Question

An automobile sack is lifted by a hydraulic jack that consists of two pistons. The larger piston is 1m in diameter and small piston is 10cm in diameter. If W be the weight of the car, the minimum force is needed on small piston to lift the car is

Answer 1

To lift the car, the pressure must be atleast same in both pistons.

Diameter of larger piston = 1m = 100cm
Radius of larger piston, R = 100/2 cm = 50cm
Area(A₁) = πR² = π×50² = 2500π cm²

Diameter of smaller piston = 10cm
Radius of smaller piston, r = 10/2 cm = 5cm
Area(A₂) = πR² = π×5² = 25π cm²

Car of weight W is on larger piston,
Let the force on smaller piston required to lift it = F
Since pressure on both pistons is same,

$\frac{F}{A_2} = \frac{W}{A_1} \\ \\ \Rightarrow F=W \times \frac{A_2}{A_1} \\ \\ \Rightarrow F=W \times \frac{25 \pi }{2500 \pi } \\ \\ \Rightarrow F= \frac{W}{100}$

Minimum force required is W/100.

Answer 2

To lift the car, the pressure must be atleast same in both pistons.

Diameter of larger piston = 1m = 100cm
Radius of larger piston, R = 100/2 cm = 50cm
Area(A₁) = πR² = π×50² = 2500π cm²

Diameter of smaller piston = 10cm
Radius of smaller piston, r = 10/2 cm = 5cm
Area(A₂) = πR² = π×5² = 25π cm²

Car of weight W is on larger piston,
Let the force on smaller piston required to lift it = F
Since pressure on both pistons is same,

\begin{lgathered}\frac{F}{A_2} = \frac{W}{A_1} \\ \\ \Rightarrow F=W \times \frac{A_2}{A_1} \\ \\ \Rightarrow F=W \times \frac{25 \pi }{2500 \pi } \\ \\ \Rightarrow F= \frac{W}{100}\end{lgathered}A2​F​=A1​W​⇒F=W×A1​A2​​⇒F=W×2500π25π​⇒F=100W​​ 

Minimum force required is W/100

credits
>Neha