An elevator accelerates upwards at a constant | vedclass

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Question

Let $a$ be the acceleration of the lift Mass of lower portion of string

$=\frac{m}{L}(L-l)$

$	herefore T-M g-\frac{m g}{L}(L-l)=\left(M+\frac{m

Vedclass · Accepted Answer

$\frac{T}{{M\ +\ m\ - \frac{{ml}}{L}}} - g$

Vedclass · Answer

Let $a$ be the acceleration of the lift Mass of lower portion of string

$=\frac{m}{L}(L-l)$

$	herefore T-M g-\frac{m g}{L}(L-l)=\left(M+\frac{m}{L}(L-1)ight) a$

$	herefore a=\frac{T}{M+m-\frac{m l}{L}}-g$

An elevator accelerates upwards at a constant rate. A uniform string of length $L$ and mass $m$ supports a small block of mass $M$ that hangs from the ceiling of the elevator. The tension at distance $l$ from the ceiling is $T$ . The acceleration of the elevator is

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