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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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| </a> | | </a> |
| <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> |
| </p> | | </p> |
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| <h3 id="back-link"><a href="../../ru#2.1">$\leftarrow$Назад</a></h3> | | <h3 id="back-link"><a href="../../ru#2.1">$\leftarrow$Назад</a></h3> |
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| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| К задаче 2.1.49 | | К задаче 2.1.49 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <h3> Решение: </h3> | | <h3> Решение: </h3> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="sol.png" | | <img src="sol.png" |
| loading="lazy" width="400" /> | | loading="lazy" width="400" /> |
| <figcaption> | | <figcaption> |
| Силы действующие на систему | | Силы действующие на систему |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <p> | | <p> |
| Применим второй закон Ньютона: | | Применим второй закон Ньютона: |
| $$OX: \;T - T \sin \alpha = m_0 a_{0x} $$ | | $$OX: \;T - T \sin \alpha = m_0 a_{0x} $$ |
| $$ T \sin \alpha = m_{a1x} $$ | | $$ T \sin \alpha = m_{a1x} $$ |
| | | |
| $$OY: \;N - mg - T \cos \alpha = m_0 a_{0y} = 0 $$ | | $$OY: \;N - mg - T \cos \alpha = m_0 a_{0y} = 0 $$ |
| $$ T \cos \alpha - mg = m_{a1y} $$ | | $$ T \cos \alpha - mg = m_{a1y} $$ |
| $$ \tan \alpha = \frac{x_0 - x_1}{y_0 - y_1} $$ | | $$ \tan \alpha = \frac{x_0 - x_1}{y_0 - y_1} $$ |
| Учитывая $\alpha = \text{const}$ | | Учитывая $\alpha = \text{const}$ |
| <div class="scroll-wrapper"> | | <div class="scroll-wrapper"> |
| $$ | | $$ |
| \frac{d}{dt} \left( \tan \alpha \right) = \frac{\left( \frac{dx_0}{dt} - \frac{dx_1}{dt} \right)(y_0 - y_1) - \left( \frac{dy_0}{dt} - \frac{dy_1}{dt} \right)(x_0 - x_1)}{(y_0 - y_1)^2} = 0 | | \frac{d}{dt} \left( \tan \alpha \right) = \frac{\left( \frac{dx_0}{dt} - \frac{dx_1}{dt} \right)(y_0 - y_1) - \left( \frac{dy_0}{dt} - \frac{dy_1}{dt} \right)(x_0 - x_1)}{(y_0 - y_1)^2} = 0 |
| $$</div> | | $$</div> |
| | | |
| <div class="scroll-wrapper"> | | <div class="scroll-wrapper"> |
| $$ | | $$ |
| \frac{d}{dt} \left( \left( \frac{dx_0}{dt} - \frac{dx_1}{dt} \right)(y_0 - y_1) \right) = \frac{d}{dt} \left( \left( \frac{dy_0}{dt} - \frac{dy_1}{dt} \right)(x_0 - x_1) \right) | | \frac{d}{dt} \left( \left( \frac{dx_0}{dt} - \frac{dx_1}{dt} \right)(y_0 - y_1) \right) = \frac{d}{dt} \left( \left( \frac{dy_0}{dt} - \frac{dy_1}{dt} \right)(x_0 - x_1) \right) |
| $$</div> | | $$</div> |
| <div class="scroll-wrapper"> | | <div class="scroll-wrapper"> |
| $$ | | $$ |
| \left(\frac{d^{2}x_{0}}{dt^{2}}-\frac{d^{2}x_{1}}{dt^{2}}\right)(y_{0}-y_{1})+\left(\frac{dx_{0}}{dt}-\frac{dx_{1}}{dt}\right)\left(\frac{dy_{0}}{dt}-\frac{dy_{1}}{dt}\right)=$$ | | \left(\frac{d^{2}x_{0}}{dt^{2}}-\frac{d^{2}x_{1}}{dt^{2}}\right)(y_{0}-y_{1})+\left(\frac{dx_{0}}{dt}-\frac{dx_{1}}{dt}\right)\left(\frac{dy_{0}}{dt}-\frac{dy_{1}}{dt}\right)=$$ |
| $$\left(\frac{d^{2}y_{0}}{dt^{2}}-\frac{d^{2}y_{1}}{dt^{2}}\right)(x_{0}-x_{1})+\left(\frac{dy_{0}}{dt}-\frac{dy_{1}}{dt}\right)\cdot\left(\frac{dx_{0}}{dt}-\frac{dx_{1}}{dt}\right) | | $$\left(\frac{d^{2}y_{0}}{dt^{2}}-\frac{d^{2}y_{1}}{dt^{2}}\right)(x_{0}-x_{1})+\left(\frac{dy_{0}}{dt}-\frac{dy_{1}}{dt}\right)\cdot\left(\frac{dx_{0}}{dt}-\frac{dx_{1}}{dt}\right) |
| $$</div> | | $$</div> |
| | | |
| | | |
| $$ | | $$ |
| (a_{0x} - a_{1x})(y_0 - y_1) = (a_{0y} - a_{1y})(x_0 - x_1) | | (a_{0x} - a_{1x})(y_0 - y_1) = (a_{0y} - a_{1y})(x_0 - x_1) |
| $$ | | $$ |
| $$ \tan \alpha (a_{0y} - a_{1y}) = a_{0x} - a_{1x}$$ | | $$ \tan \alpha (a_{0y} - a_{1y}) = a_{0x} - a_{1x}$$ |
| Из (1) | | Из (1) |
| $$a_{0y} = 0$$ | | $$a_{0y} = 0$$ |
| $$ -\tan \alpha\, a_{1y} = a_{0x} - a_{1x} $$ | | $$ -\tan \alpha\, a_{1y} = a_{0x} - a_{1x} $$ |
| Длина веревки: | | Длина веревки: |
| $$ L = x_w - x_0 + \frac{y_0 - y_1}{\cos \alpha} $$ | | $$ L = x_w - x_0 + \frac{y_0 - y_1}{\cos \alpha} $$ |
| Т.к. нить нерастяжима | | Т.к. нить нерастяжима |
| <div class="scroll-wrapper"> | | <div class="scroll-wrapper"> |
| $$ \frac{d^2 L}{dt^2} = -\frac{d^2 x_0}{dt^2} + \frac{1}{\cos \alpha} \left( \frac{d^2 y_0}{dt^2} - \frac{d^2 y_1}{dt^2} \right) = 0 $$ | | $$ \frac{d^2 L}{dt^2} = -\frac{d^2 x_0}{dt^2} + \frac{1}{\cos \alpha} \left( \frac{d^2 y_0}{dt^2} - \frac{d^2 y_1}{dt^2} \right) = 0 $$ |
| </div> | | </div> |
| | | |
| <div class="scroll-wrapper"> | | <div class="scroll-wrapper"> |
| $$ \left( \frac{d^2 x_0}{dt^2} = a_{0x}, \frac{d^2 y_0}{dt^2} = a_{0y}, \frac{d^2 y_1}{dt^2} = a_{1y} \right) $$</div> | | $$ \left( \frac{d^2 x_0}{dt^2} = a_{0x}, \frac{d^2 y_0}{dt^2} = a_{0y}, \frac{d^2 y_1}{dt^2} = a_{1y} \right) $$</div> |
| $$ a_{0x} \cdot \cos \alpha = a_{0y} - a_{1y}$$ | | $$ a_{0x} \cdot \cos \alpha = a_{0y} - a_{1y}$$ |
| Из $(1)$: $a_{0y} = 0$ | | Из $(1)$: $a_{0y} = 0$ |
| $$ a_{0x} \cos \alpha = -a_{1y} $$ | | $$ a_{0x} \cos \alpha = -a_{1y} $$ |
| Разделив уравнения из (1): | | Разделив уравнения из (1): |
| <div class="scroll-wrapper"> | | <div class="scroll-wrapper"> |
| $$ \frac{T \sin \alpha}{T \cos \alpha} = \frac{m a_{1x}}{m (g + a_{1y})} \to \tan \alpha (g + a_{1y}) = a_{1x} $$</div> | | $$ \frac{T \sin \alpha}{T \cos \alpha} = \frac{m a_{1x}}{m (g + a_{1y})} \to \tan \alpha (g + a_{1y}) = a_{1x} $$</div> |
| $$ \tan \alpha a_{1y} = a_{1x} - g \tan \alpha$$ | | $$ \tan \alpha a_{1y} = a_{1x} - g \tan \alpha$$ |
| и используя (4) | | и используя (4) |
| $$\boxed{a_{0x} = g \tan \alpha }$$ | | $$\boxed{a_{0x} = g \tan \alpha }$$ |
| Разделив уравнения из (1): | | Разделив уравнения из (1): |
| | | |
| $$ | | $$ |
| \frac{T(1 - \sin \alpha)}{T \cos \alpha} = \frac{M}{m} \cdot \frac{a_{0x}}{g + a_{1y}}$$ | | \frac{T(1 - \sin \alpha)}{T \cos \alpha} = \frac{M}{m} \cdot \frac{a_{0x}}{g + a_{1y}}$$ |
| Обратите внимание, что: | | Обратите внимание, что: |
| $$a_{1y} = -g \sin \alpha$$ | | $$a_{1y} = -g \sin \alpha$$ |
| | | |
| $$ \frac{1 - \sin \alpha}{\cos \alpha} \frac{M}{m} = \frac{g \tan \alpha}{g - g \sin \alpha} \to m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2} $$ | | $$ \frac{1 - \sin \alpha}{\cos \alpha} \frac{M}{m} = \frac{g \tan \alpha}{g - g \sin \alpha} \to m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2} $$ |
| $$ \boxed{m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2}} $$ | | $$ \boxed{m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2}} $$ |
| | | |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p> | | <p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p> |
| | | |
