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Question:

physics- A hydraulic lift is lifting a car that weighs 10 kN. The area of the piston supporting the car is A,?

A hydraulic lift is lifting a car that weighs 10 kN. The area of the piston supporting the car is A, the area of the other piston is a, and the ratio A/a is 105.7. How far must the small piston be pushed down to raise the car a distance of 1.7 cm? [Hint: Consider the work to be done.]wrong check mark m

Answer:

Wrong check mark m? does not compute. Ignore hint - it just confuses you as the mass of the pistons and frictional forces are not known. Compare volumes and assume the fluid volume does not change when it is heated up a bit. Big cylinder volume change going up 1.7cm is A*1.7 - same volume change in thin cylinder is x*a x/105.7*A eliminate A to get x105.7/1.7 cm
Use equation A1d1A2d2 Let capital A and A1 represent the piston holding the car d11.7 cm, convert to meters.017 little aother piston, A2 Given that A1/A2105.7 SOLVING FOR d2: Re-arranging equation: A1/A2d2/d1 105.7d2/.017 d21.8 m

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