question_answer 8)

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is \[425\,c{{m}^{2}}\]. What maximum pressure would smaller piston have to bear?

Answer:

The maximum force, which the bigger piston can bear, \[F=3000\,kg\,f=3000\times 9.8\,N\] Area of piston, \[A=425\,c{{m}^{2}}=425\times {{10}^{-4}}\,{{m}^{2}}\] \[\therefore \] Maximum pressure on the bigger piston, \[\times \left[ \frac{1}{1\cdot 5\times {{10}^{-3}}}-\frac{1}{3\times {{10}^{-3}}} \right]\] Since the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is \[=4\cdot 97\times {{10}^{-3}}m\]