Where a 1208 N weight must be hung on a uniform 214 m pole so that a boy at one end supports one-third as much as a man at the other end.
To solve this problem, we need to find the distance from the boy to the weight on the pole. Let's break down the steps:
1. We know that the total weight is 1208 N, and the pole is uniform, meaning it has the same mass distribution throughout its length.
2. Let's assume the distance from the boy to the weight is "x" meters. This means the distance from the man to the weight is (214 - x) meters.
3. According to the problem, the boy at one end supports one-third as much as the man at the other end. Mathematically, this can be represented as:
Boy's weight = (1/3) * Man's weight
We can set up the equation as follows:
x * (1208 N) = (214 - x) * (1/3) * (1208 N)
4. Simplifying the equation, we have:
1208x = (214 - x) * (1/3) * 1208
5. Solve for x:
1208x = (214 - x) * 402.67
Distribute (1/3) * 1208 to get:
1208x = 402.67 * 214 - 402.67x
Simplify:
1208x + 402.67x = 402.67 * 214
Combine like terms:
1610.67x = 86158.38
Divide both sides by 1610.67:
x = 86158.38 / 1610.67
x ≈ 53.526 meters
Therefore, the weight must be hung approximately 53.526 meters from the boy and (214 - 53.526) ≈ 160.474 meters from the man for the boy to support one-third as much as the man.