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?

Difficulty: Medium

Correct Answer: 1 / m

Explanation:


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:

Write MA(W) = W / (m W + C).Divide numerator and denominator by W (for W > 0): MA(W) = 1 / (m + C/W).As W increases without bound, C/W → 0.Limit as W → ∞: MA_max = 1 / m.


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

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