Fluids at Rest — Question 3

Question 3

Bottom force and liquid weight need not be equal

An open vessel of height $H$ widens upward. Its horizontal area at height $z$ above the base is $A(z)=A_0(1+z/H)$. It contains liquid of density $\rho$ up to $z=H$. Use gauge pressure so the common atmospheric contribution is removed.

1. Find the downward force exerted by the liquid on the horizontal base and the liquid weight.

2. Find the net vertical gauge force exerted by the sidewalls on the liquid. Explain how this resolves the apparent mismatch.

Question 3 - Solution

Bottom force and liquid weight need not be equal

(a) Derivation and calculation

$$p_B=\rho gH,\qquad\boxed{F_B=p_BA_0=\rho gHA_0}.$$$$V=\int_0^H A_0(1+z/H)\,dz=\tfrac32A_0H.$$$$\boxed{W=\rho gV=\tfrac32\rho gA_0H},\qquad F_B/W=2/3.$$

(b) Result and interpretation

For the liquid as the system, the base force is upward:$$F_B+F_{w,z}-W=0.$$$$\boxed{F_{w,z}=W-F_B=\tfrac12\rho gA_0H\ \text{upward}}.$$The sloping walls support the remaining one-third of the weight through their normal forces.

Physical check

The force on the base is opposite to the base force on the liquid; keeping this action-reaction distinction resolves the sign.