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| <h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
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| <p> | | <p> |
| $2.2.24$ | | $2.2.24$ |
| 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? |
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| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem 2.2.24 | | For problem 2.2.24 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw1.png" | | <img src="draw1.png" |
| loading="lazy" width="250" /> | | loading="lazy" width="250" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the system | | Forces acting on the system |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Let's consider this system as different bodies. Using Newton's Second law of motion, we can get: | | Let's consider this system as different bodies. Using Newton's Second law of motion, we can get: |
| $$ | | $$ |
| \begin{cases} | | \begin{cases} |
| \vec{F}_1 = m_1 \vec{a}_1 \\ | | \vec{F}_1 = m_1 \vec{a}_1 \\ |
| \vec{F}_2 = m_2 \vec{a}_2 | | \vec{F}_2 = m_2 \vec{a}_2 |
| \end{cases} | | \end{cases} |
| \Rightarrow | | \Rightarrow |
| \begin{cases} | | \begin{cases} |
| \vec{a}_1 = \frac{\vec{F}_1}{m_1} \\ | | \vec{a}_1 = \frac{\vec{F}_1}{m_1} \\ |
| \vec{a}_2 = \frac{\vec{F}_2}{m_2} | | \vec{a}_2 = \frac{\vec{F}_2}{m_2} |
| \end{cases} | | \end{cases} |
| \quad \text{(1)} | | \quad \text{(1)} |
| $$ | | $$ |
| Let's subtract $\vec{a}_2$ from $\vec{a}_1$: | | Let's subtract $\vec{a}_2$ from $\vec{a}_1$: |
| $$ | | $$ |
| \vec{a}_1 - \vec{a}_2 = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(2)} | | \vec{a}_1 - \vec{a}_2 = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(2)} |
| $$ | | $$ |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw2.png" | | <img src="draw2.png" |
| loading="lazy" width="180" /> | | loading="lazy" width="180" /> |
| <figcaption> | | <figcaption> |
| Subtraction of vectors | | Subtraction of vectors |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| As we can see, it looks like the derivative of relative velocity: | | As we can see, it looks like the derivative of relative velocity: |
| $$ | | $$ |
| \vec{v}_{B/A} = \vec{v}_B - \vec{v}_A \quad \text{(3)} | | \vec{v}_{B/A} = \vec{v}_B - \vec{v}_A \quad \text{(3)} |
| $$ | | $$ |
| Now, let's solve the derivative: | | Now, let's solve the derivative: |
| $$ | | $$ |
| \frac{d(\vec{v}_{B/A})}{dt} = \frac{d(\vec{v}_B)}{dt} - \frac{d(\vec{v}_A)}{dt} \Rightarrow \vec{a}_{B/A} = \vec{a}_B - \vec{a}_A \quad \text{(4)} | | \frac{d(\vec{v}_{B/A})}{dt} = \frac{d(\vec{v}_B)}{dt} - \frac{d(\vec{v}_A)}{dt} \Rightarrow \vec{a}_{B/A} = \vec{a}_B - \vec{a}_A \quad \text{(4)} |
| $$ | | $$ |
| This is why we can say that: | | This is why we can say that: |
| <div class="scroll-wrapper"> | | <div class="scroll-wrapper"> |
| $$ | | $$ |
| \vec{a}_1 - \vec{a}_2 = \vec{a}_{12} = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(5)} | | \vec{a}_1 - \vec{a}_2 = \vec{a}_{12} = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(5)} |
| $$ | | $$ |
| </div> | | </div> |
| Due to the weightlessness of the thread and Newton's Third Law of Motion $\vec{F}_1 = -\vec{F}_2 = \vec{F}$: | | Due to the weightlessness of the thread and Newton's Third Law of Motion $\vec{F}_1 = -\vec{F}_2 = \vec{F}$: |
| | | |
| Eventually: | | Eventually: |
| $$ | | $$ |
| \vec{a}_{12} = \vec{F} \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = \vec{F} \frac{m_1 + m_2}{m_1 m_2} \quad \text{(6)} | | \vec{a}_{12} = \vec{F} \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = \vec{F} \frac{m_1 + m_2}{m_1 m_2} \quad \text{(6)} |
| $$ | | $$ |
| $a_{12} = \frac{v^2}{l}$, where $l$ is the length of the thread, and $v$ is the relative velocity$\quad (7)$ | | $a_{12} = \frac{v^2}{l}$, where $l$ is the length of the thread, and $v$ is the relative velocity$\quad (7)$ |
| $$ | | $$ |
| F = \frac{m_1 m_2}{m_1 + m_2} \frac{v^2}{l} \quad \text{(8)} | | F = \frac{m_1 m_2}{m_1 + m_2} \frac{v^2}{l} \quad \text{(8)} |
| $$ | | $$ |
| </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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