Pressure in solids liquids and gases

Pressure is a scalar field defined at a surface as the magnitude of the normal force component divided by the area over which it acts:

$$p = \frac{F_\perp}{A}.$$

In SI the unit is the pascal, $1\ \text{Pa} = 1\ \text{N/m}^2$, so $1\ \text{kPa} = 10^3\ \text{Pa}$ and $1\ \text{MPa} = 10^6\ \text{Pa}$.

Pressure is not the same as force: the same force spread over a larger area gives a smaller pressure, which is why wide foundations and snowshoes work. In a fluid at rest, pressure at a point is the same in every direction (Pascal's law), so a single scalar $p$ is enough to describe it, whereas in a solid the stress on a chosen area depends on that area's orientation.

For a uniform fluid of density $\rho$ in equilibrium, applying the balance condition to a column of height $h$ and cross-section $A$ gives

$$p = p_0 + \rho g h,$$

where $p_0$ is the external pressure at the free surface and $\rho g h$ is the gauge pressure. In gases the same balance holds, but density changes with height, so the linear formula is only a good local approximation; microscopically gas pressure comes from molecules striking the walls.

A crate of mass $m = 12\ \text{kg}$ rests on a horizontal floor on a face of area $A = 0.030\ \text{m}^2$. Take $g = 9.8\ \text{m/s}^2$. Find the pressure the crate exerts on the floor. What does it become if the crate is turned onto a face of area $0.060\ \text{m}^2$?

An open tank is filled with water to a depth $h = 5.0\ \text{m}$; the water density is $\rho = 1000\ \text{kg/m}^3$, $g = 9.8\ \text{m/s}^2$, atmospheric pressure $p_0 = 101\ \text{kPa}$. Find the gauge and absolute pressure at the bottom, and the forces on a horizontal hatch of area $A = 0.50\ \text{m}^2$ in the bottom: from the water inside, from the outside air, and their resultant.

In a hydraulic lift the small piston has area $A_1 = 20\ \text{cm}^2$ and the large piston has area $A_2 = 400\ \text{cm}^2$. A load of mass $M = 600\ \text{kg}$ rests on the large piston and must be raised by $d_2 = 2.0\ \text{cm}$. The fluid is incompressible and at rest, $g = 9.8\ \text{m/s}^2$. Find: (a) the pressure in the fluid, (b) the force needed on the small piston, (c) the displacement of the small piston, (d) check the work balance.

Self-study plan with a simulation (it replaces hazardous measurements with pressure vessels and mercury).

1. Open the PhET simulation Under Pressure (Fluid Pressure and Flow) in a browser and choose water.
2. Set atmospheric pressure to $p_0 = 101\ \text{kPa}$ and record the pressure sensor readings at depths $h = 1, 2, 3, 4, 5\ \text{m}$ in a table.
3. Subtract $p_0$ and plot $p - p_0$ against $h$. Check that the points lie on a straight line and find its slope; compare it with $\rho g = 9.8\ \text{kPa/m}$.
4. Switch the fluid density, for example to $800\ \text{kg/m}^3$, repeat the measurements and check that the ratio of the slopes equals the ratio of the densities.
5. Put the sensor plate at one point and rotate it: confirm that the pressure does not depend on orientation (Pascal's law) and compare this with a solid.
6. Safe household check (no heating, no compressed gases): place a book of known mass on two flat cardboard sheets of different area and compare the computed values of $p = mg/A$.

To show that in a solid the pressure depends on the area over which the force is spread, while in a fluid at rest the pressure at a point is the same in all directions and grows linearly with depth.

A three-part sketch. (1) The same block stands first on a narrow support and then on a wide one; the force $F$, the areas $A$ and $2A$, and the formulas $p = F/A$ and $p = F/(2A)$ are labelled. (2) A point inside the fluid with equal-length arrows pointing in several directions, all labelled $p$ — an illustration of Pascal's law. (3) A graph of pressure against depth: a straight line with slope $\rho g$ that crosses the pressure axis at $p_0$.

A striker of mass $m = 70$ kg stands on one leg while striking the ball. His weight $F = mg \approx 690$ N is transmitted to the grass through six studs of area $1.0$ cm$^2$ each, so that altogether $A = 6.0\times10^{-4}$ m$^2$. The pressure under the studs is $p = F/A \approx 1.2\times10^{6}$ Pa, about 33 times larger than the $3.5\times10^{4}$ Pa under the sole of a studless boot of area $0.020$ m$^2$.

The definition of pressure in a solid applies: $p = F_\perp/A$, where $F_\perp$ is the normal force taken from the free-body diagram (the support reaction equals the weight in magnitude) and $A$ is the actual contact area. By reducing the supporting area, the studs increase the pressure by the same factor and therefore bite into the ground.

A lorry of mass $3.0\times10^{4}$ kg stands on ten wheels; its weight is $F = mg \approx 2.9\times10^{5}$ N. The contact patch of each wheel on the asphalt is $0.030$ m$^2$, so the total area is $A = 0.30$ m$^2$ and the mean pressure on the road is $p = F/A \approx 9.7\times10^{5}$ Pa, about $1.0$ MPa. The gauge simultaneously reads an air pressure inside the tyre of about $8\times10^{5}$ Pa.

The same law $p = F/A$ acts in two media: at the solid contact the tyre presses on the asphalt with a normal force, while inside the tyre the gas pressure is produced by air molecules striking the walls. It is the large contact area that lowers the pressure on the surface, which is why heavy vehicles use wide tyres and many axles.

In a $5.0$ l pressure cooker water boils not at $100$ °C but at $120$ °C: the valve keeps the inside pressure near $p = 2.0\times10^{5}$ Pa, twice the atmospheric $1.0\times10^{5}$ Pa. A valve of area $A = 8.0$ mm$^2 = 8.0\times10^{-6}$ m$^2$ opens when the force of the vapour exceeds the weight of the load: $F = \Delta p\,A = 1.0\times10^{5}\times 8.0\times10^{-6} = 0.80$ N, corresponding to a load of mass $80$ g.

The gas pressure here is the mean force of vapour-molecule impacts per unit area of the wall, that is the same $p = F/A$; this pressure is transmitted to the liquid and to the walls of the vessel. The force lifting the valve is the product of pressure and area, so a small valve area lets a small load hold a noticeable pressure.