Answer:
F=989.6 N
Explanation:
Given that
Diameter of shaft = 70 mm
Diameter of bearing sleeve =70.2 mm
So clearance h=0.1 mm
Speed V= 400 mm/s
Length of shaft = 250 mm
[tex]\nu =0.005\ \frac{m^2}{s}\ , \rho=900\ \frac{kg}{m^3}[/tex]
We know that
[tex]\mu =\rho\times\nu[/tex]
μ= 900 x 0.005 Pa-s
μ= 4.5 Pa-s
As we know that
From Newton's law of viscosity ,the shear stress given as follows
[tex]\tau =\mu \dfrac{dU}{dy}[/tex]
We also know that
Force = shear stress x area
Now by putting the values
[tex]\tau =4.5\times \dfrac{400}{0.1}[/tex]
[tex]\tau=18,000 Pa[/tex]
So force
F= 18,000 x π x 0.07 x 0.25
F=988.6 N