Two men support a uniform horizontal beam at its two ends, if one of them suddenly lets go, the force exerted by the beam on the other man will

Options

A. remain unaffected
B. increase
C. decrease
D. become unequal to the force exerted by him on the beam

Explanation

To solve the problem, we need to analyze the situation step by step. Here’s how we can approach it:

### Step 1: Understand the Initial Setup
We have a uniform horizontal beam supported at both ends by two men. The weight of the beam (mg) acts at its center. Initially, both men exert equal forces (normal reactions) on the beam to keep it in equilibrium.

### Step 2: Calculate the Initial Forces
In equilibrium, the sum of the upward forces (normal reactions from both men) must equal the downward force (weight of the beam):
\[ N + N = mg \]
Since both men exert equal forces, we can write:
\[ 2N = mg \]
Thus, the force exerted by each man is:
\[ N = \frac{mg}{2} \]

### Step 3: Analyze the Situation When One Man Lets Go
When one man suddenly lets go, the beam will start to rotate about the point where the other man is still holding it. The weight of the beam still acts at the center.

### Step 4: Set Up the Equation for Forces After One Man Lets Go
Let’s denote the normal reaction force exerted by the beam on the remaining man as \( N' \). The net force acting on the beam in the downward direction is still \( mg \), and the upward force is now just \( N' \):
\[ mg - N' = ma \]
Where \( a \) is the downward acceleration of the center of mass of the beam.

### Step 5: Calculate the Torque and Angular Acceleration
To find the relationship between the linear acceleration \( a \) and the angular acceleration \( \alpha \), we can use the torque about the point where the remaining man is holding the beam:
- The torque due to the weight of the beam is:
\[ \tau = mg \cdot \frac{L}{2} \]
- The moment of inertia \( I \) of the beam about the pivot point is:
\[ I = \frac{1}{3}mL^2 \]
Using Newton's second law for rotation:
\[ \tau = I \alpha \]
Thus:
\[ mg \cdot \frac{L}{2} = \frac{1}{3}mL^2 \alpha \]
From this, we can solve for \( \alpha \):
\[ \alpha = \frac{3g}{2L} \]

### Step 6: Relate Linear Acceleration to Angular Acceleration
The linear acceleration \( a \) of the center of mass is related to the angular acceleration by:
\[ a = R \alpha \]
Where \( R = \frac{L}{2} \) (the distance from the pivot to the center of mass). Thus:
\[ a = \frac{L}{2} \cdot \frac{3g}{2L} = \frac{3g}{4} \]

### Step 7: Substitute Back to Find \( N' \)
Now we can substitute \( a \) back into the force equation:
\[ mg - N' = m \cdot \frac{3g}{4} \]
Rearranging gives:
\[ N' = mg - \frac{3mg}{4} = mg \left(1 - \frac{3}{4}\right) = \frac{mg}{4} \]

### Step 8: Compare the Forces
Initially, the force exerted by each man was \( \frac{mg}{2} \). After one man lets go, the force exerted by the remaining man is \( \frac{mg}{4} \).

### Conclusion
Thus, the force exerted by the beam on the other man decreases from \( \frac{mg}{2} \) to \( \frac{mg}{4} \).

### Final Answer
The force exerted by the beam on the other man will **decrease**.

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