area of the surface being considered.

The radius of the gate is $42=2 m$

The depth of the center of the gate from the surface of water is $10 m$

So, if $θ$ is the angle with the horizontal made by a radius passing through a point $p$ located on the circular edge of the gate, then

The depth of the point $p$ is $10−rsinθ=10−2sinθ$ .
The width of a horizontal strip of the gate passing through the point is $2rcosθ$ .
The length of an infinitesimal section of perimeter around the point $p$ making an angle $dθ$ with the center of the disc is $rdθ$ .
The height (thickness) of the horizontal strip of disc that corresponds to this small section of perimeter is $rdθ⋅cosθ$

Therefore, the area of this infinitesimally thin strip is

$dAθ=width×height=2rcosθ×rdθcosθ$

$=2r2cos2θdθ$

and the hydrostatic pressure at the level of point $p$ is

$Pθ=ρgh=ρg(10−2sinθ)$

Therefore, the force on this thin strip is

$dF=PθdAθ=ρg(10−2sinθ)⋅2r2cos2θdθ=784000(10−2sinθ)⋅cos2θdθ$

Now the total force on the gate is the integral of $dF$ from $θ=0$ to $θ=π$ . So

$F=∫0πdF=∫0π784000(10−2sinθ)⋅cos2θdθ$

$=784000(2cos3θ3+10(θ2+sin2θ4))]0π$

$=784000(2(−23)+10(π2+0))$

$=784000×14.36¯$

$=11263467$

Therefore, the total hydrostatic force on the gate is $11263467 N$ .

Chapter 6.6, Problem 39E

Chapter 6.6, Problem 41E

Chapter 6 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

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