Answer to Question #219160 in Civil and Environmental Engineering for Samistha Pattnaik

Question

An oil film of thickness 1.5 mm is used for lubrication in between a square plate of side 0.9 m into 0.9 m and and inclined plane having an angle of inclination 20

Answer 1

We need to find the dynamic viscosity of the oil.

Now shear stress $τ = μ \frac{du }{ dy}$

Force / unit area = $μ \frac{du }{ dy}$

Now $F = 392.4 , A = 0.9 * 0.9 = 0.81 sq m, dy = 10 mm = 10 * 10^{-3} m$

$So \\ \frac{F}{A }= \frac{392.4 }{ 0.81} = 484.4\\ So \space we \space get\\ \frac{F}{A } = μ \frac{du}{ dy}\\ 484.4 = μ (\frac{0.2 }{ 0.01})\\ 484.4 = μ *20\\ μ = \frac{484.4 }{ 20}\\ μ = 24.22 Ns / m^2\\ μ = 242.2 poise$