Correct Answer - `F=13/2rhogh^(2)b`
Let the hydrostatic forces acting on the rectangular pathes of areas `A_(1)` and `A_(2)` of the gate be `F_(1)` and `F_(2)`. Then the total hydrostatic force on the gate is
`F=F_(1)+F_(2)`
where `F_(1)=P_(av_(1))A_(1)` and `F_(2)=P_(av_(2)) A_(2). P_(av_(1))` and `P_(av_(2))` and `P_(av_(2))` are the average pressures on areas `A_(1)` and `A_(2)`, respectively.
`F=P_(av_(1))A_(1)+P_(av_(2))A_(2)`
Substituting `P_(av1)=(P_(A)+P_(B))//2, A_(1)=bh_(1)=bh, P_(av2)=(P_(B)+P_(C)).2,A_(2)=bh_(2)=b(2h),` we have
`F=((P_(A)+P_(B)))/2bh+((P_(B)+P_(C)))/2 2bh`
Substituting `P_(A)=0, P_(B)=rhogh` and `P_(C)=(rho_(1)h_(1)+rho_(2)h_(2))g=[rhoh+(2rho)(2h)]=5rhogh,` we have
`F=3/2rhogh^(2)b`