Updated greek laters @ latex compiling

astrosander edited
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revision #9964 parent #9496 GitHub dbb31ba ← older newer →
en/4.1.13.md +9 −9
@@ -90,9 +90,9 @@
$$P_A = P + \rho_0 g(h+a\sin{\alpha})$$
$$P_C = P + \rho_0 gh$$
If both equations are rested,
− $$P_A-P_C = \rho ga\sin{\alpha} \;(1)$$
+ $$P_A-P_C = \rho ga\sin{\alpha} \quad(1)$$
Note that difference of pressures between any point over side $AB$ and any point over $CD$ is expressed by $(1)$ (for any two points at same $y$ and at $a$ distance one from the other). If $(1)$ is multiplied by the side area $a^2$, buoyancy force can be found for $x$-axis
− $$F_x = \rho g a^3 \sin{\alpha} \;(2)$$
+ $$F_x = \rho g a^3 \sin{\alpha} \quad(2)$$
Applying Newton's Second Law for $x$-axis, also considering friction force
$$mg\sin{\alpha} = f_r + F_x$$
As $m = \rho a^3$ and according $(2)$
@@ -114,19 +114,19 @@
Atmospheric-pressure force is
$$F_0 = \frac{Pa^2}{\cos{\alpha}}$$
but the cube doesn't "feel" that force, else a component of it ($F_{0y}$).
− $$F_{0y} = F_0\cos{\alpha} = Pa^2 \;(3)$$
+ $$F_{0y} = F_0\cos{\alpha} = Pa^2 \quad(3)$$
The pressure exerted by water mass is
− $$F_w = m_w g \cos{\alpha} = \rho_0 V_w g \cos{\alpha} \;(4)$$
+ $$F_w = m_w g \cos{\alpha} = \rho_0 V_w g \cos{\alpha} \quad(4)$$
This volume of water $V_w$ is
− $$V_w = \frac{a^2h}{\cos{\alpha}} + \frac{a^3\tan{\alpha}}{2} \;(5)$$
+ $$V_w = \frac{a^2h}{\cos{\alpha}} + \frac{a^3\tan{\alpha}}{2} \quad(5)$$
Putting $(5)$ into $(4)$
− $$F_w = \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \;(6)$$
+ $$F_w = \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \quad(6)$$
The pressure force over superior side of cube is
− $$F_p = F_{0y} + F_w \;(7)$$
+ $$F_p = F_{0y} + F_w \quad(7)$$
Substituting $(3)$ and $(6)$ into $(7)$
− $$F_p = Pa^2 + \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \;(8)$$
+ $$F_p = Pa^2 + \rho_0 g a^3\left(\frac{h}{a}+\frac{\sin{\alpha}}{2}\right) \quad(8)$$
Applying Newton Second Law for $y$-axis
− $$N = mg\cos{\alpha}+F_p = \rho a^3g\cos{\alpha}+F_p \;(9)$$
+ $$N = mg\cos{\alpha}+F_p = \rho a^3g\cos{\alpha}+F_p \quad(9)$$
and according to $(8)$
$$\boxed{N = a^3\rho_0 g \left(\frac{\rho}{\rho_0}\cos{\alpha}+\frac{1}{2}\sin{\alpha}+\frac{h}{a}\right)+Pa^2}$$
This normal force is directed in opposed direction of $y$-axis.
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