Question

A rectangular gate 1.2 m wide and 3.5 m long is inclined 35°from a smooth wall. The gate is hinged at point B, and rests against a smooth wall at A which is 2.7 m below the water surface.

a. Compute the force on the gate due to seawater pressure (s.g. = 1.03).

b. Determine the location of the force from the center of gravity of the gate.

c. determine the horizontal force P exerted by the wall at point A.

d. determine the reaction at hinge B.

Answer 1

a w ell 2.7m lo Gate I y=2m 3.5m 350 STDBS 3.5 :.y = 2.00m x=2-87 m csgs • 35 Hinge X = 2.87 m 1 1 F 3.79 3.70 - Вх f 1 c.al 3.5 By K - 1. Bu 35 .74 solutions specific coeight of tea water = 1.03 X1000-89.81 = 10104.3 N Im3 Area = (A) = 1.223.5 = 412702 @ to locate C.G.(7) = (4.7 - 2) + ( ) = 3.7 m --1 Y = 3.7 m O Resultant force on gate (F):- FodAT = 10104,384:2 * 3.7 = 157020-82 N F = 157-02 kN) - Ano.

© Force P=9 M. . about C-G.(IG) = vertical distance of cp x1-243.53 = 4,2875 m. from free surface :- Yp= ýt IG sinzo А. = 3.7 +4.2875 \ 4.2x5.7 Yp = 3.7g6 m Vertical distance bern B&cp = 4.7 - 3.799 = 0.918 m Location of F From B, = 0.910 = 1.56 m Sin35 Taking moment about B, (P x 2.0).- 157.02X159 = 0 :( P = 124-85 kN - Ano. NOW to find reaction at hinge By & By :- {fx=0 | bx -128-83 + (157.02 x 2 ) = 0 Bx=35.10 AN Ans. 3 .5 Ery=0 :- By 157002% 2 0 . -T By = 128.76 km] -- Ans. - An