A large wooden plate of area `10m^2` floating on the surface of river is made to move horizontally wilth a speed of `2ms^-1` by applying a tangential force. If the river is 1m deep and the water contact with the bed is stationary, find the tangential force needed to keep the plate moving. Coefficient of viscosity of water at the temperature of the river `=10^-2 poise.`

Explanation

To find the tangential force needed to keep the wooden plate moving horizontally at a speed of 2 m/s, we can use the concept of shear stress in fluid mechanics. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Area of the wooden plate, \( A = 10 \, m^2 \)
- Speed of the plate, \( v = 2 \, m/s \)
- Depth of the river, \( h = 1 \, m \)
- Coefficient of viscosity of water, \( \eta = 10^{-2} \, \text{poise} \)

2. **Convert Viscosity to SI Units:**
- 1 poise = 0.1 Pa·s, so:
\[ \eta = 10^{-2} \, \text{poise} = 10^{-2} \times 0.1 \, \text{Pa·s} = 10^{-3} \, \text{Pa·s} \]

3. **Calculate the Shear Stress (\( \tau \)):**
- Shear stress is given by the formula:
\[ \tau = \eta \frac{dv}{dy} \]
- Here, \( dv \) is the change in velocity (which is \( v = 2 \, m/s \)), and \( dy \) is the distance over which this change occurs (which is the depth of the river, \( h = 1 \, m \)).
\[ \tau = 10^{-3} \, \text{Pa·s} \times \frac{2 \, m/s}{1 \, m} = 2 \times 10^{-3} \, \text{Pa} \]

4. **Calculate the Total Shear Force (\( F \)):**
- The total shear force can be calculated using the formula:
\[ F = \tau \times A \]
- Substituting the values we have:
\[ F = (2 \times 10^{-3} \, \text{Pa}) \times (10 \, m^2) = 2 \times 10^{-2} \, \text{N} \]

5. **Final Result:**
- The tangential force needed to keep the plate moving is:
\[ F = 0.02 \, \text{N} \]

### Summary:
The tangential force required to keep the wooden plate moving at a speed of 2 m/s is \( 0.02 \, \text{N} \).