In a hydraulic system two pistons sit at the same height and friction is negligible. A force of 500 N acts on a master cylinder 0.500 cm in diameter, and the pressure reaches a slave cylinder 2.50 cm in diameter. What force appears at the slave cylinder?
A2.50 x 10^3 N
B1.00 x 10^2 N
C5.00 x 10^4 N
D1.25 x 10^4 N
Answer & Solution
Correct answer: D. 1.25 x 10^4 N
1. Pascal's principle applied to a hydraulic system gives force one over area one equal to force two over area two.
2. Rearranged, the output force is the input force multiplied by the ratio of the areas.
3. Each area is pi times a radius squared, so the pi cancels and only the ratio of the squared radii survives.
4. Radii are half the diameters: 0.250 cm for the master and 1.25 cm for the slave.
5. Ratio of areas = (1.25 cm)^2 / (0.250 cm)^2 = 1.5625 / 0.0625 = 25.0.
6. Output force = (25.0)(500 N) = 1.25 x 10^4 N.
7. The value 2.50 x 10^3 N is the trap for scaling by the ratio of diameters, 5.00, rather than by its square.
8. The value 1.00 x 10^2 N is the force on the pedal before the lever multiplies it, and 5.00 x 10^4 N is four times too large.
_Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.5 Pascal's Principle_
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