Final answer:
The boy can get as close as approximately 1.67 m to the right end of the beam without it falling over.
Step-by-step explanation:
To determine how close the boy can get to the right end of the beam without it falling over, we need to consider the torque acting on the system. Torque is the rotational equivalent of force. The torque caused by the weight of the beam can be found by multiplying the weight of the beam by its distance from the pivot point. The torque caused by the boy can be found by multiplying the weight of the boy by his distance from the pivot point.
In this case, the torque caused by the weight of the beam is the same on both sides of the pivot point, so it can be disregarded. The torque caused by the boy can be balanced by the torque caused by the weight of the beam located on the right side of the pivot point. Therefore, the boy can get as close to the right end of the beam as the point where the weight of the beam is equal to his weight multiplied by his distance from the pivot point divided by the length of the beam.
Let's calculate this. The weight of the boy can be found by multiplying his mass (20 kg) by the acceleration due to gravity (9.8 m/s^2). The distance from the pivot point is given as 5.0 m.
Weight of the boy = (20 kg)(9.8 m/s^2) = 196 N
Now, we can calculate the distance from the right end of the beam where the weight of the beam is equal to the weight of the boy.
(40 kg)(9.8 m/s^2)(d) = (20 kg)(9.8 m/s^2)(5 m - d)
40d = 100 - 20d
60d = 100
d = 1.67 m
Therefore, the boy can get as close as approximately 1.67 m to the right end of the beam without it falling over.