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.