Question

Two identical point masses, each of mass $M$ are connected to one another by a massless string of length $L$. A constant force $F$ is applied at the mid-point of the string. If $l$ be the instantaneous distance between the two masses, what will be the acceleration of each mass?

Answer 1

Assume that the masses are on horizontal frictionless table.No gravity related forces.

Let the simultaneous distance between the two masses $=AB=d=L.\sin \theta .F=2M{a}_y$

$a_y=\frac{F}{2M}F=2T\cos \theta =$ Net force on the system

$T=\frac{F}{2\cos \theta }{a}_x=\frac{T\sin \theta }{M}=\frac{F\tan \theta }{2M}$

$\sec \theta =\frac{L}{\sqrt{L^2-d^2}}$

Net acceleration$=a=\sqrt{(a_x{)}^2+(a_y{)}^2}$

$a=\frac{F\sec \theta }{2M}=\frac{FL}{2M\sqrt{(L^2-d^2)}}$