Answer:
The answer is below
Explanation:
The constant velocity V = 7.5 m/s, viscosity (μ) = 0.9 N·s/m2,
a) The upper and lower shear force are given as:
[tex]F{shear,upper}=\tau_{w,u} A_s\\\\F{shear,upper}=\mu A_s|\frac{du}{dy} |\\[/tex]
Let us assume that both the upper and lower surfaces are stationary, Therefore:
[tex]V_u=velocity\ of\ upper\ surface=0\ m/s,V_l=velocity\ of\ lower\ surface=0\ m/s,[/tex]
[tex]A_s=Area\ of\ plate=2\ m*0.5\ m=1\ m^2[/tex]
h1 = 3 cm = 0.03 m
h2 = 1 cm = 0.01 m
For the upper surface, du = V - [tex]V_u[/tex] = 7.5 - 0 = 7.5 m/s. dy = h1 = 0.03
[tex]F_{shear,upper}=\mu A_s*\frac{du}{dy}=0.9*1*\frac{7.5}{0.03}=225\ N[/tex]
For the lower surface, du = V - [tex]V_p[/tex] = 7.5 - 0 = 7.5 m/s. dy = h2 = 0.01
[tex]F_{shear,lower}=\mu A_s*\frac{du}{dy}=0.9*1*\frac{7.5}{0.01}=675\ N[/tex]