Two pistons of a hydraulic machine have diameters of 30.0 cm and 2.5 cm. What is the force exerted on the larger pistion when 40 kg wt. is placed on the smaller piston? If the smaller pistion moves in through 6 cm, how much does the other pistion move out ?

Explanation

For smaller piston, Area `a_(1) = pi xx (1.25)^(2) cm^(2)`
For larger piston, Area `a_(2) = pi xx (15)^(2) cm^(2)`
Mechanical advantage at the larger piston is `(a_(2))/(a_(1))`
`therefore F_(2) = (a_(2))/(a_(1)) xx F_(1) = (pi(15)^(2))/(pi(1.25)^(2))xx 40` kg wt
`=(225)/(1.25 xx 1.25) xx 40 xx 9.8 N`
= 56,448 N
This is the force exerted on the larger piston.
The liquids are considered incompressible. Therefore volume covered by the movement of smaller piston inwards is equal to that moved outwards by larger piston.
`rArr l_(1) a_(1) = l_(2) a_(2)`
`rArr l_(2) =(a_(1))/(a_(2))l_(1) = ((1.25)^(2))/((15)^(2))xx 6 cm`
= 0.042 cm
So, the distance moved out by the larger piston is 0.042 cm.