A plank is resting on a horizontal ground in the northern hemisphere of the Earth at a `45^(@)` latitude. Let the angular speed of the Earth be `omega` and its radius `r_e`. The magnitude of the frictional force on the plank will be

Options

A. `mr_e omega^(2)`
B. `(mr_e omega^(2))/( sqrt(2))`
C. `(m r_e omega^(2))/( 2)`
D. Zero

Explanation

To find the magnitude of the frictional force on the plank resting on a horizontal ground at a latitude of \(45^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Plank**:
- The weight of the plank \(mg\) acts downward.
- The normal force \(N\) acts upward.
- The centrifugal force due to Earth's rotation acts outward, which can be expressed as \(F_c = m\omega^2 r\), where \(r\) is the radius of the circular path of the plank.

2. **Determine the Radius \(r\)**:
- At a latitude of \(45^\circ\), the radius \(r\) can be related to the Earth's radius \(R_e\) using trigonometric relationships. The radius \(r\) is given by:
\[ r = R_e \cos(45^\circ) = \frac{R_e}{\sqrt{2}} \]

3. **Calculate the Centrifugal Force**:
- Substitute \(r\) into the expression for centrifugal force:
\[ F_c = m\omega^2 r = m\omega^2 \left(\frac{R_e}{\sqrt{2}}\right) = \frac{m\omega^2 R_e}{\sqrt{2}} \]

4. **Resolve Forces**:
- The centrifugal force can be resolved into two components: one acting perpendicular to the normal force and one acting parallel (tangential) to the surface.
- The component acting parallel to the surface (which will be balanced by friction) is:
\[ F_{\text{tangential}} = F_c \sin(45^\circ) = \frac{m\omega^2 R_e}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{m\omega^2 R_e}{2} \]

5. **Frictional Force**:
- The frictional force \(f\) must balance the tangential component of the centrifugal force:
\[ f = \frac{m\omega^2 R_e}{2} \]

### Final Answer:
The magnitude of the frictional force on the plank is:
\[ f = \frac{m R_e \omega^2}{2} \]