Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

Statement

$2.8.11.$ 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 $B$?

For problem $2.8.11$

Solution

Forces acting on the stick

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 $(4)$ $$\frac{l}{l_{AB}} = \frac{2\cos2\alpha}{\cos^2\alpha}$$ $$\boxed{N_A = mg\tan\alpha}$$ $$\boxed{N_B = mg\frac{\cos2\alpha}{\cos\alpha}}$$

Answer

$$F_A=mg\operatorname{tg}\alpha; F_B=mg\cos2\alpha/\cos\alpha$$

Almaskhan Arsen