Question

The large piston in a hydraulic lift has an area of 250 cm2. What force must be applied to the small piston with an area of 25 cm2 in order to raise a car of mass 1500 kg?

Answer 1

Explanation:

Given that,

Area of large piston, $A_1=250\ cm^2$

Area of small piston, $A_2=25\ cm^2$

Force applied to large piston, $F_1=m_1g=1500\times 10=15000\ N$

It is required to find the force applied to the small piston. The pressure in each chamber must be equal so, that,

$\dfrac{F_1}{F_2}=\dfrac{A_2}{A_1}$

F₂ is force applied to smaller piston

$F_2=\dfrac{F_1A_1}{A_2}\\\\F_2=\dfrac{15000\times 250}{25}\\\\F_2=150000\ N\\\\F_2=150\ kN$

So, the force of 150 kN must be applied to the small piston.