A uniform beam of mass \(M\) and length \(L\) is held in static equilibrium, and so the magnitude of the net torque about its center of mass is zero. The magnitude of the net torque on this beam at one of its ends, a distance of \(L / 2\) from the center of mass, is a) \(M g L\). c) zero. b) \(M g L / 2\). d) \(2 M g L\).

Imagine trying to spin a seesaw by pushing on one of its ends; the effectiveness of your push relies not just on your strength but also on how far you are from the seesaw’s pivot point. This concept in physics is known as torque, a measure of the rotational force that can cause an object to rotate around an axis. Torque is mathematically expressed by the equation:
\[ \text{Torque} = \text{Force} \times \text{Distance} \times \sin(\theta) \] where \( \theta \) represents the angle between the force being applied and the axis of rotation.In the context of the given exercise, torque results from the gravitational force acting on the beam. Since the force (gravitational force) and the distance (half the beam’s length, \(L/2\)) form a right angle, the \( \sin(\theta) \) in the torque formula is \( \sin(90^\circ) \), which equals 1, simplifying our calculation. Hence, at one end of the beam, the gravitational torque is \(Mg \times (L / 2)\).It's essential to understand that in static equilibrium—when an object is at rest and not rotating—the net torque must be zero. If a beam is not rotating, it means that all the applied torques are balancing each other out. For example, the torque created by the gravitational force is counteracted by an equal and opposite torque where the beam is supported, leading to a state of balance that prohibits any rotational motion.

The gravitational force is a fundamental interaction that attracts two bodies with mass. Every object on Earth experiences this force due to the planet’s mass, pulling them towards its center. For a beam of mass \(M\), the gravitational force exerted on it can be calculated using the equation:
\[ \text{Gravitational force} = M \times g \] where \(g\) is the acceleration due to Earth’s gravity, approximately \(9.81 \text{m/s}^2\). This force acts at the center of mass of the beam, and it is this force that creates the potential for torque when applied at a distance from an axis of rotation, as mentioned earlier.In static equilibrium, there's a crucial balance between opposing forces and torques. For the beam in our problem, the gravitational force is perfectly counterbalanced by a support force, which prevents the beam from falling. As a result, the net force and the net torque on the beam both are zero. The beam remains stationary not because there aren't any forces acting on it, but because those forces are canceling each other out. When discussing equilibrium in physics, it's important to recognize this distinction between having no forces and having a balance of forces.

While static equilibrium overall requires both force and torque to be in balance, rotational equilibrium focuses specifically on the latter. An object is in rotational equilibrium when there is no net torque acting on it, which means it won’t start rotating spontaneously. This is a state where all the torques that could cause rotation are perfectly counteracting each other.Drawing on the beam scenario, since it is not rotating, we know it is in rotational equilibrium. Despite the gravitational force exerting a torque on the beam, somewhere along the beam, a support exerts an equal and opposite torque. Considering the beam as a whole, the torques at each end must cancel out to maintain equilibrium, and therefore, the net torque is zero. If these conditions weren’t met and one torque outweighed the other, the beam would start to rotate, breaking the state of rotational equilibrium. Remembering this interplay between torques is essential for visualizing and solving static equilibrium problems in physics exercises.