Answer:
The viscous shear stress is 968 Pa (or can be written as 968 [tex]N/m^2[/tex]) and the drag force is 968 N.
Explanation:
Viscous shear stress
For a parallel flow of a Newtonian fluid, the shear stress is proportional to the gradient of the velocity,
[tex]\tau_{yx}= \mu \cfrac{du}{dy}[/tex]
Considering a large square plate moving on a thin film of thickness h, the velocity profile is
[tex]u(y) = U\cfrac{y}{h}[/tex]
Thus its derivative will be just
[tex]\cfrac{du}{dy}= \cfrac U h[/tex]
So replacing on the viscous shear stress formula we get
[tex]\tau_{yx}= \mu \cfrac {U}h[/tex]
We can then replace the given information
[tex]\tau_{yx}= 0.968 \times \cfrac{kg}{m\times s}\times \cfrac{0.1 \cfrac ms}{0.0001 \, m}[/tex]
Evaluating we get
[tex]\boxed{\tau_{yx}=968\, Pa}[/tex]
The viscous shear stress is 968 Pascals or [tex]N/m^2[/tex].
Drag force
Using the given equation for the drag force we have
[tex]D = \tau_{yx} A[/tex]
And since we have a large square plate of sides L we can write the area as
[tex]A = L^2[/tex]
So the drag force is
[tex]D = \tau_{yx} L^2[/tex]
Replacing values
[tex]D = 968\, Pa \times (1\, m)^2[/tex]
We get
[tex]\boxed{D= 968 \,N}[/tex]
The drag force is 968 N.
