From the definition of equilibrium, at the moment in which both children are sitting facing each other at a certain distance the torque within the seesaw will be zero.
However, if one of the children approaches the pivot, the center of mass will move towards the end of the other child, which will immediately cause the child to rise.
Another way of observing this problem is considering the distance between the two, the distance is proportional to the Torque,
[tex]\tau_1 = \tau_2[/tex]
[tex]F*d_1 = F*d_2[/tex]
Therefore by decreasing the distance - when walking towards the pivot - the torque of the child sitting at the other end will be greater because it keeps its distance.
Being said higher torque will cause the approaching child to rise.
The correct answer is B.
The child's side will rise because of shifting of weight of the child from its place.
Two children are balanced on opposite sides of a seesaw. If one child leans inward toward the pivot point, her side will rise because of shifting of weight of the child from its place. Due to movement from its place, the weight of other child increases which leads to lowering of its side.
So we can conclude that the child's side will rise because of shifting of weight of the child from its place.
Learn more about seasaw here: https://brainly.com/question/23710049
