Question

To lift an automobile of $5000\text{kg}$, a hydraulic lift with a large piston $800{\text{cm}}^2$ in area is employed. Calculate the force that must be applied to a small piston of area $5{\text{cm}}^2$ to achieve it.

• A. $612.5\text{N}$
• B. $306.25\text{N}$
• C. $153.12\text{N}$
• D. $408.56\text{N}$

Answer 1

The correct option is B $306.25\text{N}$

Force on large pistion $\left(F_1\right)=mg$
$F_1=5000\times 9.8\text{N}=4.9\times {10}^4\text{N}$
Given area of large piston, $A_1=800{\text{cm}}^2$
Pressure on the large pistion $\left(P\right)$= $\frac{F_1}{A_1}=\frac{4.9\times {10}^4}{{800\times 10}^{-4}}=6.125\times {10}^5\text{Pa}$
Since the liquid transmits pressure equally, therefore the pressure on the small piston will be equal to the pressure on the large pistion.
$\frac{F_1}{A_1}=\frac{F_2}{A_2}=P$
Given area of small pistion $A_2=5{\text{cm}}^2$
$\therefore$ Force applied on small pistion$\left(F_2\right)=P\times A_2$
$=6.125\times {10}^5\times (5\times {10}^{-4})=306.25\text{N}$