Chapter 2, Problem 87P

A thin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity of 5 m/s. The two stationary surfaces are spaced 4 cm apart, and the medium between them is filled with oil whose viscosity is $0.9N\cdot s/m^2$ . 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

FIGURE P247

To determine

The force required maintain motion at $5\text{m}/\text{s}$

The force on the plate at $1\text{cm}$ from the bottom surface and $3\text{cm}$ from the top surface.

Answer to Problem 87P

The force required maintain motion at $5\text{m}/\text{s}$ is $\text{450}\text{N}$.

The force on the plate at $1\text{cm}$ from the bottom surface and $3\text{cm}$ from the top surface is $\text{600}\text{N}$.

Explanation of Solution

Given information:

The moving velocity of the thin plate is $5\text{m}/\text{s}$, the distance between two stationary surface is $4\text{cm}$, viscosity of the medium is $0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}$, the length of the plate is $2\text{m,}$ and width of the plate is $0.5\text{m}$, height of the plate from the bottom surface is $1\text{cm,}$ and form the top surface is $3\text{cm}$.

Write the expression for area of the plate.

  $A=l\times w$....... (I)

Here the length of the plate is $l$, width of the plate $w$, and area of the plate is $A$.

Write the expression for shear force acting on the upper side of the plate.

  $F_{upper}=\mu A\frac{V}{h_1}$....... (II)

Here, force acting on the upper side of the plate is $F_{upper}$, viscosity of the medium is $\mu$, and velocity of the thin plate is $V$.

Write the expression for shear force acting on the lower side of the plate.

  $F_{lower}=\mu A\frac{V}{h_2}$....... (III)

Here, shear force acting on the lower side of the plate is $F_{lower}$.

Write the expression for force required to maintain the motion

  $F=F_{upper}+F_{lower}$....... (IV)

Here force required to maintain the motion is $F$

Write the expression for shear force acting on the upper side of the plate.

  $F_{{\left(upper\right)}^{\prime }}=\mu A\frac{V}{h_1{}^{\prime }}$....... (V)

Here, force acting on the upper side of the plate is $F_{{\left(upper\right)}^{\prime }}$.

Write the expression for shear force acting on the lower side of the plate.

  $F_{{\left(lower\right)}^{\prime }}=\mu A\frac{V}{h_{2^{\prime }}}$....... (VI)

Write the expression for force required to maintain the motion

  $F^{\prime }=F_{{\left(upper\right)}^{\prime }}+F_{{\left(lower\right)}^{\prime }}$

Calculation:

Write the height of the plate from upper surface when the plate is between the mid of the surfaces.

  $\begin{array}{c}{h}_1=\frac{4}{2}\text{cm}\\ =\left(\frac{4}{2}\text{cm}\right)\left(\frac{0.01\text{m}}{\text{cm}}\right)\\ =0.02\text{m}\end{array}$

Here, height of the plate from upper surface is $h_1$,

Write the height of the plate from bottom surface when the plate is between the mid of the surfaces.

  $\begin{array}{c}{h}_2=\frac{4}{2}\text{cm}\\ =\frac{4}{2}\text{cm}\left(\frac{0.01\text{m}}{\text{cm}}\right)\\ =0.02\text{m}\end{array}$

Here, height of the plate from bottom surface is $h_2$.

Write the height of the plate from upper surface.

  $\begin{array}{c}{h^{\prime }}_1=3\text{cm}\\ =3\text{cm}\left(\frac{0.01\text{m}}{\text{cm}}\right)\\ =0.03\text{m}\end{array}$

Here, the height of the plate from upper surface is ${h^{\prime }}_1$

Write the height of the plate from bottom surface.

  $\begin{array}{c}{h^{\prime }}_2=1\text{cm}\\ =1\text{cm}\left(\frac{0.01\text{m}}{\text{cm}}\right)\\ =0.01\text{m}\end{array}$

Here, the height of the plate from bottom surface is ${h^{\prime }}_2$.

Substitute $2\text{m}$ for $A$ and $0.5\text{m}$ for $w$ in Equation (I).

  $\begin{array}{c}A=\left(2\text{m}\right)\times \left(0.5\text{m}\right)\\ =1{\text{m}}^2\end{array}$

Substitute $0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}$ for $\mu$, $1{\text{m}}^2$ for $A$, $5\text{m}/\text{s}$ for $V$,and $0.02\text{m}$ for $h_1$ in Equation (II).

  $\begin{array}{c}{F}_{upper}=\left(0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}\right)\left(1{\text{m}}^2\right)\left(\frac{5\text{m}/\text{s}}{0.02\text{m}}\right)\\ =\frac{4.5}{0.02}\text{N}\\ =225\text{N}\end{array}$

Substitute $0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}$ for $\mu$, $1{\text{m}}^2$ for $A$, $5\text{m}/\text{s}$ for $V$,and $0.02\text{m}$ for $h_2$ in Equation (III).

  $\begin{array}{c}{F}_{lower}=\left(0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}\right)\left(1{\text{m}}^2\right)\left(\frac{5\text{m}/\text{s}}{0.02\text{m}}\right)\\ =\frac{4.5}{0.02}\text{N}\\ =\text{225}\text{N}\end{array}$

Substitute $\text{225}\text{N}$ for $F_{upper}$ and $\text{225}\text{N}$ for $F_{lower}$ in Equation (IV).

  $\begin{array}{c}F=\text{225}\text{N}+\text{225}\text{N}\\ =\text{450}\text{N}\end{array}$

Substitute $0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}$ for $\mu$, $1{\text{m}}^2$ for $A$, $5\text{m}/\text{s}$ for $V$,and $0.03\text{m}$ for ${h^{\prime }}_1$ in Equation (V).

  $\begin{array}{c}{F}_{upper}=\left(0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}\right)\left(1{\text{m}}^2\right)\left(\frac{5\text{m}/\text{s}}{0.03\text{m}}\right)\\ =\frac{4.5}{0.03}\text{N}\\ \text{=150}\text{N}\end{array}$

Substitute $0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}$ for $\mu$, $1{\text{m}}^2$ for $A$, $5\text{m}/\text{s}$ for $V$,and $0.01\text{m}$ for ${h^{\prime }}_2$ in Equation (VI).

  $\begin{array}{c}{F}_{lower}=\left(0.9\text{N}\cdot \text{s}/{\text{m}}^{\text{2}}\right)\left(1{\text{m}}^2\right)\left(\frac{5\text{m}/\text{s}}{0.01\text{m}}\right)\\ =\frac{4.5}{0.01}\text{N}\\ =\text{450}\text{N}\end{array}$

Substitute $\text{150}\text{N}$ for $F_{{\left(upper\right)}^{\prime }}$ and $\text{450}\text{N}$ for $F_{{\left(lower\right)}^{\prime }}$ in Equation (IV).

  $\begin{array}{c}{F}^{\prime }=150\text{N}+450\text{N}\\ =\text{600}\text{N}\end{array}$

Conclusion:

The force required maintain motion at $5\text{m}/\text{s}$ is $\text{450}\text{N}$.

The force on the plate at $1\text{cm}$ form the bottom surface and $3\text{cm}$ from the top surface is $\text{600}\text{N}$.