$4.2.13$ Two equal trunks are placed as the figure shows. The lower trunk is linked to the vertical wall by wires that form an angle of 45$^\circ$ with it. The upper trunk is half submerged in water. Find the density of the trunks.
Let's analyze the following force diagram
Consider the trunks are perfect wooden cylinders. Applying Newton Second Law for cylinder 1, on $x$-direction, $$-N'\cos{45^\circ}+N = 0 \;(1)$$ and on $y$-axis $$mg = \rho_0 g V_s + N'\sin{45^\circ} = \rho_0 g V_{s1} + N'\frac{\sqrt{2}}{2}$$ as trunk 1 is half submerged, $V_{s1} = \frac{\pi R^2 L}{2}$ and its mass is $m = \rho V$, so separating $\rho$ $$\rho = \frac{1}{2}\left(\rho_0+\frac{N'\sqrt{2}}{\pi R^2 L}\right) \;(2)$$ Applying Newton Second Law for cylinder 2, on $x$-axis $$N'\cos{45^\circ}-T\cos{45^\circ}=0$$ $$N' = T \;(3)$$ and for $y$-axis $$\rho_0 g V_{s2} = T\sin{45^\circ}+N'\sin{45^\circ}+mg$$ in this case $V_{s2} = \pi R^2 L$ and considering that $\sin{45^\circ}=\cos{45^\circ}=\frac{\sqrt{2}}{2}$, $$\rho_0 g \pi R^2 L = (T+N')\frac{\sqrt{2}}{2}+\rho g \pi R^2 L \;(4)$$ Putting $(3)$ int $(4)$ $$(\rho_0-\rho)\pi R^2 L g = N' \sqrt{2} \;(5)$$ Substituting $(5)$ into $(2)$ and developing algebraically $$\rho = \frac{2}{3}\rho_0$$ as the water's density is $\rho_0=1$ g/cm$^3$, $$\boxed{\rho=\frac{2}{3}~\rm{g/cm^3}}$$
BSc. Luis Daniel Fernández Quintana
Physics Department (FCNE)
Universidad de Oriente, Cuba