A lift of weight W is lifted by a rope with an acceleration f. If the cross-sectional area of the rope is A, the stress in the rope is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 23/01/2018 (Morning)
Detailed Explanation
When a body accelerates, it is no longer in static equilibrium. Apply Newton's second law along the direction of motion (or add an inertia force and use equilibrium — d'Alembert's principle).
Mass of lift \( m = W/g \). Taking upward as positive:
\[ T - W = \frac{W}{g}\,f \;\Rightarrow\; T = W\left(1 + \frac{f}{g}\right) \]
\[ \sigma = \frac{T}{A} = \frac{W(1 + f/g)}{A} \]
Why “W(1 + f/g)/A” Is Correct
Accelerating upward, the rope must carry the weight plus the extra force needed to accelerate the mass, so the stress is W(1 + f/g)/A, option (a). If the lift accelerated downward, it would be W(1 − f/g)/A.
Why the Other Options Are Wrong
- B. (1 − g/f)/A: Missing W and has the wrong ratio (g/f), so its dimensions are not those of stress.
- C. W(2 + f/g)/A: The factor 2 would mean twice the weight even with zero acceleration.
- D. W(2 + g/f)/A: Wrong factor and wrong ratio; with f = 0 it does not reduce to W/A.
Key Point
Rope tension, lift accelerating up: W(1 + f/g); down: W(1 − f/g).
Key Concepts for Students
- With f = 0 (uniform speed), tension = W.
- Free fall (f = g downward): tension becomes zero.
- d'Alembert's principle: add an inertia force (W/g)f opposite to the acceleration.
Practice more: this is question 3 in General Conditions of Equilibrium MCQs – Part 1. Try the whole set and check every answer instantly.