In the adjoining figure, wire PQ is smooth, ring A has a mass 1kg and block B has a mass of 2 Kg .If system is released from rest with $\theta =60°$, find tension in the string (in Newton) and the values of these accelerations ( in m/s² ) at this instant.

M and Q are two fixed fixed points. Therefore,

MQ = constant =c

l= length of string = constant .

 In triangle $MQA, (l-y{)}^2=x^2+c^2$

Differentiating w.r.t time, we get

$2(l-y)\left(-\frac{dy}{dt}\right)=2x\left(+\frac{dx}{dt}\right)+0$

or

$$\begin{gathered} (l-y)\left(+\frac{dy}{dt}\right)=x\left(-\frac{dx}{dt}\right) \\ \frac{dy}{dt}=\left(\frac{x}{l-y}\right)\left(-\frac{dx}{dt}\right) \end{gathered}$$

y is increasing with time,

$+\frac{dy}{dt}=v_2$

x is decreasing with time

$-\frac{dx}{dt}=v_1\text{ and }\frac{x}{l-y}=\cos \theta$

Substituting these values in Eq. (ii), we have

$v_2=v_1\cos \theta$

 Further differentiating Eq. (i), we have

$\frac{d^{2}y}{d{t}^2}(l-y)-{\left(\frac{dy}{dt}\right)}^2=-\left[x·\frac{d^{2}x}{d{t}^2}+{\left(\frac{dx}{dt}\right)}^2\right]$

Just after the release, $v_1,v_2,\frac{dx}{dt}$ and $\frac{dy}{dt}$ all are zero. Substituting in Eq. (iii), we have,

$\frac{d^{2}y}{d{t}^2}=\left(\frac{x}{l-y}\right)\left(-\frac{d^{2}x}{d{t}^2}\right)$

Here,

$$\begin{gathered} \frac{d^{2}y}{d{t}^2}={\alpha }_2\text{ and }-\frac{d^{2}x}{d{t}^2}=a_1 \\ \frac{x}{l-y}=\cos \theta =\cos 60°=\frac{1}{2} \end{gathered}$$

Substituting in Eq. (iv), we have

$a_2=\frac{a_1}{2}$

 For A Equation is

$T\cos 60°=m_A{a}_1$or $\frac{T}{2}=\left(1\right)a_1=a_1$

For B

$20-T=m_B{a}_2$

$20-T=2a_2$

Solving Eqs. (v), (vi) and (vii), we get

$T=\frac{40}{3} \mathrm{N}\Rightarrow a_1=\frac{20}{3} \mathrm{m}/{\mathrm{s}}^2$

and

$a_2=\frac{10}{3} \mathrm{m}/{\mathrm{s}}^2$