You are designing a hydraulic lift for an automobile garage. It will consist of

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 the figure .) To handle a full range of jobs, you must be able to lift cars up to 3000 kg , plus the 600 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 100N .

A) WHAT SHOULD BE THE DIAMETER OF THE PIPE UNDER THE PLATFORM? (m)

B) IF THE WORKER PUSHES DOWN WITH A STROKE OF 60.0cm LONG,BY HOW MUCH WILL HE RAISE THE CAR AT THE OTHER END?(cm)

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Explanation / Answer

A) cars + platform = 3600 kg to lift 1kg = 9.8 N 3600 kg x 9.8 N =35280 Newtons needed to lift cars and platform 35280 N / 100 N (worker force) =352.8 mechanical advantage is 1: 352.80 In other words, for every 1 Newton of force the worker pushes he gets the hydraulics to pick up 352.80 N. If you have a 1" pipe on the worker side the A= pi * r^2 so area = .785398 square inches you need to multiply the .785398 by the 352.80 mechanical advantage to get area needed for big lift cylinder .785398 x 352.80 = 277.088 square inches of area needed to get radius solve for r^2 277.088 sq in = 3.14 * r^2 r = 9.39 inch radius or about a 20 inch diameter cylinder, using a 1 inch dia cylinder for the drive side. B) Since the mechanical advantage of the power is 1: 352.80 then the lifting distance will be 1/352.80 of the distance of the power driven piston So if your stoke is 50 cm then ---> 60 * (1/352.80) = 0.1700 cm distance big lifting piston goes up

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