Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - Problems: 21

Answer

The pressure on the sample is $4.0\times 10^7~Pa$

Work Step by Step

The lever rotates about an axis of rotation. The sum of the torque about this axis equals zero. We can find the force on the small cylinder as: $(320~N)(2L)-F(L) = 0$ $F = 640~N$ We can find the pressure on the small cylinder as: $P = \frac{F}{A}$ $P = \frac{640~N}{\pi~(1.0\times 10^{-2}~m)^2}$ $P = 2.037\times 10^6~N/m^2$ By Pascal's Principle, the same pressure is exerted on the large cylinder. We can find the force on the large cylinder. $F = P~A$ $F = (2.037\times 10^6~N/m^2)(\pi)(5.0\times 10^{-2}~m)^2$ $F = 1.60\times 10^4~N$ We can find the pressure on the sample as: $P = \frac{F}{A}$ $P = \frac{1.60\times 10^4~N}{4.0\times 10^{-4}~m^2}$ $P = 4.0\times 10^7~Pa$ The pressure on the sample is $4.0\times 10^7~Pa$.
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