Question

Two bodies A & B of equal masses m lie on a smooth frictionless horizontal table. They are initially separated by a distance $2l$ and are connected by a thread of length $2l$. The thread is pulled at its midpoint O by a constant force F at a right angle to AB in a horizontal plane. Find the acceleration of A & B in a direction at right angles to the direction of F as a function of the displacement (x) of the midpoint O.

Answer 1

When thread is pulled with a constant force of F at an angle ${90}^0$ with respect to the masses A and B,than this force F is expressed as $F=2T\cos \theta$-------(1) where$\theta =\frac{x}{l}$

Also force $F_z$ in the direction at right angles to the direction of force $m{a}_z=T\sin \theta$ -------(2)

Putting the value of tension T from equation 1 in 2 we get ,

$a_z=\frac{T\sin \theta }{m}=\frac{F\sin \theta }{2m\cos \theta }=\frac{F\sqrt{1-{\left(\frac{x}{l}\right)}^2}}{2m\frac{x}{l}}=\frac{F}{2m}\sqrt{\frac{l^2-x^2}{x^2}}$