Question

A square metal plate 1.5m side and 1.2mm thick weighting 50N is to be lifted through a vertical gap at 25mm of infinite extent . The oil in air gap has a specific gravity of 95 percent and viscosity of 2.5N.s/m^2. if the metal plate is to be lifted at a constant speed of 0.1m/s find the force and power required.

Answer 1

Viscous force on the plate is given by

$F = \eta A \frac{dv}{dx}$

$F_v = 2.5* 1.5*1.5 * \frac{0.1}{1.2* 10^{-3}}$

$F_v = 468.75 N$

Buoyancy force on the plate is given as

$F_b = \rho V g$

$F_b = 950* 1.5*1.5*1.2*10^{-3} * 9.8$

$F_b = 25.14 N$

Weight of the plate = 50 N

Now for constant speed of plate the net force on it must be zero

SO here we can say

$F_v + W = F_b + F$

$468.75 + 50 = 25.14 + F$

$F = 493.6 N$

So it will require a constant force of 493.6 N