$2.2.44^*.$ A rope thrown over a smooth nail is dragged at a speed of $v$ through the gap. The friction force in the slot $F$, the mass of the rope length unit $\rho$. Determine the force exerted on the nail if the rope sections on opposite sides of the nail form an angle $\alpha$. At what speed will the rope move away from the nail?
$$ \rho = \frac{m}{l} = \frac{\Delta m}{\Delta l} $$ $$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos\alpha}{2}}$$
For the thread to be stretched at a constant speed $\vec{v}$, there must be a force on the first half that is modulo the frictional force $\vec{F}$: $$F'^2=F^2+F^2-2F^2\cos(\pi-\alpha)$$ $$F'=F\sqrt{2(1+\cos\alpha)}$$ $$F'=2F\cos\frac{\alpha}{2}$$
Similiarly, $$p'^2=p^2+p^2-2p^2\cos(\pi-\alpha)$$ $$p'=p\sqrt{2(1+\cos\alpha)}$$ $$p'=2p\cos\frac{\alpha}{2}$$ By definition of momentum $$p=\Delta m v$$ From where, considering that $p'$ and $F'$ acts against $Ox$ $$p'=-2\Delta m v\cos\frac{\alpha}{2}$$ $$F'=2F\cos\frac{\alpha}{2}$$ Let's write Newton's second law $$\sum \vec{F}=0$$ $$N-F'-F''=0$$ $$N-2F\cos\frac{\alpha}{2}+2\Delta m v\cos\frac{\alpha}{2}=0$$ $$F''=\frac{p'}{\Delta t}$$ From where we obtain the force $\vec{N}$ $$N=2\cos\frac{\alpha}{2}(\frac{-\Delta m v}{\Delta t}+F)$$ By definition of density $$\Delta m = \rho \Delta x$$ By definition of speed $$ v = \frac{\Delta x}{\Delta t}$$ $$ N = 2F\cos\frac{\alpha}{2} - 2\rho v^2\cos\frac{\alpha}{2}$$ $$ \boxed{N = 2\cos\frac{\alpha}{2} (F -\rho v^2)}$$ When the rope comes off $$N\leq0$$ $$F-\rho v^2\leq0$$ $${F\leq \rho v^2}$$ From where we finally get $$\boxed{v \geq \sqrt{\frac{F}{\rho}}}$$
$$N=2(F-\rho v^2)\cos\frac\alpha2;\text{ при }v\geqslant\sqrt{F/\rho}.$$
Almaskhan Arsen