a uniform pole 20 ft. long and weighing 80 lb. is suspected by a boy 2.0 ft.from end A and a man 5.0 ft. from end B .at what point should a load of 100 lb. be placed on so that the man will support twice as much as the boy?
To find the point at which the 100 lb. load should be placed on the pole, we need to consider the balancing forces and moments.
Let's denote the distance between end A and the load placement point as 'x'. The distance between the load placement point and end B would then be (20 - x) ft.
Now, let's determine the forces acting on the pole.
1. The weight of the pole itself (80 lb.) acts at its center, which is 20 ft / 2 = 10 ft from each end A and B. So, the force due to the weight of the pole is 80 lb. acting downward at the midpoint.
2. The force acting on the pole due to the boy is 2.0 ft away from end A. Since both the boy and the pole are 5 ft away from end B, the force exerted by the boy is (80 lb. * 5 ft) / 2.0 ft = 200 lb.
3. The force exerted by the man is 5 ft away from end B, and we want this force to be twice the force exerted by the boy. So, the force exerted by the man is 2 * 200 lb. = 400 lb.
4. The 100 lb. load placed at distance 'x' from end A will exert a force of 100 lb.
Considering the forces acting on the pole, we can sum them up and set them equal to zero to ensure the pole is in equilibrium:
Sum of forces = 0.
Downward forces:
80 lb. - 100 lb. = -20 lb. (-ve because it acts downward).
Upward forces:
200 lb. + 400 lb. = 600 lb.
Summing up the forces:
-20 lb. + 600 lb. = 0.
Now let's consider the moments acting on the pole.
The moment due to each force is given by the force multiplied by its perpendicular distance from a reference point. Since we are given the distances from both ends, we can take one end as the reference point (end A) and calculate the moments about it.
The moment due to the weight of the pole is:
80 lb. * 10 ft = 800 ft-lb. (Since it acts at the midpoint).
The moment due to the boy's force is:
200 lb. * 2.0 ft = 400 ft-lb.
The moment due to the man's force is:
400 lb. * (20 - 5) ft = 6000 ft-lb. (Since the man is 5 ft from end B and the total length is 20 ft).
The moment due to the load is:
100 lb. * x ft. (Since the distance from end A is 'x').
Considering the moments acting on the pole, we can sum them up and set them equal to zero to ensure the pole is in equilibrium:
Sum of moments = 0.
Summing up the moments:
800 ft-lb. + 400 ft-lb. + 6000 ft-lb. + 100 lb. * x ft. = 0.
Now, we can solve this equation to find the value of 'x'.
800 ft-lb. + 400 ft-lb. + 6000 ft-lb. + 100 lb. * x ft. = 0.
7200 ft-lb. + 100 lb. * x ft. = 0.
100 lb. * x ft. = -7200 ft-lb.
x ft. = (-7200 ft-lb.) / 100 lb.
x ft. = - 72 ft-lb / lb.
Therefore, based on the calculation, the load of 100 lb. should be placed at a distance of 72 ft. from end A to achieve the desired balance, where the man will support twice as much as the boy.
To find the point at which the load should be placed, we need to consider the torques acting on the pole.
Torque can be calculated as the product of the force and the perpendicular distance from the pivot point. In this case, the pivot point is the end A.
Let's assume that the load is placed x feet from end A. Therefore, the load is (20 - x) feet away from end B.
The torque due to the weight of the pole itself can be calculated as:
Torque_boy = Weight_boy * Distance_boy
Similarly, the torque due to the load can be calculated as:
Torque_man = Weight_man * Distance_man
Since the prompt says that the man should support twice as much weight as the boy, we can set up the equation:
Torque_man = 2 * Torque_boy
Substituting the formulas for torque into the equation, we have:
Weight_man * Distance_man = 2 * (Weight_boy * Distance_boy)
Now let's plug in the given values:
Weight_boy = 80 lb
Weight_man = 100 lb
Distance_boy = 2.0 ft
Distance_man = 5.0 ft
Substituting these values, the equation becomes:
100 * Distance_man = 2 * (80 * 2)
Simplifying the equation:
100 * Distance_man = 320
Divide both sides of the equation by 100:
Distance_man = 3.2 ft
Since the distance from end B to the load is (20 - x) ft, we can now find x by subtracting the distance the man is from end B:
x = 20 - Distance_man
x = 20 - 3.2
x = 16.8 ft
Therefore, the load should be placed 16.8 feet from end A, or 3.2 feet from end B, so that the man will support twice as much weight as the boy.