Question

A thin flat plate of surface area, 500 cm², is located between two fixed plates such that the gap below the top plate is 20 mm and the gap above the bottom plate is 30 mm. The gap between the fixed plates contain SAE 30 oil at 15.6°C. Determine the force needed to push the middle plate at a velocity of 1.0 m/s.

Answer 1

Answer:0.3166N

Explanation:

Given data

Area $\left ( A\right )=500 cm^2$

Gap below top plate$\left ( y_1\right )=20 mm$

Gap above bottom plate$\left ( y_2\right )=30 mm$

SAE 30 oil viscosity =$0.38 N-s/m^2$

Velocity of middle plate$\left ( v\right )=1 m/s$

There will viscous force on middle plate i.e. at above surface and below surface

Viscous force$\left ( F\right )=\mu \frac{Av}{y}$

$Net Force on plate F =\mu Av\left [\frac{1}{y_1} +\frac{1}{y_2}\right ]$

$F=0.38\times 500\times 10^{-4}\left [\frac{1}{20\times 10^{-3}} +\frac{1}{30\times 10^{-3}}\right ]$

$F=31.66\times 10^{-2}=0.3166 N$

this is force by oil on plate thus we need to apply atleast 0.3166N to move plate