AKM98
contestada

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.

Respuesta :

Viscous force on the plate is given by

[tex]F = \eta A \frac{dv}{dx}[/tex]

[tex]F_v = 2.5* 1.5*1.5 * \frac{0.1}{1.2* 10^{-3}}[/tex]

[tex]F_v = 468.75 N[/tex]

Buoyancy force on the plate is given as

[tex]F_b = \rho V g[/tex]

[tex]F_b = 950* 1.5*1.5*1.2*10^{-3} * 9.8[/tex]

[tex]F_b = 25.14 N[/tex]

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

[tex]F_v + W = F_b + F[/tex]

[tex]468.75 + 50 = 25.14 + F[/tex]

[tex]F = 493.6 N[/tex]

So it will require a constant force of 493.6 N

Otras preguntas