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| <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="../../#2.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.3.48^*.$ On the horizontal plane lie two bars of mass $m_1$ and $m_2$, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is $\mu$. | | $2.3.48^*.$ On the horizontal plane lie two bars of mass $m_1$ and $m_2$, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is $\mu$. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.3.48.png" | | <img src="2.3.48.png" |
| loading="lazy" width="200" /> | | loading="lazy" width="200" /> |
| <figcaption> | | <figcaption> |
| For problem $2.3.48^*$ | | For problem $2.3.48^*$ |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.3.48_1.png" | | <img src="2.3.48_1.png" |
| loading="lazy" width="300" /> | | loading="lazy" width="300" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the system | | Forces acting on the system |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| States of equilibrium for horizontal axes | | States of equilibrium for horizontal axes |
| $$F_\text{fr1} = \mu N_1 \quad (1)$$ | | $$F_\text{fr1} = \mu N_1 \quad (1)$$ |
| $$F_\text{fr2} = \mu N_2 \quad (2)$$ | | $$F_\text{fr2} = \mu N_2 \quad (2)$$ |
| Alternatively for a vertical axes | | Alternatively for a vertical axes |
| $$N_1 = m_1g \quad (3)$$ | | $$N_1 = m_1g \quad (3)$$ |
| $$N_2 = m_2g \quad (4)$$ | | $$N_2 = m_2g \quad (4)$$ |
| Substituting $(3)$ and $(4)$ to the $(1)$ and $(2)$ equations | | Substituting $(3)$ and $(4)$ to the $(1)$ and $(2)$ equations |
| $$F_\text{fr1} = \mu m_1g$$ | | $$F_\text{fr1} = \mu m_1g$$ |
| $$F_\text{fr2} = \mu m_2g$$ | | $$F_\text{fr2} = \mu m_2g$$ |
| By the law of energy conservation, force $\vec{F}$ acts against the tension of the spring and against the friction force | | By the law of energy conservation, force $\vec{F}$ acts against the tension of the spring and against the friction force |
| $$A=Fx=\Delta E$$ | | $$A=Fx=\Delta E$$ |
| Substituting the potential energy of the spring | | Substituting the potential energy of the spring |
| $$Fx = \frac{kx^2}{2} + F_\text{fr1}x$$ | | $$Fx = \frac{kx^2}{2} + F_\text{fr1}x$$ |
| $$F= \frac{kx}{2} + \mu m_1 g \quad (5)$$ | | $$F= \frac{kx}{2} + \mu m_1 g \quad (5)$$ |
| From the state equilibrium in the steady state | | From the state equilibrium in the steady state |
| $$kx = \mu m_2 g \quad (6)$$ | | $$kx = \mu m_2 g \quad (6)$$ |
| After substituting $(6)$ into $(5)$, we finally get | | After substituting $(6)$ into $(5)$, we finally get |
| $$\boxed{ F = \mu g \left( m_1 +\frac{m_2}{2} \right)}$$ | | $$\boxed{ F = \mu g \left( m_1 +\frac{m_2}{2} \right)}$$ |
| </p> | | </p> |
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