Law of the machine – maximum mechanical advantage For a lifting machine satisfying P = m W + C (law of the machine), what is the maximum possible mechanical advantage (MA) as the load W varies?

Introduction / Context:The law of the machine expresses the required effort P to lift load W as P = m W + C, where m and C lump the kinematic and frictional characteristics. Mechanical advantage MA is defined as W / P. This question asks for the maximum MA achievable over the operating range.

Given Data / Assumptions:

• Law of the machine: P = m W + C, with constants m > 0 and C ≥ 0.
• Mechanical advantage: MA = W / P.
• We consider W ≥ 0 and examine limiting behavior as W varies.

Concept / Approach:Compute MA(W) = W / (m W + C). Treat C as the frictional (or lost effort) term. For fixed m and C, MA increases with W because the constant C becomes relatively less important at larger loads. The theoretical upper bound occurs as W → ∞.

Step-by-Step Solution:

Verification / Alternative check:Differentiate MA(W) with respect to W: d(MA)/dW = C / (m W + C)^2 > 0 for C > 0, confirming monotonic increase and the asymptotic limit 1/m. For an ideal machine (C = 0), MA is identically 1/m for all W, consistent with the same bound.

Why Other Options Are Wrong:

• 1 + m, 1 − m, m: have incorrect dimensions and do not follow from MA = W/(mW + C).

Common Pitfalls:Confusing mechanical advantage with velocity ratio or efficiency. MA depends on load when C ≠ 0, whereas velocity ratio is purely kinematic.

Final Answer:1 / m