14.3.5 The Hydraulic Lift

The use of pressurized liquids to do useful work is a technology known as hydraulics. Pascal’s principle is the fundamental idea underlying hydraulic devices. If you increase the pressure at one point in a liquid by pushing a piston in, that pressure increase is transmitted to all points in the liquid. A second piston at some other point in the fluid can then push outward and do useful work.

The brake system in your car is a hydraulic system. Stepping on the brake pushes a piston into the master brake cylinder and increases the pressure in the brake fluid. The fluid itself hardly moves, but the pressure increase is transmitted to the four wheels where it pushes the brake pads against the spinning brake disk. You’ve used a pressurized liquid to achieve the useful goal of stopping your car.

One advantage of hydraulic systems over simple mechanical linkages is the possibility of force multiplication. To see how this works, we’ll analyze a hydraulic lift, such as the one that lifts your car at the repair shop. FIGURE 14.16a shows force $\vec{F}_{2}$ , due to the weight of the car, pressing down on a liquid via a piston of area $A_{2}$ . A much smaller force $\vec{F}_{1}$ presses down on a piston of area $A_{1}$ . Can this system possibly be in equilibrium?

The figure labeled ("a") illustrates a hydraulic lift.

The figure labeled (b) illustrates a hydraulic lift.

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As you now know, the hydrostatic pressure is the same at all points along a horizontal line through a fluid. Consider the line passing through the liquid/piston interface on the left in Figure 14.16a. Pressures $p_{1}$ and $p_{2}$ must be equal, thus

p_{0} + \frac{F_{1}}{A_{1}} = p_{0} + \frac{F_{2}}{A_{2}} + ρ g h

(14.9)

The atmosphere presses equally on both sides, so $p_{0}$ cancels. The system is in equilibrium if

F_{2} = \frac{A_{2}}{A_{1}} F_{1} - ρ g h A_{2}

(14.10)

NOTE Force $\vec{F}_{2}$ is the force of the heavy object pushing down on the liquid. According to Newton’s third law, the liquid pushes up on the object with a force of equal magnitude. Thus $F_{2}$ in Equation 14.10 is the “lifting force.”

Suppose we need to lift the car higher. If piston 1 is pushed down distance $d_{1}$ , as in FIGURE 14.16b, it displaces volume $V_{1} = A_{1} d_{1}$ of liquid. Because the liquid is incompressible, $V_{1}$ must equal the volume $V_{2} = A_{2} d_{2}$ added beneath piston 2 as it rises distance $d_{2}$ . That is,

d_{2} = \frac{d_{1}}{A_{2} / A_{1}}

(14.11)

The distance is divided by the same factor as that by which force is multiplied. A small force may be able to support a heavy weight, but you have to push the small piston a large distance to raise the heavy weight by a small amount.

This conclusion is really just a statement of energy conservation. Work is done on the liquid by a small force pushing the liquid through a large displacement. Work is done by the liquid when it lifts the heavy weight through a small distance. A full analysis must consider the fact that the gravitational potential energy of the liquid is also changing, so we can’t simply equate the output work to the input work, but you can see that energy considerations require piston 1 to move farther than piston 2.

EXAMPLE 14.7 Lifting a car

The hydraulic lift at a car repair shop is filled with oil. The car rests on a 25-cm-diameter piston. To lift the car, compressed air is used to push down on a 6.0-cm-diameter piston. What does the pressure gauge read when a $1300 kg$ car is $2.0 m$ above the compressed-air piston?

MODEL Assume that the oil is incompressible. Its density, from Table 14.1, is $900 kg / m^{3}$ .

SOLVE $F_{2}$ is the weight of the car pressing down on the piston: $F_{2} = m g = 12, 700 N$ . The piston areas are $A_{1} = π (0.030 m)^{2} = 0.00283 m^{2}$ and $A_{2} = π (0.125 m)^{2} = 0.0491 m^{2}$ . The force required to hold the car at height $h = 2.0 m$ is found by solving Equation 14.9 for $F_{1} :$

\begin{matrix} F_{1} & = & \frac{A_{1}}{A_{2}} F_{2} + ρ g h A_{1} \\ = & \frac{0.00283 m^{2}}{0.0491 m^{2}} \times 12, 700 N + (900 k g / m^{3}) (9.8 m / s^{2}) (2.0 m) (0.00283 m^{2}) \\ = & 782 N \end{matrix}

The pressure applied to the fluid by the compressed-air piston is

\frac{F_{1}}{A_{1}} = \frac{782 N}{0.00283 m^{2}} = 2.76 \times 10^{5} Pa = 2.7 atm

This is the pressure in excess of atmospheric pressure, which is what a pressure gauge measures, so the gauge reads, depending on its units, $276 kPa or 2.7 atm$ .

REVIEW $782 N$ is roughly the weight of an average adult man. The multiplication factor $A_{2} / A_{1} = 17$ makes it quite easy for this much force to lift the car.