Question

In a hydraulic lift as shown in figure, mass of the car
is m = 1000 kg, area of pistons A1 = 20 cm2,
A2 = 5 m2. The minimum force Frequired to lift the
car is g = 10 m/s2​

Answer 1

The minimum force F required to lift the  car is 4 N.

Explanation:

1.  On Platform 1

    Area of platform $A_{1}= 20cm^{2}=20\times 10^{-4} m^{2}$

    Force on platform $F_{1}$= have to find out

2.  On Platform 2

    Area of platform $A_{2}= 5 m^{2}$

    mass on platform (m)= 1000 kg

    So

    Force on platform $F_{2}$= 1000×g=1000×10=10000 (N)

3. Pressure at platform 1 and platform 2 will be equal from pascal law due to same height of platform 1 and platform 2.

    So

    $\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}$   ...1)

4. Equation 1) can be written as

    $F_{1}=\frac{F_{2}}{A_{2}}\times A_{1}$     ...2)

   By putting corresponding value in equation 2)

   We get

   $F_{1}=\frac{10000}{5}\times 20\times 10^{-4}$

   On solving we get force required to lift car $F_{1}$= 4 (N)