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

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Solution

$P = \frac{F}{A} = \frac{3000 \times 9.8}{425 \times 10^{- 4}}$
$= 6.92 \times 10^{5} N / m^{2}$

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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 425cm2. What maximum pressure would the smaller piston have to bear?

P=FA=3000×9.8425×10−4=6.92×105N/m2

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

$P = \frac{F}{A} = \frac{3000 \times 9.8}{425 \times 10^{- 4}}$
$= 6.92 \times 10^{5} N / m^{2}$