Two identical small masses each of mass m are connected by a light inextensible string on a smooth horizontal floor. A constant force F is applied at the mid point of the string as shown in the figure. The acceleration of each mass towards each other is,

Let the tension in the string be T at any angular position $\theta$, the acceleration of each ball along x and y axes be a and a₁ respectively. Writing the equation of motion of m, we obtain 

$\begin{array}{l}\sum F_x=ma\\ \Rightarrow T\cos \theta =ma ......\left(i\right)\\ \sum F_y=m{a}_1\end{array}$

$\Rightarrow T\sin \theta =m{a}_1 .......\left(ii\right)$

At point P, as it is accelerating with an acceleration a, therefore $F-2T\cos \theta =m_p\mathrm{a}$  where m_p= mass of the string at the point $P\cong 0$

$\begin{array}{l}\Rightarrow F=2T\mathrm{sos}\theta ........\left(iii\right)\\ \left(2\right)÷\left(1\right)\Rightarrow \tan \theta =\frac{a_1}{a}\\ \Rightarrow a_1=a\tan \theta \end{array}$

$\text{ where }a=\frac{T\cos \theta }{m}\text{ from }\left(i\right)\text{. }$

$\text{ Putting }T=\frac{F}{2\cos \theta },\text{ from (iii), we obtain }$

$a_1=\frac{F}{2m}\tan \theta$

$\text{ Putting }\theta ={30}^{\circ }\Rightarrow a_1=\frac{F}{2\sqrt{3}m}$