A uniform square plate of mass 2 2 kg and side length L is pivoted smoothly at its bottom-left corner O. It is held stationary in a vertical plane such that its lower edge is perfectly horizontal. To maintain this equilibrium, a force F is applied at the top-right corner B. The force F is directed exactly perpendicular to the diagonal OB. Taking g = 10 m/s ^2 , the magnitude of the force F is :

Step-by-step solution

For the square plate to remain stationary, the net torque about the pivot O must be zero. The weight of the uniform square plate acts downwards through its center of mass. The horizontal distance from the pivot O to the line of action of the weight is half the side length, i.e., L 2 . The clockwise torque due to gravity is: _ g = mg L 2 = (2 2 )(10) ( L 2 ) = 10 2 L The applied force F acts at corner B, and its line of action is perpendicular to the diagonal OB. The perpendicular distance from the pivot O to the li