Question

Problem 2 The rectangular gate CAB, shown in cross section, is L=1.5 m high and w=2 m wide (into the page). The water has a depth of d=1 m on one side of the gate. The gate is pinned at A and is restrained by a smooth stopper at B that bears horizontally against the lower edge of the plate. The gate has a mass per area of m=27.18 kg/m² (already accounting for the thickness of the gate). Ignoring the small height of the stopper, determine the reaction forces at B and A. The density of water is pw = 1000 kg/m3. If the depth d of the water increases above a certain height the gate will rotate. What is the value of d that will cause the gate to rotate? (Hint: this will be the value of d where the normal force at B is 0.) Answers: RA = (6540e, +800e,) N Ng = 3270 N d=L * DOA

Answer 1

Draw the free body diagram. Calculate the hydrostatic force acting on the rectangular gate.

Fx = Pw8(Area of submerged portion of gate) xă = ,8(dxw)+ () = (1000)(9.81)(1+2)() = 9810 N Calculate the location of center of pressure from free surface. , (MOI of submerged portion of gate about its CG) (Submerged area of the gate) h =+(wxd)ā" (5)*(wxd) (2)(1) Calculate the net torque about pin point A under equilibrium condition.

21. =0 NgX(AB)-(FA)(AP)=0 N>(1)-(F.)(1: - ) - No*{*)(9810)+{}_15.) -0 Nox{1}=(9810)>() #Ng =3270 N Calculate the reaction force at A. R =Rg @ +R, 6x =(F; - Ng) +(Mg) ez = (F; – Ng )ę +(m(Lxw)g)e, = (9810-3270)& +(27.18)(1.5x 2) (9.81))ez =6540& +800e Calculate the location of center of pressure from free surface. 7 (MOI of submerged portion of gate about its CG) (Submerged area of the gate) h (wa) =ħt (12) =ñ+(wxd)? 3)*(wxd)

Location of Centre of pressure should be just above the pin point for the limiting case, which cause the gate to rotate and net reaction force at A should be zero along horizontal direction which means, Rg = FM Calculate the net torque about pin point A under equilibrium condition. =d=L d=L