Dynamics of Lifting – Stress in the Rope A lift of weight W is hoisted upward with constant acceleration f. The supporting rope has uniform cross-sectional area A. Neglect rope mass. What is the tensile stress developed in the rope?
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A(W / A) * (1 + f / g)
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B(W / A) * (1 - f / g)
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CW / A
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D(W / A) * (g / f)
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E(W / A) * (1 + g / f)
Answer
Correct Answer: (W / A) * (1 + f / g)
Explanation
Introduction / Context:When accelerating a suspended load, the rope must provide both the weight support and the additional force required for acceleration. This problem connects Newton second law with normal stress definition.
Given Data / Assumptions:
- Lift weight (force) = W.
- Upward acceleration = f, gravitational acceleration = g.
- Rope mass negligible, cross-sectional area = A.
- One-dimensional vertical motion, no damping or pulley friction.
Concept / Approach:
Let T be rope tension. For upward acceleration, net upward force equals mass times acceleration. Stress sigma equals tension divided by area.
Step-by-Step Solution:
Mass of the lift: m = W / gApply Newton second law upward: T - W = m * fSubstitute m: T - W = (W / g) * fSolve for tension: T = W * (1 + f / g)Stress in the rope: sigma = T / A = (W / A) * (1 + f / g)Verification / Alternative check:
For f = 0, sigma reduces to W / A, as expected for static lifting. For downward acceleration of magnitude f, the sign changes to 1 - f/g.
Why Other Options Are Wrong:
(W / A) * (1 - f / g) is for downward acceleration. W / A ignores inertia. (W / A) * (g / f) and (W / A) * (1 + g / f) are dimensionally inconsistent for small f and unphysical at f approaching zero.
Common Pitfalls:
Confusing weight W with mass, forgetting to divide by g, or using downward sign convention incorrectly.
Final Answer:
(W / A) * (1 + f / g)