For a two-cylinder locomotive (cranks at right angles) with fraction c of reciprocating mass balanced on each side, which expression correctly represents the periodic tractive force along the line of stroke in terms of m, ω, r, and crank angle θ?

Introduction / Context

The net tractive (longitudinal) force in a locomotive varies cyclically due to the inertia of reciprocating parts in each cylinder. Partial balancing introduces rotating counterweights equal to a fraction c of the reciprocating mass, modifying the amplitude of the horizontal resultant.

Given Data / Assumptions

• Two identical cylinders with cranks at 90° (right angle).
• Reciprocating mass per side: m; balanced fraction: c.
• Crank radius r; crank speed ω; crank angle θ for one side.

Concept / Approach

The unbalanced horizontal inertia force for one cylinder varies as m·ω²·r cosθ. With cranks at 90°, the other contributes m·ω²·r sinθ (phase-shifted). Partial balancing removes fraction c of each component (via rotating counterweights), leaving (1 − c) times the original components. The net tractive-force fluctuation is proportional to their algebraic combination along the line of stroke.

Step-by-Step Solution

Verification / Alternative check

Setting c = 0 (no balancing) gives F = m·ω²·r (cosθ − sinθ), matching the unbalanced case; increasing c reduces the amplitude linearly, confirming the (1 − c) factor.

Why Other Options Are Wrong

• m·ω²·r cos θ: ignores the second cylinder and balancing.
• c·m·ω²·r sin θ: uses c instead of (1 − c) and omits the cosθ term.
• m·ω²·r (cos θ − sin θ): correct shape but corresponds to c = 0 (no balancing).
• (1 + c)m·ω²·r (cos θ + sin θ): incorrect sign and amplitude; does not model partial balancing.

Common Pitfalls

• Forgetting the 90° phase shift between the two cranks.
• Applying the balancing fraction with the wrong sign or to the wrong components.

Final Answer

(1 - c)m·ω²·r (cos θ - sin θ)