Updated greek laters @ latex compiling

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revision #10158 parent #9234 GitHub dbb31ba ← older newer →
en/4.2.13.md +5 −5
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Consider the trunks are perfect wooden cylinders. Applying Newton Second Law for cylinder 1, on $x$-direction,
− $$-N'\cos{45^\circ}+N = 0 \;(1)$$
+ $$-N'\cos{45^\circ}+N = 0 \quad(1)$$
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 $\rho$
− $$\rho = \frac{1}{2}\left(\rho_0+\frac{N'\sqrt{2}}{\pi R^2 L}\right) \;(2)$$
+ $$\rho = \frac{1}{2}\left(\rho_0+\frac{N'\sqrt{2}}{\pi R^2 L}\right) \quad(2)$$
Applying Newton Second Law for cylinder 2, on $x$-axis
$$N'\cos{45^\circ}-T\cos{45^\circ}=0$$
− $$N' = T \;(3)$$
+ $$N' = T \quad(3)$$
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 $\sin{45^\circ}=\cos{45^\circ}=\frac{\sqrt{2}}{2}$,
− $$\rho_0 g \pi R^2 L = (T+N')\frac{\sqrt{2}}{2}+\rho g \pi R^2 L \;(4)$$
+ $$\rho_0 g \pi R^2 L = (T+N')\frac{\sqrt{2}}{2}+\rho g \pi R^2 L \quad(4)$$
Putting $(3)$ int $(4)$
− $$(\rho_0-\rho)\pi R^2 L g = N' \sqrt{2} \;(5)$$
+ $$(\rho_0-\rho)\pi R^2 L g = N' \sqrt{2} \quad(5)$$
Substituting $(5)$ into $(2)$ and developing algebraically
$$\rho = \frac{2}{3}\rho_0$$
as the water's density is $\rho_0=1$ g/cm$^3$,
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