In the system shown in figure pulley is smooth. String is massless and inextensible. Find acceleration of the system and tensions $T_1$ and $T_2$ . $\left(g=10 \mathrm{m}/{\mathrm{s}}^2\right)$.

 

 

 Here, net pulling force will be

Weight of $4 \mathrm{kg}$ and $6 \mathrm{kg}$ blocks on one side - weight of $2 \mathrm{kg}$ block on the other side. Therefore,

a =

$$\begin{gathered} \frac{\text{ Net pulling force }}{\text{ Total mass }} \\ =\frac{(6\times 10)+(4\times 10)-\left(2\right)\left(10\right)}{6+4+2} \\ =\frac{20}{3} \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

For $T_1$, let us consider FBD of $2 \mathrm{kg}$block. Writing equation of motion, we get

$T_1-20=2a\text{ or }{T}_1=20+2\times \frac{20}{3}=\frac{100}{3} \mathrm{N}$ 

For $T_2$, we may consider FBD of $6 \mathrm{kg}$ block. Writing equation of motion, we get

$$\begin{gathered} 60-T_2=6a \\ {T}_2=60-6a=60-6\left(\frac{20}{3}\right)=\frac{60}{3} \mathrm{N} \end{gathered}$$