Question

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,

• A. $\frac{F}{2\sqrt{3}m}$
• B. $\frac{\sqrt{3}F}{2m}$
• C. $\frac{2}{\sqrt{3}}\frac{F}{m}$
• D. None of these

Answer 1

Answer: (A)

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 $a_1$ respectively.
Writing the equation of motion of $m$, we obtain
$\Sigma F_x=ma$
$\Rightarrow T\cos \theta =ma$ ...(1)
$\Sigma F_y=m{a}_1$
$\Rightarrow T\sin \theta =m{a}_1$ ...(2)
At point $P$, as it is accelerating with an acceleration a, therefore
$F-2T\cos \theta =m_{p}a$
where $m_p=$ mass of the string at the point $P\cong 0$
$\Rightarrow F=2T\cos \theta$ ...(3)
$(2)+(1)\Rightarrow \tan \theta =\frac{a_1}{a}$
$\Rightarrow a_1=a\tan \theta$
where $a=\frac{T\cos \theta }{m}$ from (1).
Putting $T=\frac{F}{2\cos \theta }$, from (3), we obtain
$a_1=\frac{F}{2m}\tan \theta$
Putting $\theta =30^{\circ }$
$\Rightarrow a_1=\frac{F}{2\sqrt{3}m}$