Consider a 100 m rod AB of weight 10 N pivoted to a wall at point A. Suspend an object of weight w at point Q, 70 m from point A, and a spring balance measuring a force of 26 N from point B. Now, to balance the system, what should be the weight of the suspended object? To solve this problem, first, create a model of the rod system separated from its surroundings. In a free-body diagram, the reaction forces act at point A, whereas a force of 10 N acts downward at point P, which is the center of gravity of the rod. At point B, a force of 26 N acts upward, while the weight w at point Q acts downward. Here, we need to determine the weight w of the suspended object to balance the system. Using the equilibrium conditions and selecting point A as the reference, the sum of torques in the clockwise direction is equal to the sum of torques in the anticlockwise direction. By substituting the values in the equation, the unknown weight is determined.