In the arrangement shown in the figure,a_{1}, | vedclass

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(A) To find the relation between the accelerations,we use the principle of virtual work for constrained motion,which states that the sum of the dot pr

Vedclass · Accepted Answer

$4 a_{1}+2 a_{2}+a_{3}+a_{4}=0$

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(A) To find the relation between the accelerations,we use the principle of virtual work for constrained motion,which states that the sum of the dot product of tension and acceleration for all masses is zero: $\sum \vec{T} \cdot \vec{a} = 0$.Let the tension in the lowest string be $T$. Then the tension in the string supporting $m_{3}$ and $m_{4}$ is $T$. The tension in the string supporting the second pulley is $2T$,and the tension in the string supporting the first pulley is $4T$.Assuming all accelerations are directed downwards,the work done by tension on each mass is:For $m_{1}$: $-4T a_{1}$For $m_{2}$: $-2T a_{2}$For $m_{3}$: $-T a_{3}$For $m_{4}$: $-T a_{4}$Summing these,we get: $-4T a_{1} - 2T a_{2} - T a_{3} - T a_{4} = 0$.Dividing by $-T$,we obtain the relation: $4 a_{1} + 2 a_{2} + a_{3} + a_{4} = 0$.

In the arrangement shown in the figure,$a_{1}, a_{2}, a_{3}$ and $a_{4}$ are the accelerations of masses $m_{1}, m_{2}, m_{3}$ and $m_{4}$ respectively. Which of the following relations is true for this arrangement?

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