Answer:
To determine where the 20 kg weight should be attached on the pole so that the man supports three times the weight supported by the boy, we need to consider the distribution of the pole's weight.
Let's assume that the distance from the boy to the 20 kg weight is x meters. Since the pole is 7 meters long, the distance from the man to the 20 kg weight would be (7 - 3 - x) meters.
The weight distribution on the pole can be calculated using the principle of moments, which states that the sum of the anticlockwise moments is equal to the sum of the clockwise moments in equilibrium.
The moment exerted by the boy is given by the product of the weight (10 kg) and the distance (3 m). This can be represented as: 10 kg * 3 m = 30 kg·m.
The moment exerted by the man is given by the product of the weight (10 kg) and the distance (7 - x) meters. This can be represented as: 10 kg * (7 - x) m = 70 kg·m - 10 kg·x m.
According to the problem, the man should support three times the weight supported by the boy. Therefore, we can set up the equation:
70 kg·m - 10 kg·x m = 3 * 30 kg·m
Simplifying the equation:
70 - 10x = 90
-10x = 90 - 70
-10x = 20
x = 20 / -10
x = -2
The negative value for x implies that the 20 kg weight should be attached 2 meters on the opposite side of the boy. This means that the 20 kg weight should be attached 2 meters from the man's side of the pole.
Please note that the negative sign indicates the direction (opposite side of the boy) rather than a negative distance.