Question

The small piston of hydraulic lift has an area of 0.20 m2. A car weighing 1.20 x le N sits on a rack mounted on the large piston. The large piston has an area of 0.90 m2. How large a force must be applied to the small piston to support the car??

Answer 1

Answer on Question 62027, Physics, Mechanics, Relativity

Question:

The small piston of hydraulic lift has an area of $0.20\,m^2$. A car weighing $1.2 \cdot 10^4\,N$ sits on a rack mounted on the large piston. The large piston has an area of $0.90\,m^2$. How large a force must be applied to the small piston to support the car?

Solution:

By the hydraulic press formula we have:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2},$$

here, $A_1$ is the area of the small piston, $A_2$ is the area of the large piston, $F_1$ is the force applied to the small piston, $F_2$ is the force applied to the large piston.

From this formula we can find the force applied to the small piston:

$$F_1 = F_2 \frac{A_1}{A_2} = 1.2 \cdot 10^4\,N \cdot \frac{0.20\,m^2}{0.90\,m^2} = 2.7 \cdot 10^3\,N.$$

Answer:

$$F_1 = 2.7 \cdot 10^3\,N.$$

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