Answer: The plank will tip when the torque due to the weight of the person is greater than the torque due to the weight of the plank. The torque is given by the product of the force and the perpendicular distance to the pivot point (the left support).
Let x be the distance the person walks on the overhanging part of the plank, and let y be the perpendicular distance from the right support to the center of mass of the plank. Then:
y = (1/2) * 5 = 2.5 m (since the center of mass of a uniform plank is at its midpoint)
The weight of the plank is 225 N, acting at its center of mass, so the torque due to the weight of the plank is:
T_plank = 225 N * 2.5 m = 562.5 Nm
The weight of the person is 450 kg * 9.81 m/s^2 = 4414.5 N, acting at the end of the overhanging part of the plank. The perpendicular distance from the right support to the person is (1.1 - x) m. So the torque due to the weight of the person is:
T_person = 4414.5 N * (1.1 - x) m
The plank will just begin to tip when T_person = T_plank. Solving for x:
4414.5 N * (1.1 - x) m = 562.5 Nm
4845.95 N - 4414.5 Nx = 562.5 Nm
4414.5 Nx = 4283.45 Nm
x = 0.9704 m
Therefore, the person can walk a distance of 0.9704 m on the overhanging part of the plank before it just begins to tip.
Step-by-step explanation: