previous 15 of 21 next A2 (0.20 m) So a force of 810 N can lift a car weighing 9
ID: 1795605 • Letter: P
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previous 15 of 21 next A2 (0.20 m) So a force of 810 N can lift a car weighing 9800 N Let's see how Pascal's law can be used to amplify a force. Suppose the hydraulic lift shown in (Figure 1) has a small cylindrical piston with radlus 5.0 cm and a larger piston with radlus 20 cm. The mass of a car placed on the larger plston's platform is 1000 kg. (a) Assuming that the two pistons are at the same height, what force must be applied to the small piston to lift the car? (b) How far must the small piston move down to lift the car through a height of 0.10 m? Part (b): If the small piston moves a distance di, the volume of fluid displaced is V diA1. Because the total volume of must equal the volume ½ displaced by the large piston. It follows that di Ai-da A2, t the car is hoisted up a distance d2 piston moves a distance di given by 010 m, tre emal mds = r(0.20 m)' (0.10 m) (0.05 Im) 1.6 m ALTERNATIVE SOLUTION Part (b) may also be solved by using conservation of energy. If we ignore friction, the amount of work done by the applied force must equal the work done to lift the car. Thus (610 N)d, (9800 N) (0.10 m) = d, = 1.6 m REFLECT The hydraulic lift is a force-amplifying device. The amplification factor is the ratio of the areas of the pistonslt seems as though wo get something for nothing. But the much greater weight is lifted through a much smaller distance, and the two quantities of work are the same in magnitude. The device amplifies force, but it can't amplify energy Figure 1 of 1 Part A Practice Problem: Acting on a pissoe with a large area the pressure creates a force that can support a car Engineers are designing a hydraulic lift that can be used to repair trucks with a mass of 7400 kg. What ratio of piston areas would you suggest for the design if the maximum available external force is to be 3200 N? Express your answer to three significant figures. A2/A = My Answers Give Up 3At any siven height, the pressure p is the same everywhere in the fluid (Pascal's law). SubmitExplanation / Answer
For hydraulic lift ,pressure remains constant in both pistons ,therefore
F2/F1 = A2/A1
=>A2/A1 =mg /F1=7400*9.8/3200
A2/A1 =22.7
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