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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.44^*.$ Determine the steady-state velocity of a body on an inclined plane that changes one direction of its velocity $u$ to the opposite direction with high frequency. The direction of movement of the plane is shown in the figure. Coefficient of friction $\mu$, angle of inclination of the plane $\alpha$, $\tan\alpha < \mu$. | | $2.1.44^*.$ Determine the steady-state velocity of a body on an inclined plane that changes one direction of its velocity $u$ to the opposite direction with high frequency. The direction of movement of the plane is shown in the figure. Coefficient of friction $\mu$, angle of inclination of the plane $\alpha$, $\tan\alpha < \mu$. |
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| For problem $2.1.44^*$ | | For problem $2.1.44^*$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| Forces acting on the body | | Forces acting on the body |
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| Since the speed changes quickly, the body does not have time to move in the horizontal direction and all the time moves in the direction of $\vec{v}$ | | Since the speed changes quickly, the body does not have time to move in the horizontal direction and all the time moves in the direction of $\vec{v}$ |
| </p> | | </p> |
| <p> | | <p> |
| Since $v=\text{const}$, there is no acceleration in the direction along the speed $\vec{v}$ | | Since $v=\text{const}$, there is no acceleration in the direction along the speed $\vec{v}$ |
| $$ mg\sin\alpha = \mu mg \cos\alpha\cos\beta$$ | | $$ mg\sin\alpha = \mu mg \cos\alpha\cos\beta$$ |
| | | |
| $$ \cos\beta = \frac{\tan\alpha}{\mu} $$ | | $$ \cos\beta = \frac{\tan\alpha}{\mu} $$ |
| From geometric considerations, the moduli of the vectors $\vec{v}$ and $\vec{u}$ are related by the relation | | From geometric considerations, the moduli of the vectors $\vec{v}$ and $\vec{u}$ are related by the relation |
| $$ v=\frac{u}{\tan\beta} $$ | | $$ v=\frac{u}{\tan\beta} $$ |
| | | |
| $$ v=u \frac{\frac{\tan\alpha}{\mu}}{\sqrt{1-\frac{\tan^2\alpha}{\mu^2}}} $$ | | $$ v=u \frac{\frac{\tan\alpha}{\mu}}{\sqrt{1-\frac{\tan^2\alpha}{\mu^2}}} $$ |
| | | |
| $$ \boxed{v = u \frac{\tan\alpha}{\sqrt{\mu^{2} − \tan^{2} \alpha}}} $$ | | $$ \boxed{v = u \frac{\tan\alpha}{\sqrt{\mu^{2} − \tan^{2} \alpha}}} $$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$v = u \frac{\, \text{tg} \alpha}{\sqrt{\mu^{2} − \text{tg}^{2} \alpha}}$$ | | $$v = u \frac{\, \text{tg} \alpha}{\sqrt{\mu^{2} − \text{tg}^{2} \alpha}}$$ |
