Question

A hydraulic automobile lift has input and output pistons with diameters of $10cm$ and $30cm$ . The lift is used to hold up a car with a weight of $1.44\times 10^{4}N$ .
What is the force on the input piston?

• A. $1.6\times 10^{3}N$
• B. $1.5\times 10^{3}N$
• C. $1.4\times 10^{3}N$
• D. $1.8\times 10^{3}N$

Answer 1

Answer: (A)

Here, the diameter of the input piston, $d=10cm=0.1m$
The diameter of output piston, $D=30cm=0.3m$
The weight lifted by the output piston, $F=1.44\times 10^{4}N$
Let a and A be the areas of a cross-section of the input and output pistons respectively. If $f$ is the force applied on the input piston, then
$\frac{F}{A}=\frac{f}{a}$ or $\frac{F}{\frac{\pi D^2}{4}}=\frac{f}{\frac{\pi d^2}{4}}$
Or $f=F\times \frac{d^2}{D^2}=1.44\times 10^4\times \frac{0.1^2}{0.3^2}=1.6\times 10^{3}N$