Answer to Question #262522 in Mechanical Engineering for Angel

Question

A simple hydraulic jack has a pump piston diameter 38.13 mm , and a load piston 190.65 mm . Calculate:

a) the effort force needed to lift a load of 1300 Newton

b) The distance moved by the load if the effort moves a distance 285

Answer 1

$A_1 = \frac{\pi}{4}d^2 = \frac{\pi}{4}38.13^2=1141.88 mm^2$

$A_2 = \frac{\pi}{4}d^2 = \frac{\pi}{4}190.65^2=28547.19 mm^2$

a). The effort force needed to lift a load $F_1$

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$F_1=\frac{F_2}{A_2}\times A_1$

$F_1=\frac{300}{28547.19}\times 1141.88 =11.99 N$

b) The distance moved by the load $l_2$

$F_1\times l_1=F_2\times l_2$

$l_2=\frac{F_1}{F_2}\times l_1$

$l_2=\frac{11.99}{1300}\times 285 =2.63 mm$