At a roadside garage, a mechanic slides a small jack under a car. He pumps its handle with one hand, pushing with perhaps the weight of a school bag. Slowly the corner of a car weighing more than a tonne rises off the ground. His hand goes up and down many times for every centimetre the car rises.
Adding pressure to a liquid
In the last two lessons, the pressure at a depth \(h\) in a liquid was \(P = P_{\text{top}} + \rho g h\). Here \(P_{\text{top}}\) is the pressure at the surface. Now close the liquid in a vessel with a piston resting on its surface, and push on the piston. The push raises the pressure at the surface by \(\Delta P\).
What happens deeper down? A liquid is very hard to squeeze. So the push hardly changes its volume, the depths of its points, or its density. In \(P = P_{\text{top}} + \rho g h\), only \(P_{\text{top}}\) has changed. So the pressure at every point rises by the same \(\Delta P\), whatever its depth or place.
A change in pressure at one point of an enclosed fluid at rest reaches every point of the fluid, and its walls, undiminished.
Before the push, the gauges in Fig 4.1 read the pressure of the water above them. A is \(0.25\,\text{m}\) below the piston and reads \(2.5\,\text{kPa}\). B and C are deeper, at \(5.0\) and \(7.5\,\text{kPa}\). The push adds \(200/0.01 = 20\,000\,\text{Pa}\) under the piston. Every reading rises by exactly that, to \(22.5\), \(25.0\) and \(27.5\,\text{kPa}\). The differences between the gauges do not change.
"Undiminished" is true once the liquid is at rest again. The news of the push travels through the liquid as a wave of pressure, at the speed of sound in it, about \(1500\,\text{m/s}\) in water. A liquid does compress a very little, and the chapter on Stress, Strain and Hooke's Law measures how little, by the bulk modulus.
A closed bottle is completely full of water. You squeeze its side and raise the pressure there by \(5000\,\text{Pa}\). How does the pressure at the bottom of the bottle change?
A small force holds a large one
Join a narrow cylinder to a wide one with a pipe, fill them with oil, and fit a piston in each. This is a hydraulic lift. Push down on the small piston, of area \(A_1\), with a force \(F_1\). The pressure under it rises by \(F_1/A_1\). By Pascal's law the same pressure pushes up on the large piston, of area \(A_2\). So the force on the large piston is \(F_2 = (F_1/A_1)A_2\).
The force is multiplied by the ratio of the areas. A push of \(200\,\text{N}\) on \(10\,\text{cm}^2\) raises the pressure by \(2 \times 10^5\,\text{Pa}\). On a large piston of \(500\,\text{cm}^2\), that pressure pushes with \(2 \times 10^5 \times 0.05 = 10^4\,\text{N}\). It can hold up a load of \(1000\,\text{kg}\).
But not work for nothing
The oil cannot be squeezed, so the volume it leaves the small cylinder with equals the volume it brings to the large one. If the small piston moves down \(d_1\) and the large one up \(d_2\), then \(A_1 d_1 = A_2 d_2\). The large piston moves \(A_1/A_2\) as far as the small one. So the work done on the small piston, \(F_1 d_1\), equals the work done on the load, \(F_2 d_2\).
In a hydraulic lift the small piston has an area of \(10\,\text{cm}^2\) and the large one \(500\,\text{cm}^2\). A force of \(200\,\text{N}\) is applied to the small piston. Take \(g = 10\,\text{m/s}^2\). Find the force on the large piston and the mass it can lift. The small piston moves down \(50\,\text{cm}\); how far does the large one rise? Compare the work at the two pistons.
Pascal's law: the pressure added by the small piston reaches the large piston undiminished. The oil cannot be squeezed, so equal volumes move at the two pistons.
\(F_2 = F_1 \times A_2/A_1\), and \(A_1 d_1 = A_2 d_2\).
\(F_2 = 200 \times 50 = 10^4\,\text{N}\), which holds \(10^4/10 = 1000\,\text{kg}\). Then \(d_2 = 50 \times 10/500 = 1\,\text{cm}\). Work in \(= 200 \times 0.50 = 100\,\text{J}\); work out \(= 10^4 \times 0.01 = 100\,\text{J}\).
The force is multiplied fifty times, and the distance divided fifty times. The work is \(100\,\text{J}\) at both pistons.
A car of weight \(15\,000\,\text{N}\) stands on the large piston of a garage lift, of radius \(20\,\text{cm}\). The small piston has a radius of \(2\,\text{cm}\). What force on the small piston holds the car up, in newtons?
In a working hydraulic lift, which quantities are the same at the small piston and at the large piston?
A hydraulic lift trades distance for force. It never gives more work out than is put in; with friction, it gives a little less. Any problem where the large piston rises as far as the small one falls has gone wrong somewhere.
Pascal described the idea of the hydraulic press in the 1650s, but nobody could build one: the liquid leaked round the pistons. In 1795 the English lock-maker Joseph Bramah patented a press that worked. His workshop foreman, Henry Maudslay, gave the large piston a leather collar that the pressure itself pressed tight against the cylinder. The harder the press worked, the better it sealed.
The garage jack
A hydraulic jack is a small hydraulic lift. Take a small piston \(1.5\,\text{cm}\) across and a large one \(6\,\text{cm}\) across. The areas are in the ratio \((6/1.5)^2 = 16\). To lift a corner of a car pressing down with \(12\,000\,\text{N}\), the small piston needs \(12\,000/16 = 750\,\text{N}\). The handle is a lever that multiplies the push again, say ten times. The mechanic's hand then pushes with only about \(75\,\text{N}\).
The price is distance. To raise the car \(5\,\text{cm}\), the small piston must push \(16 \times 5 = 80\,\text{cm}\) of its stroke through the cylinder. It does so in many short strokes. A valve lets oil from a reservoir into the small cylinder on each upstroke and stops it flowing back. So the large piston climbs a little with every stroke and stays up between them.
Now the jack of the opening. The mechanic's push, the weight of a school bag, lifts a tonne because two machines multiply it: the lever and the oil. His hand moves far more than the car rises, in exactly the ratio that keeps the work the same.
Brakes
The brakes of a car are hydraulic. The pedal pushes a small piston in a master cylinder. Pipes full of brake oil carry the pressure to a wheel cylinder at each wheel. Suppose the pedal lever pushes the master piston, of \(2\,\text{cm}^2\), with \(200\,\text{N}\). The pressure rises by \(10^6\,\text{Pa}\). A wheel cylinder of \(6\,\text{cm}^2\) then presses the brake pad with \(600\,\text{N}\). By Pascal's law, every wheel gets the same pressure, so the car stops straight.
A mechanic says a car's brakes feel "spongy" because there is a bubble of air in the brake pipe. Explain why an air bubble makes the pedal go down a long way before the brakes grip.
Pascal's law passes a pressure on through a liquid because the liquid hardly compresses, so a push on the master piston at once raises the pressure everywhere. Air is easily compressed. When the pedal is pushed, the bubble shrinks first, and much of the pedal's travel goes into squeezing it. The pressure at the wheels rises slowly, so the pedal sinks far before the brakes grip. The mechanic bleeds the pipes to let the air out.
The yellow earth-movers seen on every road project in India dig and lift with hydraulic cylinders. An engine drives a pump that raises oil to a high pressure. Valves, worked by levers in the cab, send that pressure into long cylinders on the arm and the bucket. A cylinder of large area turns the pressure into a force of many tonnes, enough to tear up a road.
Heavy buses and trucks use compressed air in their brakes, not oil. The idea is the same, a pressure sent along a pipe to every wheel. That is why a bus hisses when the driver lets the brakes off: air escapes from the brake cylinders.
- Using the ratio of the diameters for the ratio of the forces.The forces go as the areas, and the areas as the squares of the diameters.
- Thinking a hydraulic lift multiplies energy.Equal volumes move at both pistons, so the work in equals the work out. Only the force is multiplied.
- Forgetting that the large piston moves a shorter distance.\(A_1 d_1 = A_2 d_2\): the large piston moves the area ratio less.
- Assuming Pascal's law holds for a gas as for a liquid.A gas compresses. The pressure still spreads through it at rest, but a push first squeezes the gas, as in spongy brakes.