A playground seesaw is exactly 3.00 m long. A little girl who weighs 425 N is sitting on one end. If a little boy weighing 575 N wants to sit on the other side, how far must he sit from the fulcrum to keep the seesaw perfectly balanced?

575x=425(3-x) if the seesaw is not centered.

1.275

To find the distance at which the little boy must sit to keep the seesaw balanced, we can use the principle of torque. Torque is the rotational equivalent of force, given by the product of force and the perpendicular distance from the fulcrum.

In this case, since the seesaw is balanced, the torques on both sides of the fulcrum must be equal. The torque can be calculated by multiplying the force exerted by the distance from the fulcrum.

The little girl exerts a force of 425 N at a distance of x1 from the fulcrum, and the little boy exerts a force of 575 N at a distance of x2 from the fulcrum. Therefore, we have:

Torque of the little girl = Torque of the little boy

(Force of the little girl) × (Distance from the fulcrum to the girl) = (Force of the little boy) × (Distance from the fulcrum to the boy)

425 N × x1 = 575 N × x2

To solve for x2, we can rearrange the equation:

x2 = (425 N × x1) / 575 N

Now, plug in the values. We know that the length of the seesaw is 3.00 m. Since the fulcrum is at the center, the distance from the fulcrum to the girl is half the length of the seesaw, which is 3.00 m / 2 = 1.50 m.

Substitute the values:

x2 = (425 N × 1.50 m) / 575 N

Now, calculate:

x2 = 1.106 m

Therefore, the little boy must sit approximately 1.106 meters from the fulcrum to keep the seesaw perfectly balanced.