Question

A thin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity of V = 7.5 m/s. The two stationary surfaces are spaced 4 cm apart, and the medium between them is filled with oil whose viscosity is 0.9 N·s/m2. The part of the plate immersed in oil at any given time is 2 m long and 0.5 m wide. If the plate moves through the mid-plane between the surfaces, determine the force required to maintain this motion. What would your response be if the plate was 1 cm from the bottom surface (h2) and 3 cm from the top surface (h1)?

Answer 1

Answer:

The answer is below

Explanation:

The constant velocity V = 7.5 m/s, viscosity (μ) = 0.9 N·s/m2,

a) The upper and lower shear force are given as:

$F{shear,upper}=\tau_{w,u} A_s\\\\F{shear,upper}=\mu A_s|\frac{du}{dy} |$

Let us assume that both the upper and lower surfaces are stationary, Therefore:

$V_u=velocity\ of\ upper\ surface=0\ m/s,V_l=velocity\ of\ lower\ surface=0\ m/s,$

$A_s=Area\ of\ plate=2\ m*0.5\ m=1\ m^2$

h1 = 3 cm = 0.03 m

h2 = 1 cm = 0.01 m

For the upper surface, du = V - $V_u$ = 7.5 - 0 = 7.5 m/s.  dy = h1 = 0.03

$F_{shear,upper}=\mu A_s*\frac{du}{dy}=0.9*1*\frac{7.5}{0.03}=225\ N$

For the lower surface, du = V - $V_p$ = 7.5 - 0 = 7.5 m/s.  dy = h2 = 0.01

$F_{shear,lower}=\mu A_s*\frac{du}{dy}=0.9*1*\frac{7.5}{0.01}=675\ N$