A 450.8 N child and a 101.0 N child sit on either end of a 2.2 m long seesaw.

How far from the 450.8 N child should the pivot be placed to ensure rotational equilibrium? Disregard the mass of the seesaw. Answer in units of m
part 2 of 2
Suppose a 367 N child sits 0.193 m from the 450.8 N child. How far from the pivot must a 261 N child sit to maintain rotational equilibrium? Answer in units of m

To ensure rotational equilibrium, we need to consider the torque on both ends of the seesaw. Torque is defined as the product of the force applied and the distance from the pivot point.

For part 1:
We have a 450.8 N child on one end and a 101.0 N child on the other end. Let's assume the pivot point is located at a distance x from the 450.8 N child.

The torque produced by the 450.8 N child is given by:
Torque1 = Force1 * Distance1

And the torque produced by the 101.0 N child is given by:
Torque2 = Force2 * Distance2

Since we want to ensure rotational equilibrium, the sum of the torques on each side should be zero:
Torque1 + Torque2 = 0

Substituting the given values, we have:
(450.8 x) + (-101.0 * (2.2 - x)) = 0

Simplifying the equation, we get:
450.8x - 101.0(2.2 - x) = 0

Expanding and rearranging the terms, we have:
450.8x - 222.2 + 101.0x = 0
551.8x - 222.2 = 0

Solving for x, we get:
x = 222.2 / 551.8
x ≈ 0.402 m

Therefore, the pivot should be placed approximately 0.402 m away from the 450.8 N child to ensure rotational equilibrium.

For part 2:
We have a 367 N child sitting at a distance of 0.193 m from the 450.8 N child. Let's assume the pivot point is located at a distance y from the 450.8 N child.

The torque produced by the 367 N child is given by:
Torque3 = Force3 * Distance3

And the torque produced by the 261 N child should be equal in magnitude but opposite in direction, so:
Torque4 = -Force4 * Distance4

Again, we want the sum of the torques to be zero for rotational equilibrium:
Torque3 + Torque4 = 0

Substituting the given values, we have:
(367 * 0.193) + (-261 * (2.2 - y)) = 0

Simplifying the equation, we get:
70.791 + (-261 * (2.2 - y)) = 0

Expanding and rearranging the terms, we have:
70.791 - 574.2 + 261y = 0
261y - 503.409 = 0

Solving for y, we get:
y = 503.409 / 261
y ≈ 1.93 m

Therefore, the 261 N child should sit approximately 1.93 m away from the pivot to maintain rotational equilibrium.

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