Question 2.5: The gate in Fig. E2.5a is 5 ft wide, is hinged at point B, a...

By geometry the gate is 10 ft long from A to B, and its centroid is halfway between, or at elevation 3 ft above point B. The depth h_{CG} is thus 15 – 3 = 12 ft. The gate area is
5(10) = 50 ft^{2} . Neglect p_{a}as acting on both sides of the gate. From Eq. (2.26) the hydrostatic force on the gate is
F = p_{CG} A = \Delta h_{CG} A =(64 lbf/ft^3)(12 ft)(50 ft^2) = 38,400 lbf
First we must find the center of pressure of F. A free-body diagram of the gate is shown in Fig. E2.5b. The gate is a rectangle, hence
I_{xy}=0 and I_{xx}= \frac{bL^3}{12}= \frac{(5 ft)(10 ft)^3}{12}=417 ft^4
The distance l from the CG to the CP is given by Eq. (2.29) since pa is neglected.
l=-y_{cp}=+\frac{I_{xx}\sin \Theta }{h_{CG}A}=\frac{(417ft^{4})(\frac{6}{10} )}{(12ft)(50ft^{2})}=0.417 ft
The distance from point B to force F is thus 10 = l = 5 = 4.583 ft. Summing the moments counterclockwise about B gives
PL \sin \theta-F(5-l)=P(6 ft) = (38,400 lbf )(4.583 ft) = 0

or P =29,300 lbf
With F and P known, the reactions B_{x}and B_{z} are found by summing forces on the gate:
F_{x}= 0 = B_{x} = F \sin \theta -P = B_{x} + 38,400 lbf (0.6) = 29,300 lbf
or B_{x}=6300 lbf
F_{z} = 0 = B_{z} – F \cos \theta =B_{z} – 38,400 lbf (0.8)
or B_{z} =30,700 lbf
This example should have reviewed your knowledge of statics.