Question

The diameter of the two piston of a hydraulic pressure are 17 mm & 340 mm . What multiplication of force is possible ? How much force muat be excerted at a smaller piston to support an automobile weighing 5000kg on the bigger piston ? How far muat be smaller piston be displaced for the larger piston to have a displacement of 2 cm ?

Answer 1

Dear Student ,
Here in this case look the diameter of the two piston are 17 mm and 340 mm .
So the radii of the two piston are , $r_1=\frac{17}{2}mmand{r}_2=\frac{340}{2}mm$
$$\begin{gathered} Now,themultiplicationofforceis, \\ \frac{A_2}{A_1}=\frac{\pi {\left({\displaystyle \frac{17}{2}}\times {10}^{-3}\right)}^2}{\pi {\left(\frac{340}{2}\times {10}^{-3}\right)}^2}=2·5\times {10}^{-3} \\ Nowifthemassofthecaris5000kgthen,theforceonthelargerpistonis, \\ {F}_2=\frac{A_2}{A_1}{F}_1=2·5\times {10}^{-3}\times 5000\times 10=125N \\ So,P_2=\frac{F_2}{A_2}=\frac{125}{\pi \times (0·17{)}^2}=1376·77Pa \\ Nowletthesmallerppistonbeplaced{d}_{1}dis\tan ceapartso, \\ {d}_1=\frac{A_2}{A_1}{d}_2=2·5\times {10}^{-3}\times 2\times {10}^{-2}=5\times {10}^{-5}m \end{gathered}$$
Regards