In many Indian homes the water comes from a tank on the roof. There is no pump between the tank and the taps. Yet the tap on the ground floor gushes, while the tap on the top floor, just below the tank, only trickles. The water in the tank is the same for both.
A column of liquid
Take a liquid of density \(\rho\) at rest. Measure the depth \(y\) downward from its free surface. Picture a thin horizontal slab of the liquid, of area \(A\), between the depths \(y\) and \(y + dy\). The liquid above pushes down on the top face with a force \(PA\). The liquid below pushes up on the bottom face with \((P + dP)A\). The slab also has a weight.
\[ (P + dP)A = PA + (\rho A\,dy)\,g \]
Why. the slab is at rest, so the upward push on its bottom balances the push on its top and its weight, mass \(\rho A\,dy\).
\[ \dfrac{dP}{dy} = \rho g \]
Why. cancel \(PA\) and divide by \(A\,dy\). The pressure grows downward at the rate \(\rho g\).
\[ \int_{P_0}^{P} dP = \int_0^{h} \rho g\,dy \]
Why. integrate from the free surface, where \(y = 0\) and the pressure is \(P_0\), down to the depth \(h\).
\[ P = P_0 + \rho g h \]
Read it. \(\rho\) and \(g\) are constant for a liquid, which is hard to compress.
The rate \(\rho g\) is worth remembering for water. It is \(1000 \times 10 = 10^4\,\text{Pa}\) for every metre, or \(100\,\text{Pa}\) for every centimetre. Ten metres of water therefore add \(10^5\,\text{Pa}\), about one atmosphere. A diver \(10\,\text{m}\) down feels twice the pressure felt at the surface.
In \(P = P_0 + \rho g h\), \(h\) is the depth below the free surface. It is not the height above the bottom. A point \(2\,\text{cm}\) above the base of a \(30\,\text{cm}\) column is \(28\,\text{cm}\) deep.
A probe going down
The gauge of the last lesson can be lowered on a cable into a beaker, as in Fig 2.1. A graph of the gauge pressure, \(P - P_0\), against the depth draws itself as the probe goes down. The first chip puts \(10\,\text{cm}\) of oil of density \(800\,\text{kg/m}^3\) on \(20\,\text{cm}\) of water. The second fills the beaker with water alone.
The graph in Fig 2.1 is a straight line because the density is constant. Its slope is \(\rho g\), so the lighter oil gives a gentler slope. The pressure at the boundary is \(800 \times 10 \times 0.10 = 800\,\text{Pa}\). Below it the water adds its own \(\rho g h\). At the bottom, \(800 + 1000 \times 10 \times 0.20 = 2800\,\text{Pa}\). Each layer adds its own share.
Take \(g = 10\,\text{m/s}^2\) and atmospheric pressure \(P_0 = 10^5\,\text{Pa}\). Find the gauge pressure and the absolute pressure \(10\,\text{m}\) below the surface of a lake. At what depth is the absolute pressure three times atmospheric?
The surface is open to the air, so the pressure there is \(P_0\). The gauge pressure is \(\rho g h\), the part added by the water.
\(P - P_0 = \rho g h\), and \(P = P_0 + \rho g h\).
At \(10\,\text{m}\): \(\rho g h = 1000 \times 10 \times 10 = 10^5\,\text{Pa}\), so \(P = 2 \times 10^5\,\text{Pa}\). For \(P = 3P_0\): \(\rho g h = 2 \times 10^5\,\text{Pa}\), so \(h = 20\,\text{m}\).
The gauge pressure at \(10\,\text{m}\) is \(10^5\,\text{Pa}\), and the absolute pressure is \(2 \times 10^5\,\text{Pa}\). The absolute pressure is three times atmospheric at \(20\,\text{m}\).
A jar holds \(10\,\text{cm}\) of water with \(5\,\text{cm}\) of kerosene, density \(800\,\text{kg/m}^3\), floating on it. Take \(g = 10\,\text{m/s}^2\). Find the gauge pressure at the bottom of the jar, in Pa.
Along a level line
Now picture a thin horizontal cylinder of liquid joining two points at the same level. Its weight acts straight down, so it has no horizontal part. The only horizontal forces on the cylinder are the pushes on its two end faces. The cylinder is at rest, so those pushes balance, and the pressures at its two ends are equal.
In a liquid at rest, all points at the same level have the same pressure, if a path through that one liquid joins them.
Every point of a free surface open to the air is at the same pressure, \(P_0\). Points at the same pressure in a liquid at rest lie at the same depth. So every point of the free surface is at one height: a liquid at rest has a level surface. This holds even when the liquid fills several vessels joined at the bottom. The water stands at one height in all of them, whatever their shapes.
The masons at a building site use this every day. They fill a long clear plastic tube with water and hold its two ends against two walls. The water stands at the same height at both ends, and they mark the two walls there. The marks are exactly level, even across a room or round a corner.
Now the taps of the opening. Suppose the water in the roof tank stands \(12\,\text{m}\) above the ground-floor tap and \(3\,\text{m}\) above the top-floor tap. When the taps are closed, the water in the pipes is at rest. The gauge pressure at a tap is \(\rho g h\), where \(h\) is its depth below the water in the tank. At the ground floor it is \(1.2 \times 10^5\,\text{Pa}\), and at the top floor only \(3 \times 10^4\,\text{Pa}\), four times less. The winding path of the pipes does not matter; only the depth does.
The rule needs one connected liquid. Take a U-tube with oil in one arm and water in the other. A point in the oil and a point in the water, at one level, have different pressures. Compare only at a level where the same liquid joins the two points.
Explain, using pressure, why the free surface of a liquid at rest in a wide vessel is horizontal.
Two points at the same level in a liquid at rest have equal pressures, because a horizontal cylinder of liquid joining them has only its end pushes acting sideways, and these must balance. Every point of the free surface is at atmospheric pressure. Points at the same pressure lie at the same depth, so every point of the surface is at the same height, and the surface is horizontal.
The three vessels have equal square bases, \(10\,\text{cm}\) on a side, and hold water \(20\,\text{cm}\) deep. B holds the most water and C the least. On which base does the water press with the largest force?
The gauge pressure on each base is \(\rho g h = 1000 \times 10 \times 0.20 = 2000\,\text{Pa}\). The bases are equal, \(0.01\,\text{m}^2\) each, so the water presses on each with \(20\,\text{N}\). Yet A holds \(20\,\text{N}\) of water, B about \(47\,\text{N}\) and C about \(12\,\text{N}\). This surprise is the hydrostatic paradox.
The walls explain it. In B the walls slope outward. The water presses on them at right angles, partly downward, and the walls push back on the water, partly upward. So the walls hold up about \(27\,\text{N}\) of B's water, and the base carries only the column above it. In C the walls lean inward. They push down on the water and add about \(8\,\text{N}\) to the push on the base.
Stand each vessel on a weighing scale. The scale reads the weight of the vessel and all its water, because the walls pass their share down to the base of the vessel. The paradox is only about the force of the water on the bottom face, not about the weight on the scale.
The water in vessel B weighs about \(47\,\text{N}\), but it presses on the base with only \(20\,\text{N}\). Where does the rest of the weight go?
The Flemish engineer Simon Stevin worked on dykes, sluices and windmills in the Low Countries, where the sea is held back by walls of earth. In 1586 he published a book on the weight of water. It showed that the force of water on the bottom of a vessel depends only on the depth and the area of the bottom. Stevin proved it by imagining parts of the water turned solid, which changes nothing while all is at rest.
Liquids in layers
Liquids that do not mix settle in layers, the densest at the bottom. Crossing each layer adds its own \(\rho g h\). So the gauge pressure at the bottom of a stack of layers is \(\rho_1 g h_1 + \rho_2 g h_2 + \dots\). Fig 2.1 showed it for oil on water: \(800 + 2000 = 2800\,\text{Pa}\).
The force on a base
The base of a tank is level, so the pressure is the same all over it. The force of the water on the base is then the gauge pressure times the area. Take a tank with a base \(2\,\text{m}\) by \(1\,\text{m}\), holding water \(0.5\,\text{m}\) deep. The gauge pressure at the base is \(1000 \times 10 \times 0.5 = 5000\,\text{Pa}\). The force is \(5000 \times 2 = 10^4\,\text{N}\), the weight of the \(1000\,\text{kg}\) of water it holds.
The force on a wall
A wall is harder, because the pressure changes down its height. Take a vertical wall of width \(L\) in water of depth \(H\). The gauge pressure on it grows evenly, from zero at the surface to \(\rho g H\) at the bottom. So the average pressure on the wall is \(\tfrac12 \rho g H\), and the area is \(LH\). The force is the average pressure times the area, \(\tfrac12 \rho g L H^2\).
For the tank above, the \(2\,\text{m}\) side wall carries \(\tfrac12 \times 1000 \times 10 \times 2 \times 0.5^2 = 2500\,\text{N}\). The square of the depth matters most for a dam. A check dam \(10\,\text{m}\) long holding water \(4\,\text{m}\) deep feels \(8 \times 10^5\,\text{N}\). At \(8\,\text{m}\) deep the force is four times as large, \(3.2 \times 10^6\,\text{N}\). The pressure is also largest at the bottom, so a dam is built much thicker at its base than at its top.
The front glass of an aquarium is \(0.6\,\text{m}\) wide, and the water in it is \(0.4\,\text{m}\) deep. Take \(g = 10\,\text{m/s}^2\). What force does the water exert on the glass, in newtons?
The water behind a dam rises from \(5\,\text{m}\) to \(10\,\text{m}\) deep. Which statements are true?
The Bhakra dam on the Sutlej is very much thicker at its base than at its top. The pressure of the water is greatest at the bottom, so the concrete must be strongest there. Near the top, where the pressure is small, a thinner wall is enough. The shape of every large dam is the graph of \(P = \rho g h\), turned into concrete.
- Measuring \(h\) from the bottom of the vessel.\(h\) is the depth below the free surface. Measure down from the surface.
- Expecting more liquid to mean more pressure on the base.The pressure depends on the depth alone. Sloping walls carry the difference in weight.
- Comparing pressures at one level across two different liquids.Same level means same pressure only inside one connected liquid.
- Using \(\rho g H \times LH\) for the force on a wall.The pressure on a wall grows from zero at the top, so use the average, \(\tfrac12 \rho g H\).