In a hydraulic system with an input ram of diameter 0.75 inches carrying 100 lb, what is the approximate output force on a connected 2.0 inch diameter output ram (ideal hydraulics)?

In a hydraulic system with an input ram of diameter 0.75 inches carrying 100 lb, what is the approximate output force on a connected 2.0 inch diameter output ram (ideal hydraulics)?

• A. 56.50 pounds
• B. 710.74 pounds
• C. 226.35 pounds
• D. 177.41 pounds

Answer: (B)

Pascal’s law shows that in an ideal hydraulic system the fluid pressure is the same on both pistons, so the output force scales with the ratio of the piston areas. The input piston has diameter 0.75 in, so its area is A_in = π(d^2)/4 = π(0.75)^2/4 ≈ 0.4418 in^2. The output piston has diameter 2.0 in, so A_out = π(2)^2/4 = π ≈ 3.1416 in^2. The force scales with the area ratio, so F_out = F_in × (A_out/A_in) = 100 × (3.1416 / 0.4418) ≈ 100 × 7.111 ≈ 711 lb. With more precise calculation this comes to about 710.7–711 lb, matching the given option.

Pascal’s law shows that in an ideal hydraulic system the fluid pressure is the same on both pistons, so the output force scales with the ratio of the piston areas. The input piston has diameter 0.75 in, so its area is A_in = π(d^2)/4 = π(0.75)^2/4 ≈ 0.4418 in^2. The output piston has diameter 2.0 in, so A_out = π(2)^2/4 = π ≈ 3.1416 in^2. The force scales with the area ratio, so F_out = F_in × (A_out/A_in) = 100 × (3.1416 / 0.4418) ≈ 100 × 7.111 ≈ 711 lb. With more precise calculation this comes to about 710.7–711 lb, matching the given option.