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The small piston of a hydraulic lift has an area of 0.20 m2 . A car weighing 1.2 x 104 N sits on a rack mounted on the large piston. The large piston has an area of 0.90 m2. How large force must be applied to the small piston to support the car?Group of answer choices2.7 x 10^3 N1.2 x 10^4 N2.1 x 10^3 N2.7 x 10^2 N

Question

The small piston of a hydraulic lift has an area of 0.20 m2 . A car weighing 1.2 x 104 N sits on a rack mounted on the large piston. The large piston has an area of 0.90 m2. How large force must be applied to the small piston to support the car?Group of answer choices2.7 x 10^3 N1.2 x 10^4 N2.1 x 10^3 N2.7 x 10^2 N

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Solution

The principle at work here is Pascal's principle, which states that a change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid. This is the principle that allows hydraulic lifts to work.

The formula for pressure is P = F/A, where P is pressure, F is force, and A is area.

We know the weight of the car (which is the force due to gravity), and the area of the large piston. We can use these to find the pressure in the fluid:

P = F/A = (1.2 x 10^4 N) / (0.90 m^2) = 1.33 x 10^4 N/m^2.

This pressure is transmitted undiminished to the small piston. So, we can set up the equation:

P = F/A

and solve for F, the force on the small piston:

F = P * A = (1.33 x 10^4 N/m^2) * (0.20 m^2) = 2.7 x 10^3 N.

So, a force of 2.7 x 10^3 N must be applied to the small piston to support the car.

This problem has been solved

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