Question

The small piston of a hydraulic lift has an area of 0.20m sq . A car weighing 1.20×10 raise to power 4 N sits on a rack mounted on the large piston . The large piston has an area of 0.90m sq. How large a force must be applied to the small piston to support the car ?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Pressure},\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}} \\ {\mathrm{P}}_1={\mathrm{P}}_2 \\ \frac{{\mathrm{F}}_1}{{\mathrm{A}}_1}=\frac{{\mathrm{F}}_2}{{\mathrm{A}}_2} \\ {\mathrm{F}}_1=\frac{1.2\times {10}^4}{0.9}\times 0.2 \\ {\mathrm{F}}_1=2.7\times {10}^3\mathrm{N} \end{gathered}$$