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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#2.8">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <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 $B$?
+</p>
+<center>
+ <figure>
+ <img src="2.8.11.png"
+ loading="lazy" width="200" />
+ <figcaption>
+ For problem $2.8.11$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="2.8.11_1.png"
+ loading="lazy" width="300" />
+ <figcaption>
+ Forces acting on the stick
+ </figcaption>
+ </figure>
+</center>
+<p>
+Equilibrium condition for vertical and horizontal axes respectively
+$$mg = N_A \sin2\alpha +N_B\cos\alpha\quad(1)$$
+$$N_A \cos2\alpha = N_B\sin\alpha\quad(2)$$
+Moment of forces with respect to point $A$
+$$mg\cos\alpha \cdot \frac{l}{2} = N_B\cdot l_{AB}\quad(3)$$
+Alternatively, moment of forces with respect to point $B$
+$$mg\cos\alpha \cdot \left(l_{AB} - \frac{l}{2}\right) = N_A\sin\alpha\cdot l_{AB}\quad(4)$$
+Substituting in $(2)$, $(3)$ and $(4)$
+$$\frac{l}{l_{AB}} = \frac{2\cos2\alpha}{\cos^2\alpha}$$
+$$\boxed{N_A = mg\tan\alpha}$$
+$$\boxed{N_B = mg\frac{\cos2\alpha}{\cos\alpha}}$$
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$F_A=mg\operatorname{tg}\alpha; F_B=mg\cos2\alpha/\cos\alpha$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
+ </p>
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