   Chapter 8.3, Problem 14E

Chapter
Section
Textbook Problem

A vertical dam has a semicircular gate as shown in the figure. Find the hydrostatic force against the gate. To determine

To find: The hydrostatic force against the gate.

Explanation

Given:

The depth of vertical dam is 12 m.

The water level of vertical dam is 10 m.

The radius of the semi circle gate is 2 m.

Calculation:

Draw the semicircular gate as shown in Figure 1.

Refer to Figure 1.

Calculate the width of the ith strip as follows:

wi=24x2

Calculate the depth of the ith strip is as follows:

di=10x

Consider the acceleration due to gravity as g=9.8m/s2.

Consider the density of water as ρ=1,000kg/m3.

Calculate the area of the ith strip as follows:

Ai=widx (1)

Substitute 24x2 for wi in Equation (1).

Ai=24x2dx

Calculate the pressure of the ith rectangular strip:

Pi=ρgdi (2)

Substitute 1,000kg/m3 for ρ, 9.8m/s2 for g, and (10x) for di in Equation (2).

Pi=1,000×9.8×(10x)=9,800(10x)

Calculate the hydrostatic force on one end of the tank:

F=abPiAi (3)

Substitute 0 for a, 2 for b, 9,800(10x) for Pi and 24x2dx for Ai in Equation (3).

F=029,800(10x)24x2dx=19,60002(104x2x4x2)dx=196,000024x2dx19,60002x4x2dx (4)

Consider F1 and F2 as follow:

F1=196,000024x2dx (5)

F2=19,60002x4x2dx (6)

Substitute F1 for 19,6000024x2dx and F2 for 19,60002x4x2dx in Equation (4).

F=F1F2 (7)

Consider x=2sinθ (8)

Differentiate both sides of the Equation (8).

dx=2cosθdθ

Substitute (2sinθ) for x and (2cosθdθ) for dx in Equation (5).

F1=196,000024(2sinθ)2(2cosθdθ)=196,000×20244sin2θcosθdθ=392,000024(1sin2θ)cosθdθ=392,000022cos2θcosθdθ

=392,000×202cosθcosθdθ=392,000×202cos2θdθ=392,000×20212(1+cos2θ)dθ=392,00002(1+cos2θ)dθ (9)

Integrate Equation (8)

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