Question

A heavy iron bar, of weight W is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle \(\theta\) with the horizontal. The weight experienced by the person is:

• A. \(\frac{W}{2}\)
• B. W
• C. \(W cos\theta\)
• D. \(W sin\theta\)

Answer 1

The Correct Option is A

The problem at hand involves analyzing the distribution of weight when a heavy iron bar is balanced such that one end is on the ground and the other is resting on a person's shoulder. Here is how we can solve this:

Understanding the Situation:

• The iron bar is static, balanced between the ground and the person's shoulder, making an angle \(\theta\) with the horizontal.
• This is a standard physics problem involving equilibrium and the distribution of forces.

Principle Involved:

• When a uniform rod is balanced at two points, the weight of the rod is distributed equally at those two points when the rod is symmetrically placed.
• Therefore, the total weight \(W\) is shared equally between the contact points, which are the ground and the shoulder of the person.

Solution Steps:

1. Consider the iron bar as a uniform rod. The gravitational force acts at its center of mass, which is at the midpoint of the rod.
2. Since the rod is symmetric and in equilibrium, the supported forces by the ground and the person must be equal to balance the gravitational force acting on it.
3. This results in each support (ground and shoulder) experiencing half of the total weight of the bar.
4. Hence, the weight experienced by the person is \(\frac{W}{2}\).

Conclusion:

• The correct answer is: \(\frac{W}{2}\) as the weight of the rod is equally distributed between the ground and the person.

Thus, the weight experienced by the person is option: \(\frac{W}{2}\).

Answer 2

Step 1: Concept of center of mass The weight of the bar is uniformly distributed. Since the bar is uniform, the center of mass lies at the midpoint of the bar.

The total weight \( W \) of the bar is supported by two points:

• One end is on the ground.
• The other end is on the shoulder of the person.

Step 2: Distribution of weight In a symmetric situation like this, the total weight is evenly distributed between the two points of contact.

\[ \text{Weight experienced by the person} = \frac{W}{2}. \]

Final Answer: \( \frac{W}{2} \).