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

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SUMMARY

The discussion focuses on calculating the weight required to keep a rectangular water gate closed, specifically a plate measuring 1.8 m by 2.0 m. The specific weight of water is established at 9810 N/m³, and the angle of inclination, θ, is determined to be 53.13° using a tangent ratio of 4:3. The total hydrostatic force acting on the gate is calculated to be 98884.8 N, and the moment equilibrium equation is set up to solve for the weight of the gate. The user seeks clarification on the placement of θ in their free body diagram.

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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
hCG=(2m)+(1m)sin(53.13)=2.8m
FH2O=(9.807)(1.6)(3.6)=98884.8 N

Ixx=((1.8m)(2m)3)/12 =1.2m4

ycp=(-Ixx)sinθ)/(hCGA)= - 0.09508m

MA=0=(By)(2)cos(53)-W(1)cos(53)+FH2O(1)cos(53)-FH2O(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!
 

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

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