Atmospheric pressure and hydraulics

Put the tip of a syringe in water and raise its piston: water follows it. Yet the piston does not touch the water outside. What pushes the water, and how can the same pressure help lift a heavy load?

Air is invisible, but its particles move and strike surrounding surfaces, exerting pressure from all directions. Earth holds the atmosphere in place, and its lower layers support the weight of the layers above. Go higher: less air remains overhead, so atmospheric pressure usually decreases. This connects two descriptions: particle collisions and the balance of air layers. Air density changes with altitude, so we cannot treat the whole atmosphere as a uniform liquid column.

In a barometer tube sealed at the top, the space above the liquid is almost a vacuum; the lower end opens into a basin exposed to air. Why does the whole column not drain out? The atmosphere presses on the basin's surface and supports it. At equilibrium, pатм ≈ ρgh if the vapour pressure above the column is negligible. Higher atmospheric pressure supports a taller column. In the same atmosphere, a less dense liquid would need a taller column: the same formula tells us why.

A liquid column is not the only way to measure atmospheric pressure. In an aneroid barometer, pressure deforms a sealed metal capsule containing rarefied air; a mechanism transfers the change to a pointer. A barometer measures atmospheric pressure. A manometer compares a system's pressure with a chosen reference pressure. Here the mercury column is a historical measurement diagram, not a home experiment. Both instruments turn pressure into an observable change.

Raise a pump's piston: the available volume inside increases and the pressure beneath the piston falls. The outside pressure on the water surface is now greater than the inside pressure, so it pushes water into the pump. Valves open the required path and prevent backflow on the next stroke. Saying that the piston 'pulls water' hides the important cause: a pressure difference. Reducing the outside pressure weakens this way of moving water; empty space is not itself a force pulling the liquid.

Connect two pistons of different areas with a liquid. The small piston produces an increase in pressure, transmitted to the large piston by Pascal's principle. Under the large piston, that same pressure acts on more equal patches of area, so the total force is greater. Call the forces F₁ and F₂ and the areas S₁ and S₂. This device is a hydraulic machine. It does not multiply pressure between the pistons: the force gain comes from the difference in area.

For ideal pistons at the same height, the added pressures are equal: F₁/S₁ = F₂/S₂. Thus F₂/F₁ = S₂/S₁: doubling the large piston's area at the same pressure doubles its force. Forces are measured in N and areas in m². We assume slow motion and equal atmospheric pressure outside, neglecting friction and piston masses. If the pistons are at different heights, we must include the familiar hydrostatic pressure difference.

If S₂ = 4S₁, then F₂ = 4F₁. But how much liquid is needed to raise the large piston? For displacements s₁ and s₂, equal displaced volumes give S₁s₁ = S₂s₂. The large piston therefore travels four times less far. Multiply force by distance: the gain in one is offset by the loss in the other. Later we will call this product work; already we can see that the machine creates no extra energy.

Ideal pistons at the same height have areas of 2 cm² and 20 cm². An additional force of 30 N acts on the small piston. Find the force on the large piston.

The area ratio is 20/2 = 10. Identical units cancel, so there is no need to convert the areas here.

With equal added pressure, the force on the large piston is 10 times greater: F₂ = 30·10 = 300 Н.

The large piston travels 10 times less far than the small one because the liquid volume is conserved.

300 N; a tenfold force gain.

The atmosphere pushes on surfaces from all directions; a pressure difference explains why liquid rises in a pump.

A barometer converts atmospheric pressure into a column height or deformation of a sensing element.

In an ideal hydraulic press, F₂/F₁ = S₂/S₁, but the larger piston travels a correspondingly shorter distance.

Pressure differs between the bottom and top of an immersed object. The next lesson explains buoyant force.