91Ó°ÊÓ

Gauge pressure is related to standard atmospheric pressure. When the atmospheric pressure on a system becomes zero, then gauge pressure will be equal to the absolute or total pressure.

The given data can be listed as,

• The diameter of hydraulic lift is, $\mathrm{d}=0.30\mathrm{m}$
• The mass of car is,$\mathrm{m}=1200\mathrm{kg}.$

The force applied on piston is given by,

$\mathrm{F}=\mathrm{pA}$

Here, p is the pressure, A is the area.

The force diagram for the piston is given by,

The net force can be expressed as,

$$\begin{gathered} \mathrm{pA}-\mathrm{w}-{\mathrm{p}}_{\mathrm{atm}}\mathrm{A}=0 \\ \mathrm{p}-{\mathrm{p}}_{\mathrm{atm}}=\frac{\mathrm{w}}{\mathrm{A}} \\ \mathrm{p}-{\mathrm{p}}_{\mathrm{atm}}=\frac{\mathrm{mg}}{{\mathrm{Ï€}}^2} \\ \mathrm{p}-{\mathrm{p}}_{\mathrm{atm}=\frac{\mathrm{mg}}{\mathrm{Ï€}{\left({\displaystyle \frac{\mathrm{d}}{2}}\right)}^2}} \end{gathered}$$

Substitute all values in the above equation.

$$\begin{gathered} \mathrm{p}-{\mathrm{p}}_{\mathrm{atm}}=\frac{1200\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left({\displaystyle \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}}\right)}{\mathrm{π}{\left({\displaystyle \frac{0.30\mathrm{m}}{2}}\right)}^2} \\ =\frac{11772\mathrm{N}}{0.07068{\mathrm{m}}^2}\left(\frac{1\mathrm{Pa}}{1\mathrm{N}/{\mathrm{m}}^2}\right) \\ =1.665×{10}^5\mathrm{Pa} \\ ≈1.7×{10}^5\mathrm{Pa} \end{gathered}$$

Thus, the gauge pressure required to lift the car is$1.7×{10}^5\mathrm{Pa}$.

It can be expressed in atmospheric pressure as,

Thus, the gauge pressure required to lift the car is .

$$\begin{gathered} \mathrm{p}-{\mathrm{p}}_{\mathrm{atm}}=1.7×{10}^5\mathrm{Pa}\left(\frac{1\mathrm{atm}}{1.013×{10}^5\mathrm{Pa}}\right) \\ =1.678\mathrm{atm} \\ \mathrm{Thus},\mathrm{the}\mathrm{gauge}\mathrm{pressure}\mathrm{required}\mathrm{to}\mathrm{lift}\mathrm{the}\mathrm{car}\mathrm{is}1.678\mathrm{atm}. \end{gathered}$$