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?

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:

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)