A uniform pole 7m long and weighing 10kg is supported by a boy of 2m from one end and a man 3m from the other end.at what point must a 20kg weight attached so that the man would support thrice as much weight as the boy?
Not known
To solve this problem, we need to find the point where a 20kg weight should be attached to the pole so that the man supports three times the weight as the boy.
Let's first calculate the weight supported by the boy and the man separately.
Weight supported by the boy:
The total weight of the pole is 10kg, and its center of mass is at the midpoint of the pole, which is (7m)/2 = 3.5m from either end.
The distance between the boy and the center of mass is 3.5m - 2m = 1.5m.
Therefore, the weight supported by the boy = (weight of the pole) × (distance to center of mass) / (length of the pole)
= (10kg) × (1.5m) / (7m)
≈ 2.14kg
Weight supported by the man:
The distance between the man and the center of mass is 3.5m - 3m = 0.5m.
Let's assume the distance between the point of attachment of the weight and the center of mass is 'd'.
Therefore, the weight supported by the man = (weight of the pole + weight of the attached weight) × (distance to center of mass) / (length of the pole)
= (10kg + 20kg) × (0.5m + d) / (7m)
According to the problem, the weight supported by the man must be three times the weight supported by the boy. So, we can write the equation as follows:
3 × weight supported by the boy = weight supported by the man
3 × 2.14kg = (30kg) × (0.5m + d) / (7m)
Now, we can solve this equation to find the value of 'd'.
Multiply both sides of the equation by (7m) to eliminate the denominator:
3 × 2.14kg × 7m = 30kg × (0.5m + d)
44.94kg × 7m = 15m × (0.5m + d)
314.58kgm = 7.5m + 15md
314.58kg = 7.5 + 15d
Rearranging the equation:
15d = 314.58kg - 7.5
15d = 307.08kg
d = 307.08kg / 15
d ≈ 20.47m
Therefore, the 20kg weight should be attached approximately 20.47 meters from the boy's side of the pole to ensure that the man supports three times the weight as the boy.