Final answer:
The reactions at the support points P and Q for a uniform log of wood in static equilibrium are 240 N and 80 N, respectively. Point P is located 3 m from end A and supports more weight because it is closer to the log's center of gravity compared to point Q which is 2 m from end B.
Step-by-step explanation:
To solve for the reactions at points P and Q, we must first understand that the log is in static equilibrium, meaning that the sum of the torques (moment of forces) must be zero for it to be at rest. Point P is 3 meters from end A and point Q is 2 meters from end B. The entire log is 9 meters long, meaning P is 6 meters from B and Q is 7 meters from A.
First, we can calculate the torque about point P, assuming clockwise is positive, to find the force at Q. We know that the weight of the log acts at its center of gravity, which would be at 4.5 meters from either end. Since the log is uniform, it must be at the midpoint. Therefore:
Torque at P due to weight of log = weight of log × distance from P to center of gravity
= 320 N × 1.5 m = 480 N·m
Torque at P due to reaction at Q = reaction at Q × distance from P to Q
= Reaction at Q × 6 m
For equilibrium: Torque at P due to weight = Torque at P due to reaction at Q
480 N·m = Q × 6 m
Q = 80 N
Now, we can find the reaction at P by using the fact that the sum of vertical forces must also be zero. The only vertical forces acting on the log are its weight and the reactions at P and Q:
Weight of the log = Reaction at P + Reaction at Q
320 N = P + 80 N
P = 240 N
Therefore, the reaction at point P is 240 N and the reaction at point Q is 80 N, which corresponds to option B.