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 ho. Determine the force exerted on the nail if the rope sections on opposite sides of the nail form an angle lpha. At what speed will the rope move away from the nail?

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    <h3 id="back-link"><a href="/#2.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.2.44^*.$ 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?

    <h3>Solution</h3>
    <p>
        </p>

$$\rho = \frac{m}{l} = \frac{\Delta m}{\Delta l}$$

$$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos\alpha}{2}}$$

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
$$\boxed{v \geq \sqrt{\frac{F}{\rho}}}$$

    <h4>Answer</h4>
    <p>
        $$N=2(F-\rho v^2)\cos\frac\alpha2;\text{ при }v\geqslant\sqrt{F/\rho}.$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Almaskhan Arsen<br>
    </p>


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