Interactive Solution 11.35 presents one method for modeling this problem. Multiple-Concept Example 8 also presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of 8.04 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 8.61 x 10-3 m and 0.141 m, respectively. What input force F is needed to support the 23000-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?

a) Fi = 85.76 N

b) Fi = 87.8 N

Explanation:

Given:-

- Density of hydraulic oil, ρ = 804 kg/m^3

- The radius of input piston, ri = 0.00861 m

- The radius of output piston, ro = 0.141 m

Find:-

What input force F is needed to support the 23000-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

(b) the bottom surface of the output plunger is 1.10 m above that of the input plunger?

Solution:-

For part a.

- We see that both plungers are equal levels or there is no pressure due to elevation head. So we are only dealing with static pressure exerted by the hydraulic oil on both plungers to be equal. This part is an application of Pascal's Law:

                                  Pi = Po

                                  Fi / Ai = Fo / Ao

                                  Fi = Ai / Ao * Fo

                                  Fi = (ri/ro)^2 * Fo

                                  Fi = ( 0.00861 / 0.141 )^2 * 23000

                                  Fi = 85.76 N

For part b.

- We see that both plungers are at different levels so there is pressure due to elevation head. So we are only dealing with static pressure exerted by the hydraulic oil on both plungers plus the differential in heads. This part is an application of Bernoulli's Equation:

                                  Pi = Po + ρ*g*h

Where,                     h = Elevation head = 1.10m

                                  Fi = (ri/ro)^2 * Fo + ρ*g*h*π*ri^2

                                  Fi = 85.76 + (804*9.81*1.1*3.142*0.00861^2)

                                  Fi = 87.8 N

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