a. To calculate the torque exerted by the weight of the plank about the pivot point at the edge of the dock, we need to use the formula for torque:
Torque = Force * Distance
The force we need to consider is the weight of the plank (90 N). The distance is the part of the plank that extends over the water (1 m).
Torque = 90 N * 1 m = 90 N-m
So, the torque exerted by the weight of the plank about the pivot point is 90 N-m.
b. To find how far from the edge of the dock the boy can move until the plank is just on the verge of tipping, we need to consider the balance of torques. The total torque on the plank must be zero for it to be in equilibrium.
The torque exerted by the weight of the plank about the pivot point (as calculated in part a) is 90 N-m.
The torque exerted by the boy's weight (150 N) can be calculated by considering the boy's weight acting at the center of the plank. Since the boy is at the center of the plank and the plank is 4 m long, the distance from the pivot point to the boy is 2 m.
Torque by the boy = 150 N * 2 m = 300 N-m
For the plank to be in equilibrium, the total torque must be zero:
Total torque = Torque by the plank + Torque by the boy = 0
Therefore, the torque exerted by the boy's weight is equal and opposite to the torque exerted by the weight of the plank:
300 N-m = -90 N-m
To find how far the boy can move until the plank is just on the verge of tipping, we need to calculate the distance (x) from the edge of the dock where the boy's weight is acting.
Torque by the boy = Boy's weight * Distance from the edge
300 N-m = 150 N * x
x = 300 N-m / 150 N
x = 2 m
So, the boy can move 2 meters from the edge of the dock until the plank is just on the verge of tipping.