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Question 2.9: A parabolic gate AB is hinged at A and latched at B as shown...

A parabolic gate AB is hinged at A and latched at B as shown in Fig. 2.32. The gate is 3 m wide. Determine the components of net hydrostatic force on the gate exerted by water.

2.9
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The hydrostatic force on an elemental portion of the gate of length ds (Fig. 2.32) can be written as

d F=9.81 \times 10^{3} \times(1.5-z) d s \times 3

The horizontal and vertical components of the force dF are

d F_{H}=9.81 \times 10^{3} \times 3(1.5-z) \times d s \cos \theta

 

=9.81 \times 3 \times(1.5-z) \times 10^{3} d z

 

and

d F_{V}=9.81 \times 10^{3} \times 3(1.5-z) \times d s \sin \theta

 

=9.81 \times 3 \times(1.5-z) \times 10^{3} d x

 

Therefore, the horizontal component of hydrostatic force on the entire gate

 

F_{H}=\int_{0}^{1.5} 9.81 \times 3 \times(1.5-z) \times 10^{3} d z

 

=9.81 \times 10^{3} \times \frac{1.5 \times 1.5}{2} \times 3 N = 33.11 kN

 

The vertical component of force on the entire gate

 

F_{V}=\int_{0}^{1.5} 9.81 \times 3 \times(1.5-z) \times 10^{3}\left(\frac{2}{3} z\right) d z

 

\left(\text { Since } x=\frac{1}{3} z^{2} \text { for the gate profile, } d x=\frac{2}{3} z d z\right)

 

=\frac{2}{3} \times 9.81 \times 10^{3} \times \frac{(1.5)^{3}}{6} \times 3 N = 11.04 kN

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