How Much Weight Is Needed to Keep the Water Gate Closed?

[jdawg]

Homework Statement

A rectangular plate AB is 1.8 [m] wide and 2.0 [m] long. As shown below the plate is used to keep water (blue region) confined to the region. Find the weight of the gate necessary to keep the water enclosed. Assume the hinge at A is frictionless.

Homework Equations

The Attempt at a Solution

I used 9810 N/m^3₃ for the specific weight of water.
tanθ=4/3
θ=53.13°
A=(2m)(1.8m)=3.6m²
h_CG=(2m)+(1m)sin(53.13)=2.8m
F_H2O=(9.807)(1.6)(3.6)=98884.8 N

I_xx=((1.8m)(2m)³)/12 =1.2m⁴

y_cp=(-I_xx)sinθ)/(h_CGA)= - 0.09508m

M_A=0=(By)(2)cos(53)-W(1)cos(53)+F_H2O(1)cos(53)-F_H2O(1)sin(53)

I'm not really sure if I'm correct up to this point. Is my placement of theta in the free body diagram correct?
Thanks!

 

[SteamKing]

It's not clear from the problem statement how you know the gate has a slope of 4:3 when it is closed.