Question

A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connected to a larger cylinder with a 24-cm diameter. If a 50-kg woman puts all her weight on the handle of the smaller piston, what weight could the other piston lift?

Answer 1

Answer:

Approximately $3.1 \times 10^4 \; \rm N$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm kg \cdot N^{-1}$.)

Explanation:

Let $A_1$ denote the first piston's contact area with the fluid. Let $A_2$ denote the second piston's contact area with the fluid.

Similarly, let $F_1$ and $F_2$ denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

$F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N$.  

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to $\pi$ times its radius, squared:

• $\displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}$.
• $\displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}$.

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

$\displaystyle P = \frac{F}{A}$.

For the pressures on the two pistons to match:

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}$.

$F_1$, $A_1$, and $A_2$ have all been found. The question is asking for $F_2$. Rearrange this equation to obtain:

$\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}$.

Evaluate this expression to obtain the value of $F_2$, which represents the force on the piston with the larger diameter:

$\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}$.