A 437.7 N child and a 237.1 N child sit on either end of a 2.5 meter long seesaw.
How far from the 437.7 N child should the pivot be placed to ensure rotational equilibrium? disregard the mass of the seesaw. answer in units of m.
I will be happy to critique your thinking.
The sum of the moments about the piviot has to be zero.
Let the heavier child be x meters from the fulcrum. The lighter child will be 2.5 - x meters from the fulcrum.
For equilibrium,
437.7 x = 237.1 (2.5 -x)
Solve for x.
674.8 x = 592.75
x = 0.88 m
To ensure rotational equilibrium, the total torque on the seesaw must be zero. Torque is the product of force and the perpendicular distance from the pivot point.
Let's assume the pivot point is placed at a distance 'd' from the 437.7 N child.
The torque produced by the 437.7 N child is given by Torque1 = Force1 * Distance1 = 437.7 N * d
The torque produced by the 237.1 N child is given by Torque2 = Force2 * Distance2 = 237.1 N * (2.5 - d)
For rotational equilibrium, Torque1 should be equal to Torque2:
437.7 N * d = 237.1 N * (2.5 - d)
Now, let's solve this equation to find the value of 'd'.
437.7 N * d = 237.1 N * (2.5 - d)
437.7d = 237.1 * 2.5 - 237.1d
437.7d + 237.1d = 237.1 * 2.5
674.8d = 592.75
d = 592.75 / 674.8
d ≈ 0.878 m
Therefore, the pivot should be placed approximately 0.878 meters from the 437.7 N child to ensure rotational equilibrium.