A wooden box lying at rest on an inclined surface of a wet wood is held at static equilibrium by a constant force F applied perpendicular to the incline. If the mass of the box is 1kg, the angle of inclination is `30^(@)` and the coefficient of static friction between the box and the inclined plane is 0.2, the minimum magnitude of F is (Use g= `10m//s^2` )

Options

A. A. 0 N , as `30^(@)` is less than angle of repose
B. B.`ge 1 N`
C. C.`ge 3.3 N`
D. D.`ge 16.3 N`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the forces acting on the box
1. The weight of the box (W = mg) acts downward.
2. The normal force (N) acts perpendicular to the inclined surface.
3. The applied force (F) acts perpendicular to the incline.
4. The frictional force (f) acts parallel to the incline, opposing the motion.

### Step 2: Break down the weight into components
- The weight of the box can be resolved into two components:
- Perpendicular to the incline: \( W_{\perp} = mg \cos \theta \)
- Parallel to the incline: \( W_{\parallel} = mg \sin \theta \)

### Step 3: Write the equations for static equilibrium
- In the direction perpendicular to the incline:
\[ N = F + mg \cos \theta \]
- In the direction parallel to the incline, the frictional force must balance the component of weight:
\[ f = mg \sin \theta \]
- The frictional force can also be expressed in terms of the normal force:
\[ f = \mu N \]

### Step 4: Substitute the expressions
- From the frictional force equation:
\[ mg \sin \theta = \mu N \]
- Substitute \( N \) from the first equation into this:
\[ mg \sin \theta = \mu (F + mg \cos \theta) \]

### Step 5: Rearranging the equation
- Rearranging gives:
\[ mg \sin \theta = \mu F + \mu mg \cos \theta \]
- Rearranging for \( F \):
\[ F = \frac{mg \sin \theta - \mu mg \cos \theta}{\mu} \]

### Step 6: Substitute the values
- Given:
- \( m = 1 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( \theta = 30^\circ \)
- \( \mu = 0.2 \)

- Calculate \( mg \sin 30^\circ \) and \( mg \cos 30^\circ \):
\[ mg \sin 30^\circ = 1 \times 10 \times \frac{1}{2} = 5 \, \text{N} \]
\[ mg \cos 30^\circ = 1 \times 10 \times \frac{\sqrt{3}}{2} \approx 8.66 \, \text{N} \]

### Step 7: Plug values into the equation for F
\[ F = \frac{5 - 0.2 \times 8.66}{0.2} \]
\[ F = \frac{5 - 1.732}{0.2} = \frac{3.268}{0.2} = 16.34 \, \text{N} \]

### Final Answer
The minimum magnitude of the force \( F \) is approximately **16.34 N**.

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