Respuesta :
The lift and drag if the shear stress is neglected, than the values of F0 = 1.1955 KN & F1 = 8.014 KN.
The average pressure and shear stress acting on the surface of the 1-m-square flat plate.
P1 = 2.3 * 3 = 6.9 KN/m^2
P2 = -1.2 KN/m^2
T1 = 7.6 * 10^-2 * 2 = 0.152 KN/m^2
T2 = 0.058 KN/m^2
Now,
Drag force
F0 = F1sinα - F2sinα + FS1cosα+FS2cosα
= P1Asinα - P2sinα + T1Acosα + T2Acosα
= Asinα(P1-P2) + Acosα(T1 + T2)
= 1.sin(T)[6.9-(-1.2)] + 1. cos (T)[0.152 + 0.058]
= 0.98714 + 0.20843
= 1.1955 KN
Lift force
F1 = F1cosα - F2cosα - FS1sinα - FS2sinα
= P1cosα - P2Acosα - T1Asinα - T2Asinα
= Acosα(P1 - P2) - A.sinα(T1 + T2)
= 1. cos(T)(6.9 - (-1.2)) - 1. sin(T)(0.152 + 0.058)
= 8.014 KN
Hence the answer is the lift and drag if the shear stress is neglected, than the values of F0 = 1.1955 KN & F1 = 8.014 KN.
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