Question
A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000kg$. The area of cross-section of the piston carrying the load is $425cm^2$. What maximum pressure would the smaller piston have to bear?
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Answer
The maximum mass of a car that can be lifted, $m=3000 \mathrm{~kg}$. Area of cross-section of the load-carrying piston, $\mathrm{A}=$ $425 \mathrm{~cm}^2=425 \times 10^4 \mathrm{~m}^2$ The maximum force exerted by the load, $\mathrm{F}=\mathrm{mg}=3000 \times 9.8=29400 \mathrm{~N}$ The maximum pressure exerted on the load-carrying piston, $\mathrm{P}=\mathrm{F} / \mathrm{A} \frac{29400}{(425 \times 10-4)}=6.917 \times 10^5 \mathrm{~Pa}$ Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is $6.917 \times$ $10^5 \mathrm{~Pa}$.
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