Physics - NLM Question with Solution | TestHub

The weight of the chain is balanced by the vertical components of the reaction forces at the two suspension points. The magnitude of each of these components must be $F_0=\rho Lg/2$, where $\rho$ is the mass of the chain per unit length, and $L=40$ cm. If the chain made an angle of $\theta$ with the vertical at a suspension point, the horizontal component of the chain's tension would be $F_0\text{tan}\theta$. Because, in practice, $\theta =45^o$, the horizontal component is the same as the vertical one; it is also constant along the chain, since there are horizontal forces acting on the chain only at its ends.

At the chain's lowest point, the tension is purely horizontal, and, as just shown, its magnitude is $F_0$. So, around the lowest point, consider a small piece of the chain that subtends an angle $2\phi$ at the centre of the osculating circle, whose radius is the value $r_1$ we seek. The length of the piece is $2\phi r_1$ and it is pulled down by a gravitational force $2\phi r_1\rho g$. This force is balanced by the upward net force of $2F_0\text{sin}\phi$ due to the tension in the chain. In the limit of small angles, when $\text{sin}\phi \approx \phi$, the equilibrium equation

$$2\phi r_1\rho g=2F_0\text{sin}\phi$$

leads to

$$F_0=\rho g{r}_1$$

Since $F_0=\rho Lg/2$, we have the simple result that $r_1=L/2=20$ cm.