Question

in hydraulic lift of mass m of the car is m=1000kg,area of piston A1=20cm square A2=5m square.the minimum force F required to lift the car is g= 10 m/s​

Answer 1

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

Explanation:

1. Given data

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

  Area of load  $(A_{2})=5 m^{2}$

  Value  of load  $(F_{2})$= m×g= 1000×10=10000 N

2. From Pascal Law

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

  where

  $F_{1}$ = Effort apply on piston

3. From equation 1)

   $\frac{F_{1}}{20\times 10^{-4}}=\frac{10000}{5}$   ...2)

   on solving equation 2)

   we get

   $F_{1}= 4 N$