FAA Aviation Mechanic General (AMG)Physics, Math and Fluid LinesHard
A hydraulic actuator piston has a full bore area of 3.0 square inches on the blind side. The piston rod has a cross-sectional area of 0.75 square inch. When system pressure of 1,000 psi is applied to retract the actuator (pressure acting on the rod-side annular area), what retract force is produced?
- A2,250 lb
- B750 lb
- C3,000 lb
- D1,500 lb
Show answer & explanationAnswer & explanation
Correct answer: A. 2,250 lb
The effective retract-side area is the bore area minus the rod area: 3.0 − 0.75 = 2.25 sq in. Force = Pressure × Area = 1,000 × 2.25 = 2,250 lb.
Why the other options are wrong
- B. This uses only the rod area itself instead of the annular area.
- C. This uses the full bore area of 3.0 sq in without subtracting the rod area.
- D. This assumes an effective area of 1.5 sq in, an incorrect subtraction.
Differential Piston Area (Annular Area)
On the rod side of a double-acting actuator, the effective area is reduced by the rod's cross-sectional area, resulting in less force than the blind side for the same pressure.
- Effective rod-side area = Bore area − Rod area
- Force = Pressure × Effective Area
- Retract force is typically less than extend force at the same pressure due to reduced area
Memory trick: 'Rod steals some square inches' — subtract it before multiplying by pressure.