A square block weighing 1.1 KN and 230 mm on an edge slides down an incline on a film of oil 6 micrometer thick. ?

Question

 Assuming a linear velocity profile in the oil and neglecting air resistance, what is the terminal velocity of the block? The viscosity of oil is 7 mPa-s. Angle of inclination is 20 degrees.

Answer 1

F = (mu)Au/y 

= (0.007 Pa-s)(0.23 m)^2*(u)/(6 * 10^(-6) m)

= 61.717 (N-s/m) * u.

But F = mg sin(20) = (1100 N)*(0.34202) = 376.2 N.

And hence u = 376.2 N/[61.717 Ns/m] = 6.096 m/s, round to 6.1 m/s.