Force on the side wall of the vessel cannot be directly determined as at different depths pressures are different. To find this we consider a strip of width dx at a depth x from the surface of the liquid as shown in figure, and on this strip the force due to the liquid is given as:
⇒ dF = (xρg) x (bdx).This force is acting in the direction normal to the side wall.
Net force can be evaluated by integrating equation; F = ₀∫h dF = ₀∫h xρgbdx => F = ρgbh²/ 2.
Average Pressure on Side Wall: the absolute pressure on the side wall cannot be evaluated because at different depths on this wall pressure is different. The average pressure on the wall can be given as: <p>av = F/ bh = ½ [(ρgbh)²/ bh] = ½ ρgh.
Equation show that the average pressure on side vertical wall is half of the net pressure at the bottom of the vessel.
Torque on the Side Wall due to Fluid Pressure: As shown figure, due to the force dF, the side wall experiences a torque about the bottom edge of the side which is given as: dτ = dF x (h – x) = xρgb dx (h – x).
Thus net torque is τ = ₀∫h dτ = ₀∫h ρgb (hx – x²)dx = ρgb [h³/ 2 – h³/ 3] = 1/6 ρgbh³.