Question

LEARN MORE REMARKS In this problem, we didn't address the effect of possible differences in the helghts of the pistons. If the column of fluid is higher in the small piston, the fluid weight assists in supporting the car, reducing the necessary applied force. If the column of fluid is higher in the large piston, both the car and the extra fluid must be supported, so additional applied force is required. QUESTION if the radius of the output piston is doubled, the output force increases by a factor of: **** ********** * ***** * PRACTICE IT Use the worked example above to help you solve this problem. In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radius of r1 = 4.55 cm. This pressure is transmitted by an incompressible liquid to a second piston of radius 14.5 cm. (a) What force must the compressed air exert on the small piston in order to lift a car weighing 13,300 N? Neglect the weights of the pistons. (b) What air pressure will produce a force of that magnitude? Pa (c) Show that the work done by the input and output pistons is the same. (Do this on paper. Your Instructor may ask you to turn in this work.) EXERCISE HINTS: GETTING STARTED I'M STUCK! A hydraulic lift has pistons with diameters 8.20 cm and 37.0 cm, respectively. If a force of 825 N is exerted at the input (smaller) piston, what maximum mass can be lifted at the output piston? kg PE= Need Help? Read It Talk to a Tutor 10. -14 points SerCP9 9.AE.008.

Answer 1

practice it

a)

F1 / F2 = (d1/d2) ^2

F1 / 13300 = (4.55 / 14.5)^2

F1 = 1309.6 N

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b)

P = F1 / ( pi r1^2)

P = 1309.6 / ( 3.14 0.0455^2)

P = 2.01* 10^5 Pa

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exercise

similarly

F2 = 825* ( 37 / 8.2)^2

F2 = 16796.92 N

m = F2/g = 1713.97 kg

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