Final answer:
To find the normal reaction from the floor on a stick, we apply static equilibrium principles. By summing the moments around the point where the rope is tied and setting them equal, we solve for the normal force. The magnitude of the normal reaction from the floor on the stick is found to be 3.924 N.
Step-by-step explanation:
To compute the normal reaction from the floor on the stick, we can apply the principles of static equilibrium, where the sum of the forces and the sum of the moments (torques) around any point must equal zero. In this case, because the stick is at rest and one end is on a smooth floor, the forces acting on the stick include its weight, the tension in the rope, and the normal force from the floor.
For simplicity, we can consider the stick to be in equilibrium under the action of its weight (acting through its center of gravity, which is at the midpoint of the stick), the tension in the rope, and the normal reaction at the point of contact with the floor.
Summing moments around the point where the rope is tied, the normal force exerts a moment equal to the normal force times lever arm, which is the distance from the rope to the contact point with the floor (15 cm, since the rope is tied 5 cm from the top of a 20-cm stick).
The weight of the stick exerts a moment about that point as well, which is the weight times its lever arm (10 cm, the distance from the rope to the stick's center of gravity).
By setting the clockwise moment equal to the counterclockwise moment, we find:
N × 0.15 m = mg × 0.10 m,
where N is the normal force, m is the mass of the stick (0.600 kg), g is the acceleration due to gravity (9.81 m/s²), and the lever arms are in meters.
Substituting the values and solving for N gives:
N = (0.600 kg × 9.81 m/s² × 0.10 m) / 0.15 m = 3.924 N.
Therefore, the magnitude of the normal reaction from the floor on the stick is 3.924 N.