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A hydraulic lift has a small piston with a diameter of 8 cm and a large piston with a diameter of 37.6 cm. What force must be applied to the small piston in order to lift a car on the large piston that weighs 4,803 N?

Answer :

Final answer:

To lift the car on the large piston, the force required on the small piston is approximately 601,875 N.

Explanation:

To find the force required to lift the car, we can use Pascal's law, which states that the pressure exerted at any point of an enclosed fluid is transmitted equally in all directions.

First, we can calculate the ratio of the areas of the two pistons using the formula A1/A2 = (D1)^2/(D2)^2, where D1 and D2 are the diameters of the small and large pistons, respectively. In this case, A1/A2 = (8 cm)^2/(37.6 cm)^2.

Next, we can use the formula for pressure P = F/A, where P is the pressure, F is the force, and A is the area. Since the pressure is equal in both pistons, we can set up the equation P1 = P2, which gives us F1/A1 = F2/A2. Rearranging, we can solve for F1, which is the force on the small piston.

Finally, we substitute the known values into the equation and solve for F1:

  1. Calculate the ratio of the areas: A1/A2 = (8 cm)^2/(37.6 cm)^2 = 0.004
  2. Set up the equation for pressure: F1/A1 = (4,803 N)/(0.004)
  3. Solve for F1: F1 = (4,803 N)/(0.004)

The force that must be applied to the small piston is approximately 601,875 N.

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