Archimedes' Principle — Question 10

Question 10

A dry contact can remove the force that Archimedes requires

A rectangular block of height $h$ and horizontal area $A$ is held motionless with its bottom sealed against a horizontal tank floor. Its top is at depth $d$ in water, while a dry pocket under its bottom has absolute pressure $p_d$. The side faces are vertical. Surface pressure is $p_0$, liquid density is $\rho$, and surface tension is neglected.

1. Find the net vertical pressure force on the block before water reaches its underside. Is it necessarily upward?

2. If a thin water layer is allowed underneath while the block remains at the same height, find the new net pressure force. Identify the assumption needed for the usual displaced-fluid result.

Question 10 - Solution

A dry contact can remove the force that Archimedes requires

(a) Derivation and calculation

The upper surface pressure is $p_t=p_0+\rho gd$. Vertical sides contribute no vertical pressure force:$$F_z=p_dA-p_tA.$$$$\boxed{F_z=[p_d-p_0-\rho gd]A}.$$This force can be downward. The floor contact force is separate from the pressure resultant.

(b) Result and interpretation

Once liquid reaches the underside,$$p_b=p_0+\rho g(d+h),\qquad p_t=p_0+\rho gd.$$$$\boxed{F_z=(p_b-p_t)A=\rho ghA=\rho gV\ \text{upward}}.$$The displaced-fluid result requires the appropriate hydrostatic pressure over the immersed boundary; a sealed dry contact removes the required underside contribution.

Physical check

After wetting, the common surface pressure and top-depth terms cancel, leaving the weight of displaced liquid.