A rigid, massless L-shaped rod OAB is pivoted smoothly at the bend O. The arm OA has a length of 40 cm and is held horizontally. The arm OB has a length of 20 cm and extends vertically upwards from O. Three forces act on the rod to keep it in rotational equilibrium: a 50 N force acts downwards at end A, a 25 N force acts upwards at the exact midpoint of OA, and an unknown horizontal force F acts to the left at end B.

Step-by-step solution

For the rod to be in rotational equilibrium, the net torque about the pivot O must be zero. The force of 50 N at A acts at a perpendicular distance of 40 cm = 0.4 m from O. It produces a clockwise torque: ₁ = 50 0.4 = 20 N m The force of 25 N at the midpoint of OA acts at a perpendicular distance of 20 cm = 0.2 m from O. It produces a counter-clockwise torque: ₂ = 25 0.2 = 5 N m The force F at B acts horizontally to the left at a perpendicular distance equal to the length of OB, which is 20 cm = 0.2 m . It produces