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| <header style="text-align:center;"> | | <header style="text-align:center;"> |
| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</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="../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../#2.1">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| | | |
| <p> | | <p> |
| $2.1.49$ | | $2.1.49$ |
| A load is suspended from the free end of the thread attached to the wall and thrown over the roller. The roller is fixed on a bar of mass $m_0$, which can slide along a horizontal plane without friction. At the initial moment, the thread with the load is deflected from the vertical by an angle α and then released. Determine the acceleration of the bar if the angle formed by the thread with the vertical does not change during the movement of the system. What is the weight of the cargo? | | A load is suspended from the free end of the thread attached to the wall and thrown over the roller. The roller is fixed on a bar of mass $m_0$, which can slide along a horizontal plane without friction. At the initial moment, the thread with the load is deflected from the vertical by an angle α and then released. Determine the acceleration of the bar if the angle formed by the thread with the vertical does not change during the movement of the system. What is the weight of the cargo? |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem 2.1.49 | | For problem 2.1.49 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="sol.png" | | <img src="sol.png" |
| loading="lazy" width="400" /> | | loading="lazy" width="400" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the system | | Forces acting on the system |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Newton's second law of motion: | | Newton's second law of motion: |
| $$OX: \;T - T \sin \alpha = m_0 a_{0x} $$ | | $$OX: \;T - T \sin \alpha = m_0 a_{0x} $$ |
| $$ T \sin \alpha = m_{a1x} $$ | | $$ T \sin \alpha = m_{a1x} $$ |
| | | |
| $$OY: \;N - mg - T \cos \alpha = m_0 a_{0y} = 0 $$ | | $$OY: \;N - mg - T \cos \alpha = m_0 a_{0y} = 0 $$ |
| $$ T \cos \alpha - mg = m_{a1y} $$ | | $$ T \cos \alpha - mg = m_{a1y} $$ |
| $$ \tan \alpha = \frac{x_0 - x_1}{y_0 - y_1} $$ | | $$ \tan \alpha = \frac{x_0 - x_1}{y_0 - y_1} $$ |
| Since $\alpha = \text{const}$ | | Since $\alpha = \text{const}$ |
| <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) |
| $$ | | $$ |
| $$ \tan \alpha (a_{0y} - a_{1y}) = a_{0x} - a_{1x}$$ | | $$ \tan \alpha (a_{0y} - a_{1y}) = a_{0x} - a_{1x}$$ |
| From (1) | | From (1) |
| $$a_{0y} = 0$$ | | $$a_{0y} = 0$$ |
| $$ -\tan \alpha\, a_{1y} = a_{0x} - a_{1x} $$ | | $$ -\tan \alpha\, a_{1y} = a_{0x} - a_{1x} $$ |
| Length of the rope: | | Length of the rope: |
| $$ L = x_w - x_0 + \frac{y_0 - y_1}{\cos \alpha} $$ | | $$ L = x_w - x_0 + \frac{y_0 - y_1}{\cos \alpha} $$ |
| Since the rope is inextensible | | Since the rope is inextensible |
| <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> |
| $$ a_{0x} \cdot \cos \alpha = a_{0y} - a_{1y}$$ | | $$ a_{0x} \cdot \cos \alpha = a_{0y} - a_{1y}$$ |
| From $(1)$: $a_{0y} = 0$ | | From $(1)$: $a_{0y} = 0$ |
| $$ a_{0x} \cos \alpha = -a_{1y} $$ | | $$ a_{0x} \cos \alpha = -a_{1y} $$ |
| By dividing the equations from (1): | | By dividing the equations from (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> |
| $$ \tan \alpha a_{1y} = a_{1x} - g \tan \alpha$$ | | $$ \tan \alpha a_{1y} = a_{1x} - g \tan \alpha$$ |
| and using (4) | | and using (4) |
| $$\boxed{a_{0x} = g \tan \alpha }$$ | | $$\boxed{a_{0x} = g \tan \alpha }$$ |
| By dividing the equations from (1): | | By dividing the equations from (1): |
| | | |
| $$ | | $$ |
| \frac{T(1 - \sin \alpha)}{T \cos \alpha} = \frac{M}{m} \cdot \frac{a_{0x}}{g + a_{1y}}$$ | | \frac{T(1 - \sin \alpha)}{T \cos \alpha} = \frac{M}{m} \cdot \frac{a_{0x}}{g + a_{1y}}$$ |
| Note that: | | Note that: |
| $$a_{1y} = -g \sin \alpha$$ | | $$a_{1y} = -g \sin \alpha$$ |
| | | |
| $$ \frac{1 - \sin \alpha}{\cos \alpha} \frac{M}{m} = \frac{g \tan \alpha}{g - g \sin \alpha} \to m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2} $$ | | $$ \frac{1 - \sin \alpha}{\cos \alpha} \frac{M}{m} = \frac{g \tan \alpha}{g - g \sin \alpha} \to m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2} $$ |
| $$ \boxed{m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2}} $$ | | $$ \boxed{m = \frac{M \sin \alpha}{(1 - \sin \alpha)^2}} $$ |
| | | |
| </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> |
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