A uniform pole 7m long and weighing 10kg is supported 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.
To solve this problem, we need to consider the torques acting on the pole.
The torque is the product of the force and the distance from the pivot point. In this case, the pivot point is the boy supporting the pole.
Let's label the forces and distances:
- The weight of the pole (10kg) exerts a downward force at its center, which we'll call Fp and its distance from the boy is 7/2 = 3.5m
- The boy exerts an upward force (Fb) at a distance of 2m from the boy
- The man exerts an upward force (Fm) at a distance of 3m from the boy
- The additional weight (20kg) will be attached at a distance (x) from the boy
According to the lever principle, for the pole to remain in equilibrium (not rotate), the sum of the clockwise torques must equal the sum of the counterclockwise torques.
Clockwise torques:
- Force due to the weight of the pole: Fp * (distance from pivot) = 10 * 9.8 * 3.5 Nm
Counterclockwise torques:
- Boy's force: Fb * 2m
- Man's force: Fm * 3m
- Additional weight's force: 20 * 9.8 * x Nm
Since we want the man to support thrice, we can set up the equation:
Fm * 3m = 10 * 9.8 * 3.5 Nm + Fb * 2m + 20 * 9.8 * x Nm
Simplifying and rearranging:
Fm = [(10 * 9.8 * 3.5) + (Fb * 2m) + (20 * 9.8 * x)] / 3m
Now we can plug in the given values and solve for x.