Can someone help me answer this question or set it up?

Question

A23 kg boy sits 1.5 m from the axis of rotation of a seesaw. At what distance from the axis of rotation must a 21kg boy be positioned on the other side of the axis to balance the seesaw?

Answer 1

Fd = Fd for the sides of the seesaw

23 * 1.5 = 21 * x
x = ((23 kg) * 1.5 m) / (21 kg) = 1.64285714 meters
should be with 2 significant digits: 1.6 m

Answer 2

It's a matter of balancing the CCW torque against the CW torque. Tccw = Tcw. Torque = F*d, the product of the distance to the pivot and the force perpendicular to the bar.

If the force isn't perpendicular you need to get involved with some trig, but not for this problem.

Answer 3

see-saws, or levers are sometimes known as force magnifiers!
the further away from the axis you are the less force required to balance the other side.

set up as a right hand side (RHS) and left hand side (LHS) problem

the 'moments' on each side must be balanced (equal) and
'moment' = force x distance from axis

see if that helps - don't want to do it for you as it is straight forward once you realise whats asked.
have you looked over your notes/textbook?

Answer 4

F1 X D1 = F2 X D2
F1 = 23 kg-wt
D1 = 1.5 m
F2 = 21 kg-wt
Now,
D2 = F1 x D1 / F2
= 23 x 9.8 x 1.5 / 21 x 9.8
= 1.64 m