An elevator accelerates upwards at a constant | vedclass

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(A) Let $a$ be the upward acceleration of the elevator.Consider the system consisting of the block of mass $M$ and the lower part of the string of len

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $a$ be the upward acceleration of the elevator.Consider the system consisting of the block of mass $M$ and the lower part of the string of length $(L-l)$.The mass of this lower part of the string is $m' = \frac{m}{L}(L-l)$.The total mass of the system is $M_{sys} = M + \frac{m}{L}(L-l)$.The forces acting on this system are the tension $T$ acting upwards and the weight $(M + m')g$ acting downwards.Applying Newton's second law: $T - (M + m')g = (M + m')a$.Substituting $m' = \frac{m}{L}(L-l)$:$T - (M + \frac{m}{L}(L-l))g = (M + \frac{m}{L}(L-l))a$.$T = (M + m - \frac{ml}{L})(a + g)$.Solving for $a$:$a = \frac{T}{M + m - \frac{ml}{L}} - g$.

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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