A metal plate `0.04m^(2)` in area is lying on a liquid layer of thickness `10^(-3)m` and coefficient of viscosity 140 poise. Calculate the horizontal force needed to move the plate with a speed of 0.040 m/s.

Answer

To solve the problem of calculating the horizontal force needed to move a metal plate on a liquid layer, we will use the formula derived from Newton's law of viscosity. Here is the step-by-step solution:

### Step 1: Identify the given values
- Area of the plate, \( A = 0.04 \, m^2 \)
- Thickness of the liquid layer, \( \Delta x = 10^{-3} \, m \)
- Coefficient of viscosity, \( \eta = 140 \, \text{poise} \)
- Speed of the plate, \( v = 0.040 \, m/s \)

### Step 2: Convert the coefficient of viscosity to SI units
The coefficient of viscosity in SI units is given by:
\[ 1 \, \text{poise} = 0.1 \, \text{Pa.s} \]
Thus,
\[ \eta = 140 \, \text{poise} = 140 \times 0.1 \, \text{Pa.s} = 14 \, \text{Pa.s} \]

### Step 3: Calculate the velocity gradient
The velocity gradient \( \frac{dv}{dx} \) is defined as the change in velocity per unit distance. Since the plate moves with a constant speed, we can express this as:
\[ \frac{dv}{dx} = \frac{v}{\Delta x} \]
Substituting the values:
\[ \frac{dv}{dx} = \frac{0.040 \, m/s}{10^{-3} \, m} = 40 \, s^{-1} \]

### Step 4: Apply the formula for force
According to Newton's law of viscosity, the force \( F \) required to move the plate is given by:
\[ F = \eta \cdot A \cdot \frac{dv}{dx} \]
Substituting the known values:
\[ F = 14 \, \text{Pa.s} \cdot 0.04 \, m^2 \cdot 40 \, s^{-1} \]

### Step 5: Calculate the force
Now we can compute the force:
\[ F = 14 \cdot 0.04 \cdot 40 \]
\[ F = 14 \cdot 1.6 \]
\[ F = 22.4 \, N \]

### Final Answer
The horizontal force needed to move the plate is \( F = 22.4 \, N \).

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