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| <h3 id="back-link"><a href="../../#2.8">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.8">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.8.11.$ In a smooth fixed hemisphere, a stick of mass $m$ freely lies so that its angle with the horizon is equal to $\alpha$, and its end extends beyond the edge of the hemisphere. With what forces does the wand act on the hemisphere at the points of contact between $A$ and $B$? | | $2.8.11.$ In a smooth fixed hemisphere, a stick of mass $m$ freely lies so that its angle with the horizon is equal to $\alpha$, and its end extends beyond the edge of the hemisphere. With what forces does the wand act on the hemisphere at the points of contact between $A$ and $B$? |
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| For problem $2.8.11$ | | For problem $2.8.11$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| <img src="2.8.11_1.png" | | <img src="2.8.11_1.png" |
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| Forces acting on the stick | | Forces acting on the stick |
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| <p> | | <p> |
| Equilibrium condition for vertical and horizontal axes respectively | | Equilibrium condition for vertical and horizontal axes respectively |
| $$mg = N_A \sin2\alpha +N_B\cos\alpha\quad(1)$$ | | $$mg = N_A \sin2\alpha +N_B\cos\alpha\quad(1)$$ |
| $$N_A \cos2\alpha = N_B\sin\alpha\quad(2)$$ | | $$N_A \cos2\alpha = N_B\sin\alpha\quad(2)$$ |
| Moment of forces with respect to point $A$ | | Moment of forces with respect to point $A$ |
| $$mg\cos\alpha \cdot \frac{l}{2} = N_B\cdot l_{AB}\quad(3)$$ | | $$mg\cos\alpha \cdot \frac{l}{2} = N_B\cdot l_{AB}\quad(3)$$ |
| Alternatively, moment of forces with respect to point $B$ | | Alternatively, moment of forces with respect to point $B$ |
| $$mg\cos\alpha \cdot \left(l_{AB} - \frac{l}{2}\right) = N_A\sin\alpha\cdot l_{AB}\quad(4)$$ | | $$mg\cos\alpha \cdot \left(l_{AB} - \frac{l}{2}\right) = N_A\sin\alpha\cdot l_{AB}\quad(4)$$ |
| Substituting in $(2)$, $(3)$ and $(4)$ | | Substituting in $(2)$, $(3)$ and $(4)$ |
| $$\frac{l}{l_{AB}} = \frac{2\cos2\alpha}{\cos^2\alpha}$$ | | $$\frac{l}{l_{AB}} = \frac{2\cos2\alpha}{\cos^2\alpha}$$ |
| $$\boxed{N_A = mg\tan\alpha}$$ | | $$\boxed{N_A = mg\tan\alpha}$$ |
| $$\boxed{N_B = mg\frac{\cos2\alpha}{\cos\alpha}}$$ | | $$\boxed{N_B = mg\frac{\cos2\alpha}{\cos\alpha}}$$ |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$F_A=mg\operatorname{tg}\alpha; F_B=mg\cos2\alpha/\cos\alpha$$ | | $$F_A=mg\operatorname{tg}\alpha; F_B=mg\cos2\alpha/\cos\alpha$$ |
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| <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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