Final answer:
To ensure the boy carries only one-third of the total load on an 8 m uniform pole, the 90 kg load should be placed approximately 1.78 m from the boy's end of the pole. This position is calculated using the principle of moments, considering the moments must balance for equilibrium, although this exact distance is not one of the student's provided options.
Step-by-step explanation:
The question asks where a load of 90 kg should be suspended from the boy's end of an 8 m pole so that the boy carries only one-third of the total load. The pole and the man together carry the rest of the load. To solve this, we can use the principle of moments (torque), which states that for the system to be in equilibrium, the clockwise moments must be equal to the anticlockwise moments about a pivot point.
The total mass being carried is the mass of the pole (60 kg) plus the mass of the load (90 kg), making it 150 kg. The boy should carry one-third of this, which is 50 kg, and the man should carry the rest, which is 100 kg.
Let the distance from the boy's end (where the boy is acting as the pivot) to the point where the load should be placed be x meters. The moment about the boy will be:
60 kg * 4 m (center of pole) + 90 kg * x m = 50 kg * 8 m (boy's share of the load at the end of the pole).
240 kg*m + 90 kg * x m = 400 kg*m
90 kg * x m = 400 kg*m - 240 kg*m
90 kg * x m = 160 kg*m
x = 160 kg*m / 90 kg
x = 1.777... m, which rounds off to 1.78 m (not listed in the options given by the student, indicating a possible typo).