Final answer:
The reaction at support A is 1060 N and the reaction at support B is -1060 N.
Step-by-step explanation:
To find the reactions at the supports, we need to consider the equilibrium of the rod. Since the rod is uniform, the weight of the rod can be considered to act at its center. The weight of the rod is 100 N and it acts downward. The 40 N weight suspended at 10 cm from the midpoint of the rod can be decomposed into two forces: one parallel to the rod and one perpendicular to the rod. The force parallel to the rod will contribute to the reaction at each support, while the force perpendicular to the rod will not.
Let's denote the reaction at support A as RA and the reaction at support B as RB. We can set up an equilibrium equation for the horizontal direction:
RA + RB = 0. The sum of the reactions in the horizontal direction must be zero, since there is no net force acting on the rod in that direction.
Next, let's consider the moments about point A. We can calculate the moment due to the weight of the rod and the moment due to the suspended weight:
(10 cm * RA) + (90 cm * 100 N) + (40 N * 40 cm) = 0. The clockwise moments must balance out the counterclockwise moments for the rod to be in equilibrium.
Simplifying the equation, we get:
10 RA + 9000 + 1600 = 0
10 RA = -10600
RA = -1060 N
Since the reaction at support A cannot be negative, we can conclude that the reaction at support A is 1060 N and the reaction at support B is -1060 N.