Скачать или смотреть Problem 3.12| Intro to Mechanics| Kleppner and Kolenkow|JEE|NEET|11&12

In this video we solve Problem 3.12 (Capstan) from Intro to Mechanics by Kleppner & Kolenkow.

Problem statement:
A device called a capstan is used aboard ships to control a rope under great tension. The rope is wrapped around a fixed drum for several turns. The load pulls the rope with force TA, while the sailor holds it with a much smaller force TB. Show that
TB = TA · e^(−μθ),
where μ is the coefficient of friction and θ is the total angle (in radians) subtended by the rope on the drum.

Concise derivation (shown step-by-step in the video):
• Consider a small element of rope subtending dθ. The difference in tensions across the element is dT.
• Normal force on that element ≈ T dθ (by geometry), so the frictional force is dF = μ (T dθ).
• For equilibrium of the small element dT = −dF = −μ T dθ, so dT/T = −μ dθ.
• Integrating from TA to TB over total angle θ gives ln(TB/TA) = −μ θ, hence TB = TA · e^(−μθ).

Assumptions & notes:
• θ is in radians.
• μ is the (static) coefficient of friction between rope and drum.
• Result assumes the rope is on the verge of slipping (friction used to its limit) and the drum is fixed.

Why this matters:
This elegant result explains how a relatively small holding force can control a very large load — widely used in sailing, winches, and many engineering applications. Understanding the derivation develops skill in applying equilibrium to differential elements — a key technique in advanced mechanics and competitive-exam problems.

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