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Question: An elevator accelerates upward at a constant rate. A uniform string of length \[\;L\;\]and mass\[M\]...

An elevator accelerates upward at a constant rate. A uniform string of length   L  \;L\;and massMMsupports a small block of massMMthat hangs from the ceiling of the elevator. The tension at distancellfrom the ceiling isTT. The acceleration of the elevator is
A. TM+mmlLg\dfrac{T}{{M + m - \dfrac{{ml}}{L}}} - g
B. T2M+mmlLg\dfrac{T}{{2M + m - \dfrac{{ml}}{L}}} - g
C. TM+mlLg\dfrac{T}{{M + \dfrac{{ml}}{L}}} - g
D. T2Mm+mlLg\dfrac{T}{{2M - m + \dfrac{{ml}}{L}}} - g

Explanation

Solution

Here the elevator moves with a constant acceleration upwards. The elevator consists of a small block and a string inside it. Therefore the string and the block also move with the same constant acceleration. They have said that a small part of the string with length ll has tension TT. We need to find out the equation of motion of both the parts of the string in order to find the acceleration of the elevator.

Complete step by step solution:

First, we need to find the equation of motion of the lower part. That is the part of the string whose length is LlL - lalong with the small block of the mass MM
We know the total mass of the string is mm.
The mass of the string of length LlL - l is given by,
mLl=mL(Ll){m_{L - l}} = \dfrac{m}{L}(L - l)
Here,mLl{m_{L - l}} is the mass of the part of the string whose length is LlL - l
mm is the total mass of the string inside the elevator.
LL is the total length of the string inside the elevator.
We know the equation of motion,
Tmg=maT - mg = ma
We are using TT in the place of a force because tension is a kind of force that acts on the string. We can draw the free body diagram of above mentioned lower part as

The mass of the part of the string whose length is LlL - l is mLl=mL(Ll){m_{L - l}} = \dfrac{m}{L}(L - l). We also know the mass of the small block isMM.
Substituting this in the equation of motion,
TmL(Ll)gMg=[mL(Ll)+M]aT - \dfrac{m}{L}(L - l)g - Mg = \left[ {\dfrac{m}{L}(L - l) + M} \right]a
Rearranging the above equation to find acceleration we get,
\dfrac{{T - \left[ {\dfrac{m}{L}(L - l) + M} \right]g}}{{\dfrac{m}{L}(L - l) + M}}$$$$ = a
If we expand and write the above equation we get,
TmL(Ll)+Mg=a\dfrac{T}{{\dfrac{m}{L}(L - l) + M}} - g = a
TM+mmlLg=a\dfrac{T}{{M + m - \dfrac{{ml}}{L}}} - g = a
Therefore the correct option is A.

Note:
Note that we have found the acceleration only for the lower part and not for the elevator. But we can say that the elevator will also have the same acceleration because the string is inside the elevator therefore both string and the elevator will have the same acceleration.