Respuesta :
Explanation:
The given data is as follows.
      [tex]\mu_{1}[/tex] = 0.1 Pa.s,       [tex]\mu_{2}[/tex] = 0.2 Pa.s
      [tex]h_{1}[/tex] = [tex]h_{2}[/tex] = 1 mm = [tex]1 \times 10^{-3} m[/tex]  (as 1 m = 1000 mm)
As the velocity gradients are linear so, the shear stress will be the same throughout.
         [tex]\nu_{i}[/tex] = velocity at the interface
        [tex]\tau = \mu_{1} \frac{d\mu_{1}}{dy_{1}} = \mu_{2} \frac{d\mu_{2}}{dy_{2}}[/tex]
        [tex]\mu_{1} \times \frac{\nu_{i}}{h_{1}} = \mu_{2} \times \frac{\nu - \nu_{i}}{h_{2}}[/tex]
or, Â Â Â Â Â Â [tex]\nu_{i} = \frac{\nu}{1 + \frac{\mu_{1}h_{2}}{\mu_{2}h_{1}}}[/tex] Â Â
Now, putting the given values into the above formula as follows.
       [tex]\nu_{i} = \frac{\nu}{1 + \frac{\mu_{1}h_{2}}{\mu_{2}h_{1}}}[/tex]  Â
             = [tex]\frac{1 m/s}{1 + \frac{0.1 \times 1}{0.2 \times 1}}[/tex]    Â
             = 0.667 m/s
Hence, force required will be F = [tex]\tau \times Area[/tex] Â Â Â Â Â
or, Â Â Â Â Â Â Â Â Â F = [tex]\mu_{1} \times \frac{\nu_{i}}{h_{1}} \times Area[/tex]
             = [tex]0.1 \times \frac{0.667}{0.001} \times 1[/tex]
             = 66.67 N
Thus, we can conclude that the fluid velocity is 0.667 m/s and force necessary to make the upper plate move at a speed U = 1 m/s is 66.67 N.
