In a vehicles lifter, the enclosed gas exerts a force $F$ on a small piston of $8 \mathrm{~cm}$ diameter.…
In a vehicles lifter, the enclosed gas exerts a force $F$ on a small piston of $8 \mathrm{~cm}$ diameter.
The pressure is transmitted to a second piston of diameter $24 \mathrm{~cm}$. If the mass of the vehicle to be lifted is $1400 \mathrm{~kg}$, then $F$ must at least be $\left(g=10 \mathrm{~ms}^{-2}\right)$
$1600 \mathrm{~N}$
$1200 \mathrm{~N}$
$1800 \mathrm{~N}$
$700 \mathrm{~N}$
Solution
Given, force on small piston, $F_1=F$
Diameter of small piston, $d_1=8 \mathrm{~cm}$
Mass of vehicle placed on large piston (lift),
$
M=1400 \mathrm{~kg}
$
Diameter of large piston, $d_2=24 \mathrm{~cm}$
By using Pascal's law,
Equal pressure will be transmitted from small piston to lift
$
\begin{aligned}
\therefore \quad p & =\frac{F_1}{A_1}=\frac{F_2}{A_2} \Rightarrow \frac{F}{\frac{\pi d_1^2}{4}}=\frac{M g}{\frac{\pi d_2^2}{4}} \\
F & =M g\left(\frac{d_1}{d_2}\right)^2=1400 \times 10 \times\left(\frac{8}{24}\right)^2 \\
& =1556 \mathrm{~N} \\
& \simeq 1600 \mathrm{~N}
\end{aligned}
$