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| 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> |
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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> |
| <p> | | <p> |
| $2.2.41.$ A uniform chain is suspended from the thread at one end so that the other end touches the surface of the table. The thread is burned out. Determine the dependence of the pressure force of the chain on the table on the length of its part that has not yet fallen. The impact of the links on the table is inelastic, the mass of the chain is $m$, its length is $l$. | | $2.2.41.$ A uniform chain is suspended from the thread at one end so that the other end touches the surface of the table. The thread is burned out. Determine the dependence of the pressure force of the chain on the table on the length of its part that has not yet fallen. The impact of the links on the table is inelastic, the mass of the chain is $m$, its length is $l$. |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| As the lower part of the thread has almost no time to acquire velocity, the $\Delta v$ tends to zero: | | As the lower part of the thread has almost no time to acquire velocity, the $\Delta v$ tends to zero: |
| $$\Delta v \to 0;\quad F_1\to0$$ | | $$\Delta v \to 0;\quad F_1\to0$$ |
| The gravitational force of the fallen part of the chain: | | The gravitational force of the fallen part of the chain: |
| $$F_g=m_x'g=\frac{mx}{L}g$$ | | $$F_g=m_x'g=\frac{mx}{L}g$$ |
| Consider an infinitesimal fraction of $\Delta x$ that has velocity $v$ and will fall while $m_x'$ is at rest | | Consider an infinitesimal fraction of $\Delta x$ that has velocity $v$ and will fall while $m_x'$ is at rest |
| $$m'_{\Delta x}=\rho\cdot \Delta x=m\cdot\frac{\Delta x}{L}$$ | | $$m'_{\Delta x}=\rho\cdot \Delta x=m\cdot\frac{\Delta x}{L}$$ |
| Change in the momentum $\Delta p'_{\Delta x}$ over the small period of time $\Delta t$ | | Change in the momentum $\Delta p'_{\Delta x}$ over the small period of time $\Delta t$ |
| $$\frac{\Delta p'_{\Delta x}}{\Delta t} =\frac{m'_{\Delta x}\cdot v}{\Delta t}=\frac{mv}{L}\frac{\Delta x}{\Delta t} = \boxed{\frac{mv^2}{L}}\quad(1)$$ | | $$\frac{\Delta p'_{\Delta x}}{\Delta t} =\frac{m'_{\Delta x}\cdot v}{\Delta t}=\frac{mv}{L}\frac{\Delta x}{\Delta t} = \boxed{\frac{mv^2}{L}}\quad(1)$$ |
| The falling time could be found as | | The falling time could be found as |
| $$x=\frac{gt^2}{2}\Rightarrow t=\sqrt{\frac{2x}{g}}$$ | | $$x=\frac{gt^2}{2}\Rightarrow t=\sqrt{\frac{2x}{g}}$$ |
| Similarly, the velocity at the lowest point | | Similarly, the velocity at the lowest point |
| $$v=gt=\sqrt{2gx}\Rightarrow v^2=2gx$$ | | $$v=gt=\sqrt{2gx}\Rightarrow v^2=2gx$$ |
| Substituting into $(1)$: | | Substituting into $(1)$: |
| $$F_2=\frac{\Delta p'_{\Delta x}}{\Delta t}=\frac{2mgx}{L}$$ | | $$F_2=\frac{\Delta p'_{\Delta x}}{\Delta t}=\frac{2mgx}{L}$$ |
| Support reaction force | | Support reaction force |
| $$N=F_1+F_2+F_g=0+\frac{mgx}{L}+\frac{2mgx}{L}$$ | | $$N=F_1+F_2+F_g=0+\frac{mgx}{L}+\frac{2mgx}{L}$$ |
| $$N=\frac{3mgx}{L}\quad(2)$$ | | $$N=\frac{3mgx}{L}\quad(2)$$ |
| The length length of its part that has not yet fallen | | The length length of its part that has not yet fallen |
| $$L_0=L-x\Leftrightarrow x=L-L_0$$ | | $$L_0=L-x\Leftrightarrow x=L-L_0$$ |
| After substituting into $(2)$ | | After substituting into $(2)$ |
| $$\boxed{N=3mg\left(1-\frac{L_0}{L}\right)}$$ | | $$\boxed{N=3mg\left(1-\frac{L_0}{L}\right)}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$F=3mg(1-x/l)$$ | | $$F=3mg(1-x/l)$$ |
| </p> | | </p> |
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| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
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