Question

In figure two identical particles each of mass m are tied together with an inextensible string. This is pulled at its centre with a constant force F. If the whole system lies on a smooth horizontal plane, then the acceleration of one particle with respect to other is

• A. $\frac{\sqrt{3}}{2}\frac{F}{m}$
• B. $\frac{1}{2\sqrt{3}}\frac{F}{m}$
  
• C. $\frac{2}{\sqrt{3}}\frac{F}{m}$
  
• D. $\sqrt{3}\frac{F}{m}$

Answer 1

The correct option is B $\frac{1}{2\sqrt{3}}\frac{F}{m}$


$2Tcos{30}^{\circ }=F$


$\therefore T=\frac{F}{\sqrt{3}}$

$a=\frac{Tcos{60}^{\circ }}{m}$

$=\frac{F}{2\sqrt{3}m}$

acceleration of each particle will be, $a=\frac{F}{2\sqrt{3}m}$

now, if we wish to calculate acceleration of one particle with respect to another,

then $a_{upper/lower}=a_{upper}-a_{Lower}=\frac{F}{2\sqrt{3}m}-\frac{-F}{2\sqrt{3}m}=\frac{2F}{2\sqrt{3}m}$