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.`

Options

A. velocity gradient is `2s^(-1)`
B. velocity gradient is `1s^(-1)`
C. force required to keep the plate moving with constant speed is 0.02N
D. force required to keep the plate moving with constant speed is 0.01N

Explanation

To solve the problem of finding the tangential force needed to keep the wooden plate moving, we can follow these steps:

### Step 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, \( d = 1 \, m \)
- Coefficient of viscosity of water, \( \eta = 10^{-2} \, \text{poise} = 10^{-2} \times 0.1 \, \text{N s/m}^2 = 10^{-3} \, \text{N s/m}^2 \)

### Step 2: Calculate the velocity gradient
The velocity gradient \( \frac{dv}{dx} \) can be calculated as follows:
- Since the plate is moving with a speed of \( 2 \, m/s \) and the water in contact with the bed is stationary, the velocity gradient is:
\[ \frac{dv}{dx} = \frac{v}{d} = \frac{2 \, m/s}{1 \, m} = 2 \, s^{-1} \]

### Step 3: Use the formula for viscous force
The formula for the viscous force \( F \) acting on the plate is given by:
\[ F = \eta \cdot A \cdot \frac{dv}{dx} \]
Substituting the known values:
\[ F = (10^{-3} \, \text{N s/m}^2) \cdot (10 \, m^2) \cdot (2 \, s^{-1}) \]

### Step 4: Calculate the force
Now, calculating the force:
\[ F = 10^{-3} \cdot 10 \cdot 2 = 0.02 \, N \]

### Conclusion
The tangential force needed to keep the plate moving is \( 0.02 \, N \).

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