Question

The hydraulic oil in a car lift has a density of 8.17x10^2 kg/m^3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 8.56x10^-3 m and 0.171m, respectively. What input force F is needed to support the 27200-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 1.20m above that of the input plunger?

Answer 1

a) Input piston radius = 0.00856m., output 0.171m.
Mechanical advantage = (0.171^2/0.00856^2) = 399.065 : 1.
F required = (27200/399.065) = 68.15 N.
b) Area of output piston = (pi r^2) = 0.09181 m^2.
Volume of oil above input piston = (0.09181*1.2) = 0.1101 m^3.
Mass of oil = (817*0.1101) = 89.95 kg.
Weight of oil = (89.95*9.81) = 882.40N.
Total weight to lift = (27200 + 882.40) = 28082.4 N.
Input force = (28082.4/399.065), = 70.37 N.