91Ó°ÊÓ

It is given that,

$$\begin{gathered} m_1=3.0\mathrm{kg} \\ {\mathrm{θ}}_1=30° \\ {m}_2=2.0\mathrm{kg} \\ {\mathrm{θ}}_2=60° \end{gathered}$$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Also, this problem deals with the resolution of vectors. Since pulley is ideal, the tension across both sides of the string is the same and also the acceleration of both the objects would be same. Thus, using Newton’s second law of motion and concept of vector resolution, the tension can be calculated by writing and solving equation of motion for individual mass.

Formula:

$F_{net}=ma$ (i)

where, $F_{net}$ is the net force, Mis mass and a is an acceleration.

Free Body Diagram:

According to Newton’s second law:

$m_{1}g\sin \left({\mathrm{θ}}_1\right)-T=m_{1}a$ (ii)

$T-m_{2}g\sin \left({\mathrm{θ}}_2\right)=m_{2}a$ (iii)

By adding equations(ii) and (iii), T will cancel out to get a in terms of masses.

$m_{1}g\sin \left({\mathrm{θ}}_1\right)-m_{1}g\sin \left({\mathrm{θ}}_1\right)=\left(m_1+m_2\right)a$

Hence,

$$\begin{gathered} a=\frac{m_{1}g\sin \left({\mathrm{θ}}_1\right)-m_{2}g\sin \left({\mathrm{θ}}_2\right)}{\left(m_1+m_2\right)} \\ a=\frac{\left(3×9.8×\sin \left(30\right)\right)-\left(2×9.8×\sin \left(60\right)\right)}{\left(3+2\right)} \\ a=-0.45\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Negative sign indicates that mass $m_1$is moving upward.

Now, by plugging this value in equation (ii),

$$\begin{gathered} 3×9.8×\sin \left(30\right)-T=3×\left(-0.45\right) \\ 14.7-T=-1.35 \\ T=16\mathrm{N} \end{gathered}$$

Hence, the tension in the cord is$T=16\mathrm{N}$