A hydraulic automobile lift is designed to li | vedclass

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Question

(C) The maximum mass of the car to be lifted is $m = 3000 \; kg$.The area of the cross-section of the load-carrying piston is $A = 425 \; cm^{2} = 425

Vedclass · Accepted Answer

$6.917 	imes 10^{5} \; Pa$

Vedclass · Answer

(C) The maximum mass of the car to be lifted is $m = 3000 \; kg$.The area of the cross-section of the load-carrying piston is $A = 425 \; cm^{2} = 425 	imes 10^{-4} \; m^{2}$.The maximum force exerted by the load is $F = mg = 3000 	imes 9.8 = 29400 \; N$.The pressure exerted on the load-carrying piston is $P = \frac{F}{A}$.Substituting the values,$P = \frac{29400}{425 	imes 10^{-4}} = 6.917 	imes 10^{5} \; Pa$.According to Pascal's Law,pressure is transmitted equally in all directions in an enclosed fluid. Therefore,the maximum pressure that the smaller piston must bear is $6.917 	imes 10^{5} \; Pa$.

$A$ 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 $425 \; cm^{2}$. What maximum pressure would the smaller piston have to bear?

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