A uniform bar is 15 metre long is balanced on a pivot placed at the mid point .a boy of a mass 55kg sits on one hand of the bar at a point 5m away from theotherendtokeepthebarhorizontallywithadiagram?

To keep the bar horizontally balanced, the torque on one side of the pivot must be equal to the torque on the other side of the pivot.

The torque on the side with the boy can be calculated as:
Torque = force x distance
Torque = (55kg x 9.8m/s^2) x 5m
Torque = 2695 Nm

Since the bar is balanced at the midpoint, the torque on the other side of the pivot must also be 2695 Nm.

Let the distance from the pivot to the other end of the bar be x. Since the bar is 15m long, x = 15m - 5m = 10m.

Now we can calculate the force needed on the other side of the pivot to balance the bar:
Force = Torque / distance
Force = 2695 Nm / 10m
Force = 269.5 N

Therefore, a force of 269.5 N must be applied on the other side of the pivot to keep the bar horizontally balanced.