Question

The hydraulic oil in a car lift has a density of 8.81 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 5.07 x 10-3 m and 0.150 m, respectively. What input force F is needed to support the 27800-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 1.20 m above that of the input plunger?

Answer 1

a.
`F_2=31.76N`

b.
`F_2=185.86N`

Step-by-step explanation:

Given:

`F_1=27800N`

`r_1=5.07x10^(-3)m`

`r_2=0.150 m`

`p=8.81x10^2 kg/m^3`

Using the equation to find the force so replacing

a.

`F_1*A_2=F_2*A_1`

`A=\pi*r^2`

`F_2=F_1*(A_2)/(A_1)=27800*(\pi*(5.07x10^(-3)m)^2)/(\pi*(0.150m)^2)`

`F_2=31.76N`

b.

`F_2=F_1+F_p`

`F_2=27800*(\pi*(5.07x10^(-3)m)^2)/(\pi*(0.150m)^2)+(8.81x10^2kg/m^3*9.8m/s^2*1.20m*\pi*(5.07x10^(-3))m^2)`

`F_2=185.86N`