A metal plate 1.25m ** 1.25 m ** 6 mm thick and weighing 90 N is placed midway in 24 mm gap between two vertical plane surface as shown below. The gap is filled with an oil of specific gravity 0.85 and dynamic viscosity 3 Ns/`m^2`. Determine force required to lift plate with a constant velocity of 0.15 m/s.

SOLUTION:

Given,

Area  of the plate, `A=1.25**1.25=1.5625 m^2`

Specific gravity of the oil, `S=0.85`

Thickness of the  plate = 6 mm

Now,

`t_1=t_2=(24-6)/2=9 mm`

Thus,

Shear Force required to drag the plate vertically =shear force due to fluid between the plate and left vertical plane + shear force due to fluid between the plate and right vertical plane

or, `F=F_1+F_2`

or, `F=tau_1*A+tau_2*A`

or, `F=....Show More

SOLUTION:

Given,

Area  of the plate, `A=1.25**1.25=1.5625 m^2`

Specific gravity of the oil, `S=0.85`

Thickness of the  plate = 6 mm

Now,

`t_1=t_2=(24-6)/2=9 mm`

Thus,

Shear Force required to drag the plate vertically =shear force due to fluid between the plate and left vertical plane + shear force due to fluid between the plate and right vertical plane

or, `F=F_1+F_2`

or, `F=tau_1*A+tau_2*A`

or, `F=A*(tau_1+tau_2)`

or, `F=A*mu*V*(1/t_1+1/t_2)`

or, `F=156.25 N`

Again,

Upward thrust or buoyant force on the  plate is,

`=V*rho*g`

`=(1.25**1.25**0.006)**0.85**1000**9.81`

`=78.17 N`

Now, the effective weight of the plate `=90-78.17 =11.83 N`

Thus, the total force required to lift the plate at velocity of 0.15 m/s is,

=F+ effective weight of the plate

`=156.25 + 11.83`

`=168.08 N`