A particle of mass m rests on a horizontal floor with which it has a coefficient of static friction `mu`. It is desired to make the body move by applying the minimum possible force F. Find the magnitude of F and the direction in which it has to be applied.

Options

A. `tan^(-1)mu,(mumg)/sqrt((1+mu^(2)))`
B. `tan^(-1)mu,(mg)/sqrt((1+mu^(2)))`
C. `tan^(-1)mu,(mumg)/sqrt((1-mu^(2)))`
D. `tan^(-1)mu,(mg)/sqrt(1-mu^(2))`

Explanation

To solve the problem of finding the minimum force \( F \) required to move a particle of mass \( m \) resting on a horizontal floor with a coefficient of static friction \( \mu \), we will follow these steps:

### Step 1: Identify the Forces Acting on the Particle
The forces acting on the particle include:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting upward.
- The applied force \( F \) at an angle \( \theta \) to the horizontal.
- The frictional force \( f \) acting opposite to the direction of motion.

### Step 2: Resolve the Applied Force into Components
The applied force \( F \) can be resolved into two components:
- Horizontal component: \( F \cos \theta \)
- Vertical component: \( F \sin \theta \)

### Step 3: Write the Equation for the Normal Force
Since there is no vertical motion, the net force in the vertical direction is zero. Therefore, we can write:
\[ N + F \sin \theta = mg \]
Rearranging gives us:
\[ N = mg - F \sin \theta \tag{1} \]

### Step 4: Write the Equation for the Frictional Force
The frictional force \( f \) can be expressed in terms of the normal force:
\[ f = \mu N \]
Substituting equation (1) into this gives:
\[ f = \mu (mg - F \sin \theta) \tag{2} \]

### Step 5: Set Up the Equation for Horizontal Forces
For the particle to just start moving, the horizontal component of the applied force must equal the frictional force:
\[ F \cos \theta = f \]
Substituting equation (2) into this gives:
\[ F \cos \theta = \mu (mg - F \sin \theta) \]

### Step 6: Rearrange the Equation
Rearranging the equation gives:
\[ F \cos \theta + \mu F \sin \theta = \mu mg \]
Factoring out \( F \):
\[ F (\cos \theta + \mu \sin \theta) = \mu mg \]
Thus, we can solve for \( F \):
\[ F = \frac{\mu mg}{\cos \theta + \mu \sin \theta} \tag{3} \]

### Step 7: Minimize the Force \( F \)
To find the minimum force, we need to maximize the denominator \( \cos \theta + \mu \sin \theta \). We can use calculus to find the maximum by differentiating:
\[ \frac{d}{d\theta} (\cos \theta + \mu \sin \theta) = -\sin \theta + \mu \cos \theta = 0 \]
This gives:
\[ \mu = \tan \theta \quad \Rightarrow \quad \theta = \tan^{-1}(\mu) \]

### Step 8: Substitute Back to Find \( F \)
Now substituting \( \theta = \tan^{-1}(\mu) \) into equation (3):
- Calculate \( \cos \theta \) and \( \sin \theta \):
\[ \sin \theta = \frac{\mu}{\sqrt{1 + \mu^2}}, \quad \cos \theta = \frac{1}{\sqrt{1 + \mu^2}} \]
Substituting these values into the denominator:
\[ \cos \theta + \mu \sin \theta = \frac{1}{\sqrt{1 + \mu^2}} + \mu \cdot \frac{\mu}{\sqrt{1 + \mu^2}} = \frac{1 + \mu^2}{\sqrt{1 + \mu^2}} = \sqrt{1 + \mu^2} \]
Thus, substituting back into equation (3):
\[ F = \frac{\mu mg}{\sqrt{1 + \mu^2}} \tag{4} \]

### Final Result
The magnitude of the minimum force \( F \) required to move the particle is:
\[ F = \frac{\mu mg}{\sqrt{1 + \mu^2}} \]