Question

Two particles each of mass 'm' are connected to a light weight string of length '2L' . a steady force 'F' is applied at the midpoint of the string (x=0) at a right angle to the initial position of the string. Show that the acceleration of each mass in the direction at 90 degrees to F is given by

Ax = (Fx)/(2m((L^2) - (x^2) ))^(1/2)

Answer 1

force acting on one of the mass m:\[Fcos \theta =m*a\]
where a is the accn of one of the mass along x axis. the reason behind writting cos theta is because, the force is acting along y-axis...but the particles move along y axis. the force acts along the string and the component of this force along the direction of displacement of the particle is fcos theta.
\[Ftotal = 2Fcos \theta \]
since there are two particles.....
now force acting on 1 particle is :\[F= fcos \theta =(mLa/x)\]
\[Ftotal= (2mLa/x)*\cos phi \]
phi is the angle made by each component of force with the y axis.
therefore, \[\cos \phi = (L ^{2}-x ^{2})^1/2 /L\]
thus, \[Ftotal= 2fcos \theta = (2mL/x)*(L ^{2}-x ^{2})^1/2 /L\]
this is equal to \[2m* a(total)\]
therefore, \[a(total) = F/(2m) ^{2}*(L ^{2}-x ^{2})^{1/2} \]\[= F/\sqrt{(2m(L ^{2}}-x ^{2}))\]