A thin plate of very large area is placed in a gap of height h with oils of viscosities `mu^'` and `mu^''` on two sides of plate. The plate is pulled at a constant velocity V. Calculate position of plate so that:

• shear force on two sides of plate is equal
• shear force required to drag plate is minimum

Assume viscous flow and neglect all end effects.

SOLUTION:

Given,

Height of the gap=h

Viscosities of the oils =`mu^'` and `mu^''`

Velocity of the plate= V

Let, y be the distance of the thin plate from one  of the surface of the gap.

Force on the upper side of the plate is,

`F_(Upper)=mu*(du)/(dy)*A=mu^'*V/(h-....Show More

SOLUTION:

Given,

Height of the gap=h

Viscosities of the oils =`mu^'` and `mu^''`

Velocity of the plate= V

Let, y be the distance of the thin plate from one  of the surface of the gap.

Force on the upper side of the plate is,

`F_(Upper)=mu*(du)/(dy)*A=mu^'*V/(h-y)*A`

Similarly, force on the lower side of the plate is,

`F_(Lower)=mu^''*V/y*A`

Case I:

Since the shear force on the two side of the plate are equal, we have,

`mu^'*V/(h-y)*A=mu^''*V/y*A`

or,`(mu^')/(h-y)=(mu^'')/y`

or, `mu^'*y=mu^''*h-(mu^'')*y`

or, ` y=(mu^''*h)/(mu^'+mu^'')`

Case II:

When the shear force required to drag the plate is minimum,

`(dF)/(dy)=0`

We have,

`F=mu^'*A*v/(h-y)+ mu^''*A*V/y`

or `(dF)/(dy)=(mu^'*A*V)/(h-y)^2-mu^''*A*V/y^2=0`

or,`mu^'/(mu^'')=((h-y)/y)^2=(h^2-2hy+y^2)/y^2`

or,`mu^'/(mu^'')=h^2/y^2+1-(2h)/y`

or, `h^2/y^2-2*h/y+(1-mu^'/(mu^''))=0`

Solving this quadratic equation for `h/y`, we get,

`h/y=(2+-root ()(4-4 (1-mu^'/(mu^''))))/2`

`=(2+-2root ()(1-1+mu^'/(mu^'')))/2`

`=1+-root ()(mu^'/(mu^''))`

Since the ratio `h/y` cannot be less than unity, we have,

`h/y=1+root ()(mu^'/(mu^''))`

or, `y=h/(1+root ()(mu^'/mu^''))`