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Q.
In the figure shown, $m_{1}=10\ kg, m_{2}=6\ kg, m_{4}=4 \ kg .$ If $T_{3}=40 \ N, T_{2}=$?
ManipalManipal 2010
Solution:
Let $\alpha$ be the acceleration of each block. Then,
$T_{3}=\left(m_{1}+m_{2}+m_{3}\right) a \ldots$ (i)
and $T_{2}=\left(m_{1}+m_{2}\right) a \ldots$ (ii)
From Eqs. (i) and (ii), we get
$T_{2}=\left(\frac{m_{1}+m_{2}}{m_{1}+m_{2} \pm m_{3}}\right) \times T_{3}$
$=\left(\frac{10+6}{10+6+4}\right) \times 40=32 \,N$
