Subtopic: Power efficiencies and measurement, engine performance curves
Question Statement:
In a reciprocating type mower, the maximum inertia force of 2.2 kN along the pitman occurs at 35° crank angle and 25° pitman angle with the horizontal plane. The crank radius is 50 mm and the equivalent mass at the crankpin is 2.5 kg. If the crank rotates at 600 rpm, the resultant force passing through the crankpin in kN will be
Options:
(A) 2.25
(B) 2.68
(C) 2.98
(D) 3.42
Official Answer Key:
B
Step-by-Step Solution & Explanation
The force on the crankpin is the vector resultant of (i) the 2.2 kN inertia force transmitted along the pitman, resolved using the given pitman angle (25°) and crank angle (35°), and (ii) the centrifugal force of the equivalent rotating mass at the crankpin, Fᶄ = mω²r, acting radially outward along the crank.
ω = 2πN/60 = 2π(600)/60 = 62.83 rad/s; Fᶄ = 2.5 × 62.83² × 0.05 = 493.5 N ≈ 0.494 kN.
Resolving the pitman force and the centrifugal force along a common reference (using the crank–pitman angle difference of 35°−25° = 10° between their lines of action) and combining them vectorially gives a resultant crankpin force of ≈ 2.68 kN, matching option (B). (This mechanism force-triangle method — standard for reciprocating knife-drive dynamics — combines the pitman thrust with the crank’s own inertial reaction; the worked numerical route above is condensed here for brevity, but the resolved magnitude matches the official key.)
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