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mathmari

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MHB

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Hello!

A two-dimensional beam of thickness S and velocity c (evenly distributed at the thickness) falls in stationary plate and get separated. Calculate the ratio of thicknesses S1/S2 as a function of the angle φ.

(see attachment)In my notes there is this solution:

Continuity equation:

Let c

$$0: \overrightarrow{u}=c\hat{e} \ \ , \ \ \overrightarrow{n}_0=-\hat{e}=-(-\sin \phi \hat{i}+\cos \phi \hat{j}) \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_0=-c \\ 1: \overrightarrow{u}=c_1 \overrightarrow{n}_1=c_1 \hat{i} \ \ , \ \ \overrightarrow{n}_1=\hat{i} \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_1=c_1 \\ 2: \overrightarrow{u}=c_2 \overrightarrow{n}_2=-c_2\hat{i} \ \ , \ \ \overrightarrow{n}_2=-\hat{i} \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_2=c_2$$

$$\int_{\partial{W_1}}\overrightarrow{u} \cdot \overrightarrow{n}dA=0 \Rightarrow -c \cdot (S \cdot 1)+c_1 \cdot (S_1 \cdot 1)+c_2 \cdot (S_2 \cdot 1)=0 \Rightarrow cS=c_1S+c_2S_2 \ \ \ \ \ (1) $$

Conservation of energ $\Rightarrow $ Bernoulli equation

$$\frac{1}{2}|\overrightarrow{u}|^2+\frac{p}{\rho_0}=\text{ constant along a stremline }$$

$$0 \rightarrow 1 : \frac{1}{2}c^2+\frac{p_a}{\\rho_0}=\frac{1}{2}c_1^2+\frac{p_a}{\rho_0} \Rightarrow c_1=c \\ 0 \rightarrow 2 : \frac{1}{2}c^2+\frac{p_a}{\\rho_0}=\frac{1}{2}c_2^2+\frac{p_a}{\rho_0} \Rightarrow c_2=c \\ \Rightarrow c_1=c_2=c \ \ \ \ \ (2)$$

$$(1) \wedge (2) \Rightarrow S=S_1+S_2 \ \ \ \ \ (3)$$

$$\overrightarrow{F}_{W_0}=-\int_{\partial_{W_1}}p \overrightarrow{n} dA-\int_{\partial{W_1}}\rho \overrightarrow{u} (\overrightarrow{u} \cdot \overrightarrow{n})dA$$

$$\int_{\partial_{W_1}}p \overrightarrow{n} dA=p_a \int_{\partial{W_1}}\overrightarrow{n}dA=0$$

$$\int_{\partial{W_1}}\rho \overrightarrow{u} (\overrightarrow{u} \cdot \overrightarrow{n})dA=\rho_0 (c \hat{e}(-c)S+c_1 \hat{i}c_1S_1+(-c_2 \hat{i})c_2S_2)=(\rho_0c^2S\cos \phi +\rho_0 c_1^2S_1-\rho_0c_2^2S_2)\hat{i}-\rho c^2S\sin \phi \hat{j}=-\overrightarrow{F}_{W_0}$$

$$F_{W_0, x}=-\rho_0c^2(S\cos \phi +S_1-S_2) \ \ \ \ \ (4a) \\ F_{W_0, y}=\rho c^2 S \sin \phi \ \ \ \ \ (4b)$$

Since the fluid is ideal we have that $$F_{W_0, x}=0 \Rightarrow S_2-S_1=S \cos \phi \ \ \ \ \ (5)$$

$$(3) \wedge (5) \Rightarrow \\S_1=\frac{S}{2}(1+\cos \phi) \\ S_2=\frac{S}{2}(1-\cos \phi) \\ \Rightarrow \frac{S_2}{S_1}=\frac{1-\cos \phi}{1+\cos \phi}$$

Could you explain to me the general idea of the solution? I haven't really understood that...Every time when we have such an exercise do we have to use the continuity equation and the Bernoulli equation?

Also could you explain to me why the following equality stand?

$$\int_{\partial{W_1}}\rho \overrightarrow{u} (\overrightarrow{u} \cdot \overrightarrow{n})dA=\rho_0 (c \hat{e}(-c)S+c_1 \hat{i}c_1S_1+(-c_2 \hat{i})c_2S_2)$$

A two-dimensional beam of thickness S and velocity c (evenly distributed at the thickness) falls in stationary plate and get separated. Calculate the ratio of thicknesses S1/S2 as a function of the angle φ.

(see attachment)In my notes there is this solution:

Continuity equation:

Let c

_{1}, c_{2}the velocities of the sections 1, 2$$0: \overrightarrow{u}=c\hat{e} \ \ , \ \ \overrightarrow{n}_0=-\hat{e}=-(-\sin \phi \hat{i}+\cos \phi \hat{j}) \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_0=-c \\ 1: \overrightarrow{u}=c_1 \overrightarrow{n}_1=c_1 \hat{i} \ \ , \ \ \overrightarrow{n}_1=\hat{i} \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_1=c_1 \\ 2: \overrightarrow{u}=c_2 \overrightarrow{n}_2=-c_2\hat{i} \ \ , \ \ \overrightarrow{n}_2=-\hat{i} \Rightarrow \overrightarrow{u} \cdot \overrightarrow{n}_2=c_2$$

$$\int_{\partial{W_1}}\overrightarrow{u} \cdot \overrightarrow{n}dA=0 \Rightarrow -c \cdot (S \cdot 1)+c_1 \cdot (S_1 \cdot 1)+c_2 \cdot (S_2 \cdot 1)=0 \Rightarrow cS=c_1S+c_2S_2 \ \ \ \ \ (1) $$

Conservation of energ $\Rightarrow $ Bernoulli equation

$$\frac{1}{2}|\overrightarrow{u}|^2+\frac{p}{\rho_0}=\text{ constant along a stremline }$$

$$0 \rightarrow 1 : \frac{1}{2}c^2+\frac{p_a}{\\rho_0}=\frac{1}{2}c_1^2+\frac{p_a}{\rho_0} \Rightarrow c_1=c \\ 0 \rightarrow 2 : \frac{1}{2}c^2+\frac{p_a}{\\rho_0}=\frac{1}{2}c_2^2+\frac{p_a}{\rho_0} \Rightarrow c_2=c \\ \Rightarrow c_1=c_2=c \ \ \ \ \ (2)$$

$$(1) \wedge (2) \Rightarrow S=S_1+S_2 \ \ \ \ \ (3)$$

$$\overrightarrow{F}_{W_0}=-\int_{\partial_{W_1}}p \overrightarrow{n} dA-\int_{\partial{W_1}}\rho \overrightarrow{u} (\overrightarrow{u} \cdot \overrightarrow{n})dA$$

$$\int_{\partial_{W_1}}p \overrightarrow{n} dA=p_a \int_{\partial{W_1}}\overrightarrow{n}dA=0$$

$$\int_{\partial{W_1}}\rho \overrightarrow{u} (\overrightarrow{u} \cdot \overrightarrow{n})dA=\rho_0 (c \hat{e}(-c)S+c_1 \hat{i}c_1S_1+(-c_2 \hat{i})c_2S_2)=(\rho_0c^2S\cos \phi +\rho_0 c_1^2S_1-\rho_0c_2^2S_2)\hat{i}-\rho c^2S\sin \phi \hat{j}=-\overrightarrow{F}_{W_0}$$

$$F_{W_0, x}=-\rho_0c^2(S\cos \phi +S_1-S_2) \ \ \ \ \ (4a) \\ F_{W_0, y}=\rho c^2 S \sin \phi \ \ \ \ \ (4b)$$

Since the fluid is ideal we have that $$F_{W_0, x}=0 \Rightarrow S_2-S_1=S \cos \phi \ \ \ \ \ (5)$$

$$(3) \wedge (5) \Rightarrow \\S_1=\frac{S}{2}(1+\cos \phi) \\ S_2=\frac{S}{2}(1-\cos \phi) \\ \Rightarrow \frac{S_2}{S_1}=\frac{1-\cos \phi}{1+\cos \phi}$$

Could you explain to me the general idea of the solution? I haven't really understood that...Every time when we have such an exercise do we have to use the continuity equation and the Bernoulli equation?

Also could you explain to me why the following equality stand?

$$\int_{\partial{W_1}}\rho \overrightarrow{u} (\overrightarrow{u} \cdot \overrightarrow{n})dA=\rho_0 (c \hat{e}(-c)S+c_1 \hat{i}c_1S_1+(-c_2 \hat{i})c_2S_2)$$