The center of pressure is located at; X_cp = c/2
We are given;
Chord = c
Angle of attack = α
Pu (s) = c₁
P₁(s) = c₂, and c₂ > c₁
First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.
N' = [tex]\int\limits^T_L ({P_{u}dx - \tau_{u} dy} \,) + \int\limits^T_L ({P_{l}dx - \tau_{l} dy} \,)[/tex]
That is the normal force.
Now the total moment about the leading edge is;
[tex]M_{LE} = \int\limits^T_L [({P_{u}dx - \tau_{u} dy} \,)x - ({P_{u}dy - \tau_{u} dx})] + \int\limits^T_L [({-P_{l}dx - \tau_{l} dy} \,)x + (-{P_{l}dy - \tau_{l} dx})y][/tex]
Now, if we ignore the y-direction and neglect shear stress and put C for T and 0 for L, then we have;
[tex]N' = \int\limits^c_0 ({P_{L} - P_{U}} \,) dx \\and M'_{LE} = \int\limits^c_0 ({P_{U} - P_{L}} \,) x.dx[/tex]
When we integrate the above, we arrive at;
[tex]N' = (P_{L} - P_{U})C \\Also, M'_{LE} = (P_{U} - P_{L})\frac{C^{2}}{2}[/tex]
From the above, we can see that the moment about the leading edge, P_u subtracts P_l which renders it negative.
Thus;
M'_LE can be expressed as the product of the normal force and the length from the leading edge to the location to the center of the pressure. Thus;
M'_LE = N' * X_cp
So, [tex]X_{CP} = \frac{-(P_{u} - P_{l})\frac{C^{2}}{2} }{(P_{L} - P_{U})C} = \frac{(P_{L} - P_{U})\frac{C^{2}}{2} }{(P_{L} - P_{U})C}[/tex]
X_cp = (C²/2) ÷ C
X_cp = C²/2C = C/2
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