A horizontal heavy uniform bar of weight W is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to

Options

A. W
B. `(W)/(2)`
C. `(3W)/(4)`
D. `(W)/(4)`

Explanation

To solve the problem, we need to analyze the situation step by step. Let's break it down:

### Step 1: Understand the Initial Setup
We have a horizontal uniform bar of weight \( W \) supported at both ends by two men (let's call them Man A and Man B). The bar is in equilibrium, meaning the forces acting on it are balanced.

### Step 2: Forces Acting on the Bar
Initially, both men exert an upward force on the bar to support its weight. Let the force exerted by Man A be \( F_A \) and by Man B be \( F_B \). Since the bar is uniform, the weight \( W \) acts at the center of the bar.

### Step 3: Apply the Equilibrium Condition
For the bar to be in equilibrium, the sum of the upward forces must equal the weight of the bar:
\[ F_A + F_B = W \]

### Step 4: Analyze the Situation When One Man Lets Go
When one of the men (let's say Man A) lets go of his end of the bar, he no longer exerts any force on it. Therefore, the force \( F_A \) becomes zero.

### Step 5: Recalculate the Forces
Now, only Man B is supporting the weight of the bar. The equation for forces now becomes:
\[ F_B = W \]
This means that Man B now has to support the entire weight of the bar alone.

### Step 6: Conclusion
Thus, the force felt by Man B on his hand changes to equal the weight of the bar \( W \).

### Final Answer
When one man lets go of his end of the rod, the other feels the force on his hand change to \( W \).

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