Same liquid height in three differently shaped vessels gives same pressure at the bottom, but the weight is different. Where does the force come from?

6 answers & replies

• [+][–]Dev Agarwal12thSample· 1mo ago

  Some unit anchors for this chapter: pressure is force per area, so the unit is the pascal, 1 Pa = 1 N m⁻², and 1 atm $\approx 1.013\times10^5$ Pa. A mercury barometer reads about 76 cm, which comes straight from $P = \rho g h$. Dams are built thicker at the bottom for the same reason, pressure grows linearly with depth.

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• [+][–]Rahul DubeyDropperSample· 1mo ago

  Coming back to your main question, Sara, this one has a name in many books: the hydrostatic paradox. The force on the base is $PA = (P_a + \rho g h)A$, fixed by depth only, not by the shape. It is not automatically the weight of the liquid in the vessel.

  The gap is made up by the slanting walls. In a vessel that widens upwards (your wide jar), the liquid presses on the sloping wall outwards and downwards, so the wall pushes back inwards and upwards on the liquid and carries part of its weight. Base force is less than the weight there. In a vessel that narrows upwards (your conical flask), the wall pushes the liquid downwards, so the base has to bear more than the weight of the liquid directly in the flask. Only for a straight-sided tube do the two match exactly. The total weight on the table is still vessel plus liquid, no rule is broken.

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  • [+][–]Sara Fernandes11thSample· 1mo agoAuthor

    oh!! the walls push. so the total weight on the table is the same, but the force on the base alone is PA, and the slanted walls cover the difference. wide jar: walls hold some of the water up. thanks

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• [+][–]Nisha Verma12thSample· 1mo ago

  Hydraulic lift: two pistons areas $A_1$ (small) and $A_2$ (large), connected by a liquid. Pascal's law: pressure applied to an enclosed fluid is transmitted undiminished everywhere. So

  $\frac{F_1}{A_1} = \frac{F_2}{A_2}\Rightarrow F_2 = F_1\frac{A_2}{A_1}$

  Force is multiplied, yes, but the volume of liquid pushed out of the small cylinder equals the volume entering the large one: $A_1 d_1 = A_2 d_2$. So the large piston moves by a distance smaller by the same factor. Work done is $F_1d_1 = F_2d_2$. No free energy, just a trade of force for distance, like a lever.

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• [+][–]Kavya NaikDropperSample· 1mo ago

  Gauge vs absolute. Absolute pressure is the total: $P = P_a + P_{gauge}$, where $P_a \approx 1.013\times10^5$ Pa (1 atm). Gauge pressure is the excess over atmospheric, which is what a tyre gauge or a blood-pressure cuff measures. If the gauge says 2 atm, the absolute pressure inside is about 3 atm. Gauge pressure at depth $h$ is just $\rho g h$.

  NEET likes this one: "pressure in a liquid at depth h (absolute)" has the $P_a$ in it, gauge doesn't.

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• [+][–]Ishita GoyalDropperSample· 1mo ago

  Two more facts to keep: (1) pressure at the same horizontal level in a connected liquid at rest is the same, which is why liquid in a U-tube or in a kettle spout has the same level on both sides, and (2) pressure is a scalar. It acts in all directions on a small element, the force direction comes from the surface it presses on. That's why pressure at depth doesn't depend on the direction you measure, just the depth.

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