Answer:
[tex]\mu =0.169704 Ns/m^2[/tex]
Explanation:
From the question we are told that
Area of plate[tex]a=1m^2[/tex]
Angle of inclination [tex]\theta=60 \textdegree[/tex]
Velocity [tex]v=0.75m/s[/tex]
Thickness of oil [tex]t=2mm[/tex]
Weight of plate [tex]w=90N[/tex]
Generally the equation for shear force is mathematically given by
[tex]F=mgsin45 \textdegree[/tex]
[tex]F=90sin45 \textdegree[/tex]
[tex]F=63.639N[/tex]
[tex]F=\mu* \frac{v}{t}[/tex]
[tex]F=\mu* \frac{0.75}{2*10^-^3}[/tex]
[tex]\mu* \frac{0.75}{2*10^-^3}=63.639N[/tex]
[tex]\mu =\frac{63.639}{375}[/tex]
Therefore viscosity is given by
[tex]\mu =0.169704 Ns/m^2[/tex]
