Two persons of equal weight are hanging by their hands from the ends of a rope passing over a frictionless pulley. They begin to climb. One person can climb at twice the speed of the other. Who gets to the top first?
Try it first: pick an option, then read the explanation.
Asked in SSC JE 03/03/2017 (Morning)
Detailed Explanation
With a light rope over a frictionless pulley, the tension T is the same everywhere in the rope. Each person is acted on by only two forces: their weight W (down) and the rope tension T (up).
For each person: \( T - W = \dfrac{W}{g}\,a \Rightarrow a = g\left(\dfrac{T}{W} - 1\right) \)
T and W are the same for both, so \(a\) is the same for both.
Why “Both get there together” Is Correct
When the ‘faster’ climber pulls harder, the tension rises — but it rises equally on both sides of the rope, lifting the other person by exactly the same amount. Starting from rest at the same height with identical accelerations, the two always move together relative to the ground and reach the top at the same time. The faster climber just pulls more rope through his hands.
Why the Other Options Are Wrong
- A. Slower climber: Both have identical acceleration, so neither falls behind.
- C. Faster climber: Climbing faster relative to the rope only pulls more rope over; it does not give a larger ground acceleration.
- D. Cannot climb at all: Both can rise: pulling on the rope increases the tension above their weight.
Key Point
Same mass + same rope tension → same acceleration → both arrive together.
Key Concepts for Students
- Frictionless pulley: tension equal on both sides.
- Newton's third law: pulling down on the rope makes the rope pull you up.
- If the masses differed, the lighter person would reach the top first.
Practice more: this is question 4 in Friction MCQs – Part 1. Try the whole set and check every answer instantly.