Question

The hydraulic oil in a car lift has a density of 8.73 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 8.10 x 10-3 m and 0.101 m, respectively. What input force F is needed to support the 24000-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.10 m above that of the input plunger?

Answer 1

a )

given F₁ = 24000-N

we have equation F = P X A

P = F / A

but A₁ = 3.14 r²

A₁ = 3.14 X 0.101²

A₁ = 0.032 m²  

and

A₂ = 3.14 r²

A₂ = 3.14 X ( 8.10 X 10^-3 )²

A₂ = 2.06 X 10^-4 m²  

now F₁ / A₁ = F₂ / A₂

24000 / 0.032 = F₂ / ( 2.06 X 10^-4 )

F₂ = 24000 X ( 2.06 X 10^-4 ) / 0.032

F₂ = 154.5 N

b )

now taking equation

P_in = P_out + ( 8.73 X 10² X 9.8 X 1.1 )

F / A₂ = ( 24000 / A₁ ) + ( 9410.94 )

F / ( 2.06 X 10^-4 ) = ( 24000 / 0.032) + ( 9410.94 )

F / ( 2.06 X 10^-4 ) = 750000 + 9410.94

F / ( 2.06 X 10^-4 ) = 759410.94

F = 759410.94 X ( 2.06 X 10^-4 )

F = 156.43 N