Question

(i) A force of 50kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25cm respectively.
(ii) The diameter of neck and bottom of a bottle are 2cm and 10cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2kgf, what force is exerted on the bottom of the bottle?

Answer 1

i) hi...actually the pascals law says that pressure applied to any point in a liquid will be transmitted equally to all points in

liquids..no matter the force and area can differ..so keeping this statement;

let force and area of smaller piston be F and A ..

force and area of larger piston be F and A ..

radius of smaller piston=5/2=2.5cm

radius of large piston=25/2

since, pressure=force/area

so, F1 / A1=F2/ A2

A1=pie(r)2=pie(2.5)2cm

A2=Pie(r)2=Pie (12.5)2 cm

50/pie 2.5*2.5=f2/Pie12.5*12.5

50/pie 2.5*2.5=f2/12.5*12.5

F2=50*12.5*12.5/6.25=1250 kgf

hence force on large piston =1250kgf..:-))

Answer 2

i) Ratio of diameter of smaller piston to bigger piston = 5:25

.•. Ratio of area of smaller piston to bigger piston = 25:625

Force applied on smaller piston, F1 = 50kgf

Let F2 be the force on the bigger piston.

By the principle of hydraulic machine,

Pressure on narrow piston = Pressure on wider piston

$\frac{F1}{A1} = \frac{F2}{A2} \\ \\ \implies \: \frac{F1}{F2} = \frac{A1}{A2} \\ \\ \implies \: \frac{50}{F2} = \frac{25}{625} \\ \\ \implies \: F2 \: = \: \cancel{50} \: \: ²\times \frac{625}{ \cancel{25} } \\ \\ \ = 1250 \: kgf \: \: \: \: \: (ans)$