A heavy iron bar of weight `W` is having its one end on the ground and the other on the shoulder of a man The rod makes an angle `theta` with the horizontal What is the weight experienced by the man ?

Options

A. `Wsin theta`
B. `Wcos theta`
C. `W`
D. `W//2`

Explanation

To solve the problem, we need to analyze the forces acting on the heavy iron bar and apply the principles of equilibrium and torque. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a heavy iron bar of weight \( W \) resting on the ground at one end and on the shoulder of a man at the other end. The bar makes an angle \( \theta \) with the horizontal.

### Step 2: Identify Forces Acting on the Bar
1. The weight of the bar \( W \) acts downwards at its center of mass, which is located at the midpoint of the bar.
2. The force exerted by the man on the bar (let's call it \( F_1 \)) acts upwards at the point where the bar rests on his shoulder.
3. The ground exerts a force (let's call it \( F_2 \)) upwards at the point where the bar touches the ground.

### Step 3: Apply the Equilibrium Condition
For the bar to be in equilibrium:
- The sum of the upward forces must equal the sum of the downward forces.

This gives us the equation:
\[ F_1 + F_2 = W \quad \text{(1)} \]

### Step 4: Analyze Torque
To find the force \( F_1 \), we can analyze the torques about the point where the bar touches the ground. The torque due to the weight of the bar will cause a clockwise rotation, while the torque due to the force \( F_1 \) will cause a counterclockwise rotation.

1. The torque due to the weight \( W \) at the center of the bar (which is at a distance \( \frac{L}{2} \) from the ground) is:
\[ \text{Torque due to } W = W \cdot \left(\frac{L}{2} \cos \theta\right) \]

2. The torque due to the force \( F_1 \) (which acts at a distance \( L \) from the ground) is:
\[ \text{Torque due to } F_1 = F_1 \cdot (L \sin \theta) \]

Setting the sum of torques to zero for equilibrium:
\[ W \cdot \left(\frac{L}{2} \cos \theta\right) = F_1 \cdot (L \sin \theta) \quad \text{(2)} \]

### Step 5: Solve for \( F_1 \)
From equation (2), we can simplify:
\[ W \cdot \frac{L}{2} \cos \theta = F_1 \cdot L \sin \theta \]
Dividing both sides by \( L \):
\[ \frac{W}{2} \cos \theta = F_1 \sin \theta \]
Now, solving for \( F_1 \):
\[ F_1 = \frac{W \cos \theta}{2 \sin \theta} \]

### Step 6: Final Expression for Weight Experienced by the Man
The weight experienced by the man is equal to the force \( F_1 \):
\[ F_1 = \frac{W \cos \theta}{2 \sin \theta} \]

### Conclusion
Thus, the weight experienced by the man is given by:
\[ F_1 = \frac{W \cos \theta}{2 \sin \theta} \]