A metal block of area $0.10 ext{ m}^2$ is connected to a $0.010 ext{ kg}$ mass via a string over a

Question

A metal block of area $0.10	ext{ m}^2$ is connected to a $0.010	ext{ kg}$ mass via a string over a massless pulley. A liquid film of thickness $0.30	ext{ mm}$ is placed between the block and the table. The block moves with a constant speed of $0.085	ext{ m/s}$. What is the coefficient of viscosity of the liquid? ($g = 9.8	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

Shear force $F = T = mg = 0.010 	imes 9.8 = 0.098	ext{ N}$. Shear stress $= \frac{F}{A} = \frac{0.098}{0.10} = 0.98	ext{ N/m}^2$. Strain rate $= \frac{v}{l} = \frac{0.085}{0.30 	imes 10^{-3}} = 283.3	ext{ s}^{-1}$. Viscosity $\eta = \frac{	ext{Stress}}{	ext{Strain rate}} = \frac{0.98}{283.3} \approx 3.45 	imes 10^{-3}	ext{ Pa s}$.