Pulling force making an angle `theta` to the horizontal is applied on a block of weight `W` placed on a horizontal table. If the angle of friction is `alpha`, then the magnitude of force is `alpha` , then the magnitude of force required to move the body is equal to

Options

A. `(W sin alpha)/( g tan (theta - alpha))`
B. `(W cos alpha)/( cos (theta - alpha))`
C. `(W sin alpha)/( cos (theta - alpha))`
D. `(W tan alpha)/( sin (theta - alpha))`

Explanation

To solve the problem, we need to analyze the forces acting on the block and derive the expression for the minimum pulling force required to move the block. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Block
- The block has a weight \( W \) acting downwards.
- A pulling force \( F \) is applied at an angle \( \theta \) to the horizontal.
- The force \( F \) can be resolved into two components:
- Horizontal component: \( F \cos \theta \)
- Vertical component: \( F \sin \theta \)

### Step 2: Set Up the Normal Force
- The normal force \( N \) acts vertically upward. According to the equilibrium of vertical forces:
\[ N + F \sin \theta = W \]
- Rearranging gives:
\[ N = W - F \sin \theta \]

### Step 3: Determine the Limiting Friction
- The limiting friction \( f_{\text{lim}} \) can be expressed using the coefficient of friction \( \mu \):
\[ f_{\text{lim}} = \mu N = \mu (W - F \sin \theta) \]
- Given that the angle of friction \( \alpha \) relates to the coefficient of friction by \( \mu = \tan \alpha \), we can substitute:
\[ f_{\text{lim}} = \tan \alpha (W - F \sin \theta) \]

### Step 4: Apply the Condition for Motion
- For the block to start moving, the horizontal component of the pulling force must be greater than or equal to the limiting friction:
\[ F \cos \theta \geq \tan \alpha (W - F \sin \theta) \]

### Step 5: Rearranging the Inequality
- Rearranging the inequality gives:
\[ F \cos \theta \geq \frac{\tan \alpha (W - F \sin \theta)}{1} \]
- Expanding this leads to:
\[ F \cos \theta \geq \tan \alpha W - \tan \alpha F \sin \theta \]
- Rearranging terms to isolate \( F \):
\[ F \cos \theta + \tan \alpha F \sin \theta \geq \tan \alpha W \]
- Factoring out \( F \):
\[ F \left( \cos \theta + \tan \alpha \sin \theta \right) \geq \tan \alpha W \]

### Step 6: Solve for the Minimum Force \( F \)
- Finally, solving for \( F \):
\[ F \geq \frac{\tan \alpha W}{\cos \theta + \tan \alpha \sin \theta} \]

### Step 7: Simplifying the Expression
- Recognizing that \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \), we can rewrite:
\[ F \geq \frac{W \sin \alpha}{\cos \theta \cos \alpha + \sin \theta \sin \alpha} \]
- The denominator can be recognized as \( \cos(\theta - \alpha) \):
\[ F \geq \frac{W \sin \alpha}{\cos(\theta - \alpha)} \]

### Final Result
Thus, the magnitude of the force required to move the body is:
\[ F = \frac{W \sin \alpha}{\cos(\theta - \alpha)} \]