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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="../../#4.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#4.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $4.1.22.$ At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle $\alpha$ to the horizon if the friction coefficient is $\mu$? | | $4.1.22.$ At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle $\alpha$ to the horizon if the friction coefficient is $\mu$? |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| <center> | | <center> |
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| <img src="4.1.22_1.png" | | <img src="4.1.22_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> |
| | | |
| The vessel with liquid goes downward with acceleration $a$ as shown in the figure</p><p> | | The vessel with liquid goes downward with acceleration $a$ as shown in the figure</p><p> |
| Given the reaction force $N=mg\cos\alpha$, the sliding friction force is found as | | Given the reaction force $N=mg\cos\alpha$, the sliding friction force is found as |
| $$F_\text{fr} = \mu N = \mu mg\cos\alpha$$ | | $$F_\text{fr} = \mu N = \mu mg\cos\alpha$$ |
| Newton's second law for the vessel | | Newton's second law for the vessel |
| $$ma = mg\sin\alpha-F_\text{fr}$$ | | $$ma = mg\sin\alpha-F_\text{fr}$$ |
| $$\boxed{a=g(\sin\alpha-\mu\cos\alpha)}\quad(1)$$ | | $$\boxed{a=g(\sin\alpha-\mu\cos\alpha)}\quad(1)$$ |
| Let us introduce the effective acceleration of free fall $\vec{g^*}$ in a non-inertial frame of reference associated with the vessel, then vectorially, it is found as | | Let us introduce the effective acceleration of free fall $\vec{g^*}$ in a non-inertial frame of reference associated with the vessel, then vectorially, it is found as |
| $$\vec{g^*}=\vec{g}-\vec{a}$$ | | $$\vec{g^*}=\vec{g}-\vec{a}$$ |
| | | |
| </p> | | </p> |
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| <img src="4.1.22_2.png" | | <img src="4.1.22_2.png" |
| loading="lazy" width="150" /> | | loading="lazy" width="150" /> |
| <figcaption> | | <figcaption> |
| Effective acceleration of free fall $\vec{g^*}$ | | Effective acceleration of free fall $\vec{g^*}$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
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