Question

A block of weight W slides down an inclined plane that is lubricated by a thin film of oil as shown in the figure below. Block Oil Plane surface The film contact area is A, and its thickness is h. Part A Assuming a linear velocity distribution in the film, derive an expression for the terminal (zero-acceleration) speed, V, of the block. O V = Wh cos e JA OV=hA cose дW V= Wh sin e MA OV - PA sine

Answer 1

Sol:- Block Wsino HO WW - Plane surface oil Assumptions: laminas flow and Linear velocity profile, No slip condition. From No-slip condition, we have u=0 at y=o And, uav at y=h : {: velocity profile is linear u - Eu velocity u- Calvo Differentiate u won't y i we get line co How, shear stress is given by; -3 { uz Dynamic Viscosity of fluid}

from ea no ③ l ; we get t=ur tangential component since this shear stress is produced by of weight of block. Hence; T= W sino - {: stress-force ? A from ea no. 4 4 0; we have W sind u v a seriode vandag : vo whicho 1 Terminal velocity ; v i = Wh sino MA (Third option is correct)