Question

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 8.68 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 5.09 x 10-3 m and 0.140 m, respectively. What input force F is needed to support the 21700-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 1.30 m above that of the input plunger?

Answer 1

Here, area of output plunger / area of input piston = (0.140^2/0.00509^2) = 756.52 : 1.

Combined weight of the car and the output plunger, m = 21700 N

(a) When the bottom surfaces of the piston and the plunger are at the same level,

the requisite force = (21700)*(1/ 756.52) = 28.68 N

(b) When the bottom surface of the output plunger is 1.30 m above that of the input plunger,

Area of output plunger = (pi r^2) = 3.141*0.140^2 = 0.0615636 m^2.
Volume of oil above input plunger = (0.0615636 x 1.30) = 0.08003268 m^3.
Density = 868 kg/m^3.

So, mass above input plunger = (868 x 0.08003268) = 69.5 kg.

So, the total weight = 21700 + (69.5 x g) = 22381.1 N

Therefore, the requisite input force = 22381.1 / 756.52 = 29.58 N