Question

Your answer is partially correct. Try again. 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 ol in a car lift has a density of 8.73 x 102 kg/m3. The weight of the input piston is negligible. The radili of the input piston and output plunger are 6.95 x 10-3 m and 0.194 m, respectively. What input force F is needed to support the 23600-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? (a) Numbe (b) Numbe Uni

Answer 1

a) Input piston radius = 0.00695m., output 0.194m.

Mechanical advantage = (0.194^2/0.00695^2) = 779.17:1.

F required = (23,600/779.17) = 30.29N.

b) Area of output piston = (pi r^2) = (pi*0.194^2) = 0.11824 m^2.

Volume of oil above input piston = (0.11824 x 1.2) = 0.1419m^3.

Mass of oil = (873 x 0.1419) = 123.87kg.

Weight of oil = (123.87 x g) = 1213.88N. (g = 9.8 used).

Total weight to lift = (23,600 + 1213.88) = 24,813.88N.

Input force = (23813.88/779.17), = 31.85N.