A uniform pole 7m long weighting 10kg is support by a boy 2m from one end and a man 3m from the other end. At what point must a 20kg weight be attached so that the man would support thrice as much weight as the boy
How can this be solved
To solve this problem, we need to consider the moments (or torques) acting on the pole.
The torque of an object is calculated by multiplying the force applied by the distance from the pivot point. In this case, the pivot point is the fulcrum, which is the point where the pole is being supported.
Let's assume that the distance from the boy to the fulcrum is x and the distance from the fulcrum to the weight is d. We can set up the equation to solve for d.
Torque exerted by the boy: (force of the boy) * (distance from the boy to the fulcrum) = (10 kg) * (9.8 m/s^2) * (2 m) = 196 N*m.
Torque exerted by the man: (force of the man) * (distance from the man to the fulcrum) = (10 kg) * (9.8 m/s^2) * (3 m) = 294 N*m.
Let's assume the force exerted by the weight is F kg * 9.8 m/s^2. The torque exerted by the weight is then (F) * (distance from the weight to the fulcrum) = F * d.
Since the man must support thrice as much weight as the boy, we can set up the equation:
3 * (torque exerted by the boy) = (torque exerted by the man) + (torque exerted by the weight).
Replacing the torques with their respective formulas, we have:
3 * 196 N*m = 294 N*m + F * d.
Simplifying the equation, we get:
588 N*m - 294 N*m = F * d,
294 N*m = F * d.
Now, we need to find the value of d when the weight is 20 kg. Let's plug in the values:
294 N*m = (20 kg) * (9.8 m/s^2) * d,
294 N*m = 196 N * d.
Simplifying further:
d = 1.5 m.
So, the 20 kg weight must be attached 1.5 meters from the fulcrum in order for the man to support thrice as much weight as the boy.