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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> |
| </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> |
| $2.1.48.$ A wedge of mass $m_2$ with an angle $\alpha$ is inserted between two identical smooth bars of mass $m_1$. Determine the acceleration of bodies. | | $2.1.48.$ A wedge of mass $m_2$ with an angle $\alpha$ is inserted between two identical smooth bars of mass $m_1$. Determine the acceleration of bodies. |
| </p> | | </p> |
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| <img src="https://savchenkosolutions.com/2/2.1.48/statement.png" | | <img src="https://savchenkosolutions.com/2/2.1.48/statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem $2.1.48$ | | For problem $2.1.48$ |
| </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="https://savchenkosolutions.com/2/2.1.48/sol.png" | | <img src="https://savchenkosolutions.com/2/2.1.48/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> |
| Let us consider a small displacement over time $dt$ from the point of view of kinematics: | | Let us consider a small displacement over time $dt$ from the point of view of kinematics: |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="https://savchenkosolutions.com/2/2.1.48/sol1.png" | | <img src="https://savchenkosolutions.com/2/2.1.48/sol1.png" |
| loading="lazy" width="280" /> | | loading="lazy" width="280" /> |
| <figcaption> | | <figcaption> |
| Small transition in time $dt$ | | Small transition in time $dt$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| From geometric considerations: | | From geometric considerations: |
| $$ dy = dx \cdot c\tan\beta$$ | | $$ dy = dx \cdot c\tan\beta$$ |
| Let us differentiate both parts of the expression twice: | | Let us differentiate both parts of the expression twice: |
| $$ \frac{dy}{d^2t} = \frac{dx}{d^2t} \cdot c\tan\beta$$ | | $$ \frac{dy}{d^2t} = \frac{dx}{d^2t} \cdot c\tan\beta$$ |
| | | |
| $$ a_1 = a_2 \cdot \tan\beta\quad(1) $$ | | $$ a_1 = a_2 \cdot \tan\beta\quad(1) $$ |
| Since the Author did not say anything about friction, example $F_{tr}=0$ | | Since the Author did not say anything about friction, example $F_{tr}=0$ |
| Next, we write Newton's 2nd law for the bar $m_1$ on the $OY$ axis: | | Next, we write Newton's 2nd law for the bar $m_1$ on the $OY$ axis: |
| $$ m_1a_1=N_1\cos\beta$$ | | $$ m_1a_1=N_1\cos\beta$$ |
| | | |
| $$ N_1=\frac{m_1a_1}{\cos\beta}\quad(a) $$ | | $$ N_1=\frac{m_1a_1}{\cos\beta}\quad(a) $$ |
| Similarly for $m_2$: | | Similarly for $m_2$: |
| $$ m_2a_2=m_2g-2N_1\,\sin\beta\quad(b) $$ | | $$ m_2a_2=m_2g-2N_1\,\sin\beta\quad(b) $$ |
| Substitute $(a)$ into $(b)$: | | Substitute $(a)$ into $(b)$: |
| $$ m_2a_2=m_2g-2m_1a_1\cdot \tan\beta$$ | | $$ m_2a_2=m_2g-2m_1a_1\cdot \tan\beta$$ |
| We substitute $(1)$ into $(c)$: | | We substitute $(1)$ into $(c)$: |
| $$ m_2a_2=m_2g-2m_1a_2\cdot \tan^2\beta$$ | | $$ m_2a_2=m_2g-2m_1a_2\cdot \tan^2\beta$$ |
| We express $a_2$: | | We express $a_2$: |
| $$ \boxed{a_2=\frac{m_2g}{m_2+2m_1\,\tan^2\beta}}\quad(d) $$ | | $$ \boxed{a_2=\frac{m_2g}{m_2+2m_1\,\tan^2\beta}}\quad(d) $$ |
| Given $(1)$, we multiply $(d)$ by $\tan\beta$: | | Given $(1)$, we multiply $(d)$ by $\tan\beta$: |
| $$ \boxed{a_1=\frac{m_2g\,\tan\beta}{m_2+2m_1\,\tan^2\beta}} $$ | | $$ \boxed{a_1=\frac{m_2g\,\tan\beta}{m_2+2m_1\,\tan^2\beta}} $$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$a_1=\frac{m_2g\,\tan\beta}{m_2+2m_1\,\tan^2\beta}$$ | | $$a_1=\frac{m_2g\,\tan\beta}{m_2+2m_1\,\tan^2\beta}$$ |
| $$a_2=\frac{m_2g}{m_2+2m_1\,\tan^2\beta}$$ | | $$a_2=\frac{m_2g}{m_2+2m_1\,\tan^2\beta}$$ |
