A uniform pole 20 ft. long and weighing 80 lbs is held by a boy 2 ft from one end while a man carries the same pole 5 ft from the other end. At what point should a load of 100 lbs be placed so that the man will carry twice as much weight as the boy.
To solve this problem, we need to use the principle of moments or torque.
The formula for moments (or torque) is given by:
Moment = Force × Distance
Let's assume that the distance from the boy's end to the load is x feet.
According to the problem, the load of 100 lbs should be placed in such a way that the man will carry twice as much weight as the boy.
Since the pole is uniform, its weight of 80 lbs should be evenly distributed along its length.
Now let's calculate the moments for the boy and the man:
For the boy:
Weight of the pole (80 lbs) × Distance from the boy's end (2 ft)
For the man:
Weight of the pole (80 lbs) × Distance from the man's end (20 ft) - Load (100 lbs) × Distance from the man's end (x ft)
According to the problem, the man should carry twice as much weight as the boy. Therefore, we can set up the following equation:
Weight of the pole × Distance from the boy's end = 2 × (Weight of the pole × Distance from the man's end - Load × Distance from the man's end)
Using the given values, we can solve for x:
80 lbs × 2 ft = 2 × (80 lbs × 20 ft - 100 lbs × x ft)
Simplifying the equation:
160 ft-lbs = 2 × (1600 ft-lbs - 100x ft-lbs)
160 ft-lbs = 3200 ft-lbs - 200x ft-lbs
Moving all terms to one side of the equation:
3200 ft-lbs - 160 ft-lbs = 200x ft-lbs
3040 ft-lbs = 200x ft-lbs
Dividing both sides of the equation by 200 ft-lbs:
x = 3040 ft-lbs / 200 ft-lbs
x = 15.2 ft
Therefore, the load of 100 lbs should be placed 15.2 ft from the man's end so that the man carries twice as much weight as the boy.
To find the point at which a load of 100 lbs should be placed on the pole so that the man carries twice as much weight as the boy, we need to consider the torque exerted by each individual.
The torque exerted by an object is calculated by multiplying the force applied to an object by the distance from the point of rotation (fulcrum or pivot). In this case, the point of rotation is the center of the pole.
Let's assign variables to the unknown distances:
- The distance of the boy from the center of the pole: x ft
- The distance of the man from the center of the pole: 20 ft - x ft (since the total length of the pole is 20 ft)
Now, let's calculate the torque exerted by the boy and the man:
Torque exerted by the boy = force applied by the boy x distance of the boy from the center.
Torque exerted by the man = force applied by the man x distance of the man from the center.
The force applied by the boy and the man can be derived from their respective positions relative to the center of the pole. Since we know the objective is for the man to carry twice the weight of the boy, we can set up an equation:
Force applied by the man = 2 x Force applied by the boy.
Given that the pole weighs 80 lbs, the load is 100 lbs, and the total weight carried by the boy and man combined is (80 + 100) lbs = 180 lbs, we can rewrite the equation:
Force applied by the man + Force applied by the boy = 180 lbs.
Substituting the expressions for torque and force, we get:
(2 x Force applied by the boy x (20 - x ft)) + (Force applied by the boy x x ft) = 180 lbs.
Now, we can solve this equation to find the value of x, which represents the distance from the center of the pole where the load should be placed.