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Problem 10.21 Constants A hydraulic press for compacting powdered samples has a large cylinder which is 10.0 cm in diameter, and a small cylinder with a diameter of 2.0 cm. A lever is attached to the small cylinder as shown in (Figure 1). The sample, which is placed on the large cylinder, has an area of 4.0 cm2 Figure < 1of1 > Provide Sample Hydraulic 100 cm fluid Small cylinder 2.0 cm

Answer 1

The magnitude of force acting on the small cylinder is equal to F_N In state equilibrium, all the torques about should add upto zero. F_N l - 290 (2l) = 0 F_N = 290 (2) = 580 N The pressure applied to the fluid (Through the small cylinder) P = 580/pi (2 times 10^-2/2)^2 rightwardsdoublearrow P = 1.85 times 10^4 Pa The force of fluid on large cylinder. F = P pi R^2 = 1.85 times 10^6 times pi (0.05)^2 rightwardsdoublearrow F = 1.45 times 10^4 Pa The pressure on the sample, P_s = F/A = 1.45 times 10^4 N/4 times 10^-4 m^2 P_s = 3.6 times 10^7 N/m^2

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