Water and liquid is filled up behind a square wall of side `l`
Find out (a). Pressure at A, B and C
(b). Forces in part AB and BC
(c). Total force and point of application of force. (Neglect atmosphere pressure in every calculation.)

Explanation

(a) As there is no liquid above `'A'`,
So pressure at `A, P_(A) = 0`
Pressure at `B, P_(B) = rhogh_(1)`
Pressure at `C, P_(C) = rhogh_(1) + 2rhogh_(2)`
(b) Force at `A = 0`
Take a strip of width `'dx'` at a depth `'x'` in part `AB`.
Pressure is equal to `rhogx`.
Force on strip `= pressure xx area`
`dF = rhopx ldx`
Total force upto `B`
`F = overset(h_(1))underset(0)(int)rhogxldx = (rhogxlh_(1)^(2))/(2) = (1000 xx 10 xx 10 xx5 xx 5)/(2)`

`= 1.25 xx 10^(8)N`
In part `BC` for force ltake a elementary strip of width `dx` in portion `BC`. Pressure is equal to
`= rhogh_(1) + 2rhog(x - h_(1))`
Force on elementry strip `= "pressure" xx "area"`
`dF = [rhogh_(1) + 2rhog(x - h_(1))]ldx`
Total force on part `BC`
`F = overset(t)underset(h_(1))(int)[rhogh_(1) + 2rhog(x - h_(1))]ldx`

`= [rhogh_(1) + 2rhog[(x^(2))/(2) - h_(1)x]]_(h_(1))^(l) l`
`= rhogh_(1)h_(2)l + 2rhogl[(l^(2) - h_(1)^(2))/(2) - h_(1)l + h_(1)^(2)]`
`rhogh_(1)h_(2)l + (2rhogl)/(2) [l^(2) + h_(1)^(2) - 2h_(1)l]`
`rhogh_(1)h_(2) + rhogl (l - h_(1))^(2)`
`rhogh_(2)l [h_(1) + h_(2)] = rhogh_(2)l^(2)`
`= 1000 xx 10 xx 5 xx 10 xx 10`
`5 xx 10^(6) N`