A uniform pole 7m long and weighing 10kg is supported by a boy 2m from one end.At what point must a 20kg weight be attached so that the man would support thrice as much weight as the boy?
A uniform pole 70 m long and weigh 10kg is supported by a boy 2m from one end and a man 3m from the order end. At what point must a 20kg weigh be attach so that the man would support twice as much weight as the boy
To solve this problem, we need to find the point on the pole where a 20kg weight must be attached to make the man support three times the weight that the boy is supporting.
Let's first analyze the situation. The pole is 7m long and weighs 10kg. The boy is 2m from one end, which means the distance between the boy and the other end of the pole is 7m - 2m = 5m.
To keep the pole in equilibrium, the torques on both sides of the fulcrum (the point where the boy is supporting the pole) must be equal. Torque is the force multiplied by the distance from the fulcrum.
Let's assume that the point where the 20kg weight is attached is x meters from the boy's end. The torques acting on the pole are as follows:
Torque exerted by the boy = Weight supported by the boy * Distance from the boy's end
Torque exerted by the man = Weight supported by the man * Distance from the man's end
Since we want the torque exerted by the man to be three times the torque exerted by the boy, we can set up the following equation:
Torque exerted by the man = 3 * Torque exerted by the boy
Using the torque formula, we can write:
(20kg * x) = 3 * (10kg * 5m)
Simplifying the equation:
20x = 3 * 50
20x = 150
Dividing both sides by 20:
x = 7.5m
Therefore, the 20kg weight must be attached 7.5m from the boy's end of the pole in order for the man to support three times the weight that the boy is supporting.