53.1: Hydraulics- The Hydraulic Press

The properties of liquids may be exploited to make it possible to lift large, heavy objects using a machine called a hydraulic press (Fig. $\PageIndex{1}$). Referring to the figure, we know by Pascal's law that pressure $P_{1}$ must be equal to pressure $P_{2}$ :

$$P_{1}=P_{2} \text {. }$$

Figure $\PageIndex{1}$: The hydraulic press. (Credit: HyperPhysics project, Georgia State University, Ref. [7]).

Therefore

$$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}$$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas of the pistons on the left and right. Solving for $F_{1}$, we find

$$F_{1}=F_{2} \frac{A_{1}}{A_{2}}$$

Since $A_{2}>A_{1}$, the force $F_{1}$ is multiplied by the factor $A_{2} / A_{1}$. One may place a heavy object like an automobile on the right, and lift it by applying a relatively small force on the left. The price for gaining this multiplication of force is that the piston on the left must be moved through a greater distance than the object on the right will be raised. To find the distance $d_{1}$ through which the piston on the left must be moved in order to lift the object on the right a distance $d_{2}$, we note that the liquid is essentially incompressible; therefore the volume change on the left must equal the volume change on the right:

$$A_{1} d_{1}=A_{2} d_{2}$$

Therefore the distance $d_{1}$ is

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

We can find $d_{1}$ in terms of the ratio of forces using Eq. $\PageIndex{3}$ to substitute for $A_{2} / A_{1}$; we get $$d_{1}=d_{2} \frac{F_{2}}{F_{1}}$$

The force in the small cylinder must be exerted over a much larger distance. A small force exerted over a large distance is traded for a large force over a small distance.

Example $\PageIndex{1}$

Figure $\PageIndex{2}$: An automobile on a hydraulic press. (Credit: HyperPhysics project, Georgia State University, Ref. [7]).

Suppose the piston on the left has a diameter of $10 \mathrm{~cm}$, and the piston on the right has a diameter of $1 \mathrm{~m}$. What force must be applied on the left to lift a 1000-kg automobile on the right? (See Fig. \(\PageIndex{2}\.)

Solution

The automobile has a weight $F_{2}=m g=(1000 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)=9.8 \times 10^{3} \mathrm{~N}$. The area $A_{1}=\pi r^{2} / 4=(\pi / 4)(0.1 \mathrm{~m})^{2}=(\pi / 4) \times 10^{-2} \mathrm{~m}^{2}$. The area $A_{2}=\pi r^{2} / 4=(\pi / 4)(1 \mathrm{~m})^{2}=\pi / 4 \mathrm{~m}^{2}$. The force $F_{1}$ is then

$$F_{1} =F_{2} \frac{A_{1}}{A_{2}}$$
$$=\left(9.8 \times 10^{3} \mathrm{~N}\right) \frac{\pi / 4 \times 10^{-2} \mathrm{~m}^{2}}{\pi / 4 \mathrm{~m}^{2}}$$
$$=98 \mathrm{~N}$$

In this case, the piston on the left must be pushed in $1 \mathrm{~m}$ to lift the car by $1 \mathrm{~cm}$.