In a hydraulic scenario with a 3 in^2 piston and a 2 in^2 rod on the rod side, at 3000 psi, what is the retracting force?

In a hydraulic scenario with a 3 in^2 piston and a 2 in^2 rod on the rod side, at 3000 psi, what is the retracting force?

• A. 3000 pounds
• B. 6000 pounds
• C. 9000 pounds
• D. 12000 pounds

Answer: (A)

The main idea is that the force in a hydraulic cylinder when retracting is pressure times the rod-side (annular) area. The rod side doesn’t have the full piston area because the rod blocks part of it, so the effective area on retract is piston area minus rod area. Here that’s 3 in² minus 2 in², giving 1 in². With 3000 psi, the retracting force is 3000 × 1 = 3000 pounds. (If it asked for extending force, you’d use the full piston area: 3000 × 3 = 9000 pounds.)

The main idea is that the force in a hydraulic cylinder when retracting is pressure times the rod-side (annular) area. The rod side doesn’t have the full piston area because the rod blocks part of it, so the effective area on retract is piston area minus rod area. Here that’s 3 in² minus 2 in², giving 1 in². With 3000 psi, the retracting force is 3000 × 1 = 3000 pounds. (If it asked for extending force, you’d use the full piston area: 3000 × 3 = 9000 pounds.)