Understanding the concept of torque is essential in physics, particularly when analyzing how an object's weight can produce a torque about its axis of rotation. Torque, denoted as \( \tau \), is calculated using the formula:

\( \tau = F \cdot r \cdot \sin(\theta) \)

where \( F \) is the force applied, \( r \) is the distance from the axis of rotation to the point where the force is applied, and \( \theta \) is the angle between the force vector and the lever arm. In the context of an object like a bar or rod, the weight force \( mg \) acts at the center of gravity, which for most objects with uniform mass distribution coincides with the center of mass.

For a bar of length \( L \), the center of gravity is located at \( \frac{L}{2} \). For example, if a cylindrical rod has a mass \( m = 20 \, \text{kg} \) and a length \( L = 4 \, \text{m} \), the weight force acts at \( 2 \, \text{m} \) from the fixed end. If an additional mass \( M = 80 \, \text{kg} \) is placed at the other end of the rod, the net torque about the axis due to both weight forces must be calculated.

To find the net torque, we consider the torque produced by both masses. The torque due to the mass \( m \) is:

\( \tau_{m} = m g r_{m} \sin(\theta_{m}) \)

where \( r_{m} = 2 \, \text{m} \) (the distance to the center of mass) and \( \theta_{m} = 53^\circ \) (the angle between the force and the lever arm). For the mass \( M \) at the end of the rod, the torque is:

\( \tau_{M} = M g r_{M} \sin(\theta_{M}) \)

where \( r_{M} = 4 \, \text{m} \) and \( \theta_{M} = 53^\circ \). The gravitational acceleration \( g \) is typically approximated as \( 10 \, \text{m/s}^2 \).

Calculating these torques, we find:

\( \tau_{m} = 20 \cdot 10 \cdot 2 \cdot \sin(53^\circ) \approx 313 \, \text{N m} \)

\( \tau_{M} = 80 \cdot 10 \cdot 4 \cdot \sin(53^\circ) \approx 2504 \, \text{N m} \)

Both torques are positive as they cause counterclockwise rotation. Therefore, the net torque \( \tau_{net} \) is the sum of the individual torques:

\( \tau_{net} = \tau_{m} + \tau_{M} = 313 + 2504 = 2817 \, \text{N m} \)

This example illustrates the importance of understanding how weight forces contribute to torque in rotational dynamics. Recognizing the center of gravity and the angles involved is crucial for accurate calculations in various physics problems.

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