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| <h2>Solutions of Savchenko Physics Textbook</h2> | | <h2>Solutions of Savchenko Physics Textbook</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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? | | $2.2.44^*.$ 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? |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="150" /> | | loading="lazy" width="150" /> |
| <figcaption> | | <figcaption> |
| For problem $2.2.44^*$ | | For problem $2.2.44^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw1.jpg" | | <img src="draw1.jpg" |
| loading="lazy" width="180" /> | | loading="lazy" width="180" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the system | | Forces acting on the system |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| $$ | | $$ |
| \rho = \frac{m}{l} = \frac{\Delta m}{\Delta l} | | \rho = \frac{m}{l} = \frac{\Delta m}{\Delta l} |
| $$ | | $$ |
| | | |
| $$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos\alpha}{2}}$$ | | $$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos\alpha}{2}}$$ |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw2.jpg" | | <img src="draw2.jpg" |
| loading="lazy" width="130" /> | | loading="lazy" width="130" /> |
| <figcaption> | | <figcaption> |
| For problem $2.2.44^*$ | | For problem $2.2.44^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| 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 $\vec{F}$: | | 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 $\vec{F}$: |
| | | |
| $$F'^2=F^2+F^2-2F^2\cos(\pi-\alpha)$$ | | $$F'^2=F^2+F^2-2F^2\cos(\pi-\alpha)$$ |
| $$F'=F\sqrt{2(1+\cos\alpha)}$$ | | $$F'=F\sqrt{2(1+\cos\alpha)}$$ |
| $$F'=2F\cos\frac{\alpha}{2}$$ | | $$F'=2F\cos\frac{\alpha}{2}$$ |
| | | |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw3.jpg" | | <img src="draw3.jpg" |
| loading="lazy" width="130" /> | | loading="lazy" width="130" /> |
| <figcaption> | | <figcaption> |
| For problem $2.2.44^*$ | | For problem $2.2.44^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| Similiarly, | | Similiarly, |
| | | |
| $$p'^2=p^2+p^2-2p^2\cos(\pi-\alpha)$$ | | $$p'^2=p^2+p^2-2p^2\cos(\pi-\alpha)$$ |
| $$p'=p\sqrt{2(1+\cos\alpha)}$$ | | $$p'=p\sqrt{2(1+\cos\alpha)}$$ |
| $$p'=2p\cos\frac{\alpha}{2}$$ | | $$p'=2p\cos\frac{\alpha}{2}$$ |
| | | |
| By definition of momentum | | By definition of momentum |
| | | |
| $$p=\Delta m v$$ | | $$p=\Delta m v$$ |
| | | |
| From where, considering that $p'$ and $F'$ acts against $Ox$ | | From where, considering that $p'$ and $F'$ acts against $Ox$ |
| | | |
| $$p'=-2\Delta m v\cos\frac{\alpha}{2}$$ | | $$p'=-2\Delta m v\cos\frac{\alpha}{2}$$ |
| $$F'=-2F\cos\frac{\alpha}{2}$$ | | $$F'=-2F\cos\frac{\alpha}{2}$$ |
| Let's write Newton's second law | | Let's write Newton's second law |
| $$\sum \vec{F}=0$$ | | $$\sum \vec{F}=0$$ |
| $$N-F'-p''=0$$ | | $$N-F'-p''=0$$ |
| $$N-2F\cos\frac{\alpha}{2}+2\Delta m v\cos\frac{\alpha}{2}=0$$ | | $$N-2F\cos\frac{\alpha}{2}+2\Delta m v\cos\frac{\alpha}{2}=0$$ |
| From where we obtain the force $\vec{N}$ | | From where we obtain the force $\vec{N}$ |
| | | |
| $$N=2\cos\frac{\alpha}{2}(\frac{-\Delta m v}{\Delta t}+F)$$ | | $$N=2\cos\frac{\alpha}{2}(\frac{-\Delta m v}{\Delta t}+F)$$ |
| By definition of density | | By definition of density |
| $$\Delta m = \rho \Delta x$$ | | $$\Delta m = \rho \Delta x$$ |
| By definition of speed | | By definition of speed |
| $$ v = \frac{\Delta x}{\Delta t}$$ | | $$ v = \frac{\Delta x}{\Delta t}$$ |
| $$ N = 2F\cos\frac{\alpha}{2} - 2\rho v^2\cos\frac{\alpha}{2}$$ | | $$ N = 2F\cos\frac{\alpha}{2} - 2\rho v^2\cos\frac{\alpha}{2}$$ |
| $$ \boxed{N = 2\cos\frac{\alpha}{2} (F -\rho v^2)}$$ | | $$ \boxed{N = 2\cos\frac{\alpha}{2} (F -\rho v^2)}$$ |
| When the rope comes off | | When the rope comes off |
| $$N\leq0$$ | | $$N\leq0$$ |
| $$F-\rho v^2\leq0$$ | | $$F-\rho v^2\leq0$$ |
| $${F\leq \rho v^2}$$ | | $${F\leq \rho v^2}$$ |
| From where we finally get | | From where we finally get |
| $$\boxed{v \geq \sqrt{\frac{F}{\rho}}}$$ | | $$\boxed{v \geq \sqrt{\frac{F}{\rho}}}$$ |
| $$p''=\frac{p'}{\Delta t}$$ | | $$p''=\frac{p'}{\Delta t}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
