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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| <p> | | <p> |
| $2.8.1$ | | $2.8.1$ |
| The figure shows structures that hold a load weighing 10 kg. The cables are represented by thin lines, the rod is represented by a double line. Determine the tension force of the cables for case a and the force acting on the rod from the side of the cable thrown over it for case b. | | The figure shows structures that hold a load weighing 10 kg. The cables are represented by thin lines, the rod is represented by a double line. Determine the tension force of the cables for case a and the force acting on the rod from the side of the cable thrown over it for case b. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="2.8.1" width="400" /> | | loading="lazy" alt="2.8.1" width="400" /> |
| <figcaption> | | <figcaption> |
| For problem 2.8.1 | | For problem 2.8.1 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's consider the following figure | | Let's consider the following figure |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw.png" | | <img src="draw.png" |
| loading="lazy" alt="2.8.1" width="500" /> | | loading="lazy" alt="2.8.1" width="500" /> |
| <figcaption> | | <figcaption> |
| Force analysis | | Force analysis |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| $a$. Applying Newton Second Law for $x$-direction | | $a$. Applying Newton Second Law for $x$-direction |
| $$T_2~\cos{30^\circ}-T_1~\cos{30^\circ}=0$$ | | $$T_2~\cos{30^\circ}-T_1~\cos{30^\circ}=0$$ |
| $$T_1=T_2 \;(1)$$ | | $$T_1=T_2 \;(1)$$ |
| for $y$-direction | | for $y$-direction |
| $$(T_1+T_2)~\sin{30^\circ} = mg \;(2)$$ | | $$(T_1+T_2)~\sin{30^\circ} = mg \;(2)$$ |
| Let be $T_1=T_2=T$, so, back to $(2)$ | | Let be $T_1=T_2=T$, so, back to $(2)$ |
| $$2T~\sin{30^\circ}=mg$$ | | $$2T~\sin{30^\circ}=mg$$ |
| $$T = \frac{mg}{2~\sin{30^\circ}} = mg = \boxed{98~\rm{N}}$$ | | $$T = \frac{mg}{2~\sin{30^\circ}} = mg = \boxed{98~\rm{N}}$$ |
| $b$. According figure, | | $b$. According figure, |
| $$F^2 = T^2+T^2 = 2T^2$$ | | $$F^2 = T^2+T^2 = 2T^2$$ |
| We consider the same tension $T$ because we suppose the threads are unextensible. | | We consider the same tension $T$ because we suppose the threads are unextensible. |
| $$F = \sqrt{2}~T \;(3)$$ | | $$F = \sqrt{2}~T \;(3)$$ |
| Applying Newton Second Law | | Applying Newton Second Law |
| $$T = mg \;(4)$$ | | $$T = mg \;(4)$$ |
| Putting $(4)$ into $(3)$ | | Putting $(4)$ into $(3)$ |
| $$F = \sqrt{2}~mg = \boxed{138~\rm{N}}$$ | | $$F = \sqrt{2}~mg = \boxed{138~\rm{N}}$$ |
| Note: Calculations were made considering $g$ = 9.8 N/kg and $\sqrt{2}$ = 1.41. | | Note: Calculations were made considering $g$ = 9.8 N/kg and $\sqrt{2}$ = 1.41. |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
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