To solve the problem, we will use the formula for viscous force derived from Newton's law of viscosity. The formula is:
\[ F = \eta \cdot A \cdot \frac{dv}{dx} \]
Where:
- \( F \) is the force required to move the plate,
- \( \eta \) is the coefficient of viscosity,
- \( A \) is the area of the plate,
- \( \frac{dv}{dx} \) is the velocity gradient (change in velocity per unit thickness).
### Step-by-Step Solution:
1. **Identify the given values:**
- Coefficient of viscosity, \( \eta = 1.0 \, \text{kg/(m·s)} \)
- Area of the plate, \( A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2 \)
- Velocity, \( v = 7 \, \text{cm/s} = 7 \times 10^{-2} \, \text{m/s} \)
- Thickness of the glycerine layer, \( dx = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
2. **Calculate the velocity gradient \( \frac{dv}{dx} \):**
\[
\frac{dv}{dx} = \frac{v}{dx} = \frac{7 \times 10^{-2} \, \text{m/s}}{1 \times 10^{-3} \, \text{m}} = 70 \, \text{s}^{-1}
\]
3. **Substitute the values into the formula:**
\[
F = \eta \cdot A \cdot \frac{dv}{dx}
\]
\[
F = 1.0 \, \text{kg/(m·s)} \cdot 0.01 \, \text{m}^2 \cdot 70 \, \text{s}^{-1}
\]
4. **Calculate the force:**
\[
F = 1.0 \cdot 0.01 \cdot 70 = 0.7 \, \text{N}
\]
### Final Answer:
The force required to move the plate with a velocity of 7 cm/s is **0.7 N**.
