Question

The heavy boy and light girl are balanced on the seesaw. If they both move inward so that each is exactly half the original distance from the pivot, then the 1) boy's end will move downward. 2) girl's end will move downward. 3) seesaw will remain in balance. If the fulcrum of the seesaw is repositioned so each can sit at an END of the board and be in balance, then if each moves inward half way the 4) boy's end will move downward. 5) girl's end will move downward. 6) seesaw will remain balanced. choose one answer from (1-3) and one answer from (4-6)

Answer 1

If they both move inward so that each is exactly half the original distance from the pivot, then the :

(1) boy's end will move downward.

Reason : Let, the boy weighs W, the girl weighs w and length of the seesaw be L. If they both move inward so that each is exactly half the original distance from the pivot, then,

moment of force on the boy's side = W x ( L / 2 );

and, moment of force on the girl's side = w x ( L / 2 ).

Since, W > w, hence, boy's side has greater moment of force. So, the boy's end will move downward.

If the fulcrum of the seesaw is repositioned so each can sit at an END of the board and be in balance, then if each moves inward half way the :

(6) seesaw will remain balanced.

Reason : Let, the boy weighs W, the girl weighs w and length of the seesaw be L. When the fulcrum of the seesaw is repositioned so each can sit at an END of the board and be in balance, then let the the pivot be at a distance x away from the boy's end and y away from the girls end, where, ( x + y ) = L.

Given, moment of force on the boy's side = moment of force on the boy's side

or, Wx = wy.

Now, if each moves inward half way, then,

moment of force on the boy's side = W x ( x / 2 );

and, moment of force on the girl's side = w x ( y / 2 ).

Since, Wx = wy, hence, moment is same on both sides.

Hence, the seesaw will remain balanced.