where must a 1200n weight be hung on a uniform, 200N pole so that a boy at one end supports one-third as much as a man at the other end?
To find the position where the 1200N weight must be hung on the uniform 200N pole, we need to consider the conditions for the boy and the man supporting the pole.
Let's denote the distance from the boy's end of the pole to the point where the weight is hung as "x" (measured in meters). The distance from the man's end of the pole to the same point is therefore (1 - x) meters.
According to the problem, the boy supports one-third of the weight compared to the man's support. Mathematically, this can be expressed as:
Boy's support = (1/3) * Man's support
We know that the boy's support is equal to the weight at his end of the pole, which is 200N. The man's support is the difference between the weight at his end (which is 1200N) and the weight at the point where the 1200N weight is hung.
Using the above information, we can set up the equation:
200N = (1/3) * (1200N - 1200N * x)
Simplifying the equation:
200N = (1/3) * 1200N * (1 - x)
Multiplying both sides by 3:
600N = 1200N * (1 - x)
Dividing both sides by 1200N:
600N / 1200N = 1 - x
0.5 = 1 - x
Subtracting 1 from both sides:
-0.5 = -x
Multiplying both sides by -1:
0.5 = x
Therefore, the 1200N weight should be hung 0.5 meters (or 50 centimeters) from the boy's end of the pole.