Answer:
The gate will open if the height of water is equal to or more than 0.337m.
Explanation:
From the diagram attached, (as seen from the reference question found on google)
The forces are given as
Force on OA
[tex]F_1=P A_1\\F_1=\rho g \bar{h} A_{OA}[/tex]
Here
[tex]\bar{h}=h+\frac{0.6}{2}\\\bar{h}=h+0.3[/tex]
[tex]A_{OA}=L\times W\\A_{OA}=1\times 0.6\\A_{OA}=0.6 m^2[/tex]
The Force is given as
[tex]F_1=0.6\rho g[h+0.3][/tex]
Force on OB
[tex]F_2=P A_2\\F_2=\rho g \bar{h} A_{OB}[/tex]
Here
[tex]\bar{h}=h+0.6+\frac{0.4}{2}\\\bar{h}=h+0.8[/tex]
[tex]A_{OB}=L\times W\\A_{OB}=1\times 0.4\\A_{OB}=0.4 m^2[/tex]
The Force is given as
[tex]F_2=0.4\rho g[h+0.8][/tex]
Now the moment arms are given as
[tex]\bar{y}_a=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_a=h+0.3+\frac{\frac{1}{12}\times 0.6^3 \times 1}{0.6 \times[h+0.3]}\\\bar{y}_a=h+0.3+\frac{0.03}{h+0.3}[/tex]
[tex]\bar{y}_b=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_b=h+0.8+\frac{\frac{1}{12}\times 0.4^3 \times 1}{0.4 \times[h+0.8]}\\\bar{y}_b=h+0.8+\frac{0.0133}{h+0.8}[/tex]
Taking moment about the point O as zero
[tex]F_1(h+0.6-\bar{y}_a)=F_2(\bar{y}_b-h+0.6)\\F_1(h+0.6-h-0.3-\frac{0.03}{h+0.3})=F_2(h+0.8+\frac{0.0133}{h+0.8}-h-0.6)\\F_1(0.3-\frac{0.03}{h+0.3})=F_2(0.2+\frac{0.0133}{h+0.8})\\0.6\rho g[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4\rho g[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.6[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.18h -0.054-0.018=0.08h+0.064+0.00533\\h=0.337 m[/tex]
So the gate will open if the height of water is equal to or more than 0.337m.
