Question

You are designing a hydraulic lift for an automobile garage. It will consist of two oil-filled cylindrical pipes of different diameters. A worker pushes down on a piston at one end, raising the car on a platform at the other end. (See (Figure 1).) To handle a full range of jobs, you must be able to lift cars up to 3000 kg , plus the 400 kg platform on which they are parked. To avoid injury to your workers, the maximum amount of force a worker should need to exert is 100 N .

Part A)What should be the diameter of the pipe under the platform?

Part B) If the worker pushes down with a stroke 40

cmlong, by how much will he raise the car at the other end?

Answer 1

Given that, Mass of Car = 3000 kg Mass of Plat form = 400 Mass of Car + Plat form = 3400 kg weight of Car + Plat form = 3400 times 9.8 N F_B = 3400 times 9.8 = 3.3320 N F_A = 100 N (Given) D_A = Diameter of pipe By pascal's law P_A = P_B = > F_A/F_B = F_B/A_B = > F_B/F_A = A_B/A_A. A_B/A_A = 33320/100 = 333.2 D^2_B/D^2_A = 333.2 = > D_B/D_A = squareroot 333.2 = 18.25 = > D_B = 18.25 D_A D_B = Diameter of plat form \r\n Given that worker pushes down with a stroke of 40 cm long, By formula, P_A h_A = P_B h_B = > h_B = P_A h_A/P_B h_B = 100 times 0.4/33320 h_B = 0.0012 m = 1.2 mm