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 cm2. What maximum pressure would the smaller piston have to bear?
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Here,
maximum mass can be lifted ( m) = 3000 Kg
cross section area (A) = 425cm²
= 4.25 × 10^-2 m²
maximum pressure on the bigger piston ,
P = F/A = mg/A
= 3000×9.8/4.25×10^-2
= 6.92 × 10^5 Pa
According to Pascal law ,
the pressure applied on an enclosed liquid is transmitted equally in all directions .
So, maximum pressure on small piston = maximum pressure on bigger piston = 6.92 × 10^5 Pa
maximum mass can be lifted ( m) = 3000 Kg
cross section area (A) = 425cm²
= 4.25 × 10^-2 m²
maximum pressure on the bigger piston ,
P = F/A = mg/A
= 3000×9.8/4.25×10^-2
= 6.92 × 10^5 Pa
According to Pascal law ,
the pressure applied on an enclosed liquid is transmitted equally in all directions .
So, maximum pressure on small piston = maximum pressure on bigger piston = 6.92 × 10^5 Pa
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