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| <h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.3.46.$ What force should be applied to the upper weight of mass $m_1$ so that the lower weight of mass $m_2$, connected to the upper one by a spring, breaks off from the floor after this force ceases? | | $2.3.46.$ What force should be applied to the upper weight of mass $m_1$ so that the lower weight of mass $m_2$, connected to the upper one by a spring, breaks off from the floor after this force ceases? |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.3.46.png" | | <img src="2.3.46.png" |
| loading="lazy" width="170" /> | | loading="lazy" width="170" /> |
| <figcaption> | | <figcaption> |
| For problem $2.3.46$ | | For problem $2.3.46$ |
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| <p> | | <p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| <img src="2.3.46_1.png" | | <img src="2.3.46_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> |
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| <p> | | <p> |
| | | |
| Forces acting on the upper body in the force-less situation ($F=0$) | | Forces acting on the upper body in the force-less situation ($F=0$) |
| In the case where no external force $F$ is applied, we have the equation: | | In the case where no external force $F$ is applied, we have the equation: |
| $$N+k\Delta x'=m_2g $$ | | $$N+k\Delta x'=m_2g $$ |
| As $N=0$: | | As $N=0$: |
| $$\Delta x' = \frac{m_2g}{k}$$ | | $$\Delta x' = \frac{m_2g}{k}$$ |
| For the case where there is a force $F$ | | For the case where there is a force $F$ |
| $$\Delta x = \frac{F+m_1g}{k}$$ | | $$\Delta x = \frac{F+m_1g}{k}$$ |
| Law of energy preservation | | Law of energy preservation |
| $$m_1g\left(l-\Delta x\right)+\frac{k\Delta x^2}{2} = m_1g\left(l+\Delta x'\right)+\frac{k{\Delta x'}^2}{2}$$ | | $$m_1g\left(l-\Delta x\right)+\frac{k\Delta x^2}{2} = m_1g\left(l+\Delta x'\right)+\frac{k{\Delta x'}^2}{2}$$ |
| $$-m_1g\left(\Delta x + \Delta x'\right)=\frac{k}{2}\left({\Delta x'}^2-{\Delta x}^2\right)$$ | | $$-m_1g\left(\Delta x + \Delta x'\right)=\frac{k}{2}\left({\Delta x'}^2-{\Delta x}^2\right)$$ |
| The expression $ {\Delta x'}^2 - {\Delta x}^2 $ can be rewritten using the difference of squares: | | The expression $ {\Delta x'}^2 - {\Delta x}^2 $ can be rewritten using the difference of squares: |
| $$2m_1g=k\Delta x - k\Delta x'$$ | | $$2m_1g=k\Delta x - k\Delta x'$$ |
| Substituting the previously obtained values | | Substituting the previously obtained values |
| $$2m_1g=F+m_1g-m_2g$$ | | $$2m_1g=F+m_1g-m_2g$$ |
| From where the final result could be obtained | | From where the final result could be obtained |
| $$\boxed{F=m_1g+m_2g}$$ | | $$\boxed{F=m_1g+m_2g}$$ |
| </p> | | </p> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$F=(m_1+m_2)g$$ | | $$F=(m_1+m_2)g$$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
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