A mass of 1 kg is attached to the middle of a rope, which is being pulled from both the opposite directions. Taking g - 10 m/s², what is the minimum pull required to completely straighten the rope ?

If A, B and C be three vectors acting at a point and their resultant is zero, then
$\frac{\mathrm{A}}{\sin \mathrm{\alpha }}=\frac{\mathrm{B}}{\sin \mathrm{\beta }}=\frac{\mathrm{C}}{\sin \mathrm{\gamma }}$
where $\mathrm{\alpha },\mathrm{\beta } \mathrm{and} \mathrm{\gamma }$ are shown in figure.

To straighten the rope, $\mathrm{\lambda }$ should be 180°. Now $\mathrm{\alpha }=\mathrm{\beta }={90}^{\circ }$
$\therefore \mathrm{A}=\mathrm{B}=\frac{\mathrm{C}}{\sin 180}=\frac{1\times 10}{0}=\mathrm{\infty }$