No you will need the weight of the seesaw. Just consider the seesaw as another massive object, with the mass coming from directly at the center of it. Does that make sense?
By giving you the weight of the object, they are indirectly giving you it's mass ([tex]W=mg[/tex]).
Hope that helps,
If you kid is in high school or less - you can ignore the seesaw. Your pivot point will be the center, which will allow you to completely cancel out the mass of the seesaw. If you wish to include it - that is fine - but you will see that it cancels out.
Assume the mass distribution of the seesaw itself to be symmetric with respect to the fulcrum; that is, that its torques balance out. The seesaw is balanced by the kids when the torque Mkid1rkid1=Mkid2rkid2, where r is the torque arm to the given kid measured from the fulcrum, and M is the mass of the given kid.
Just a quick note. everyone so far has assumed that the seesaw is symetrical and if this is the case then they are correct in assuming that its weight has no bearing on the problem however if it isnt then it would. For example:
If you take a 10kn seesaw which is 10 metres long but the fulcrum (pivot) is placed 6 metres from one end then the seesaw will fall towards the 6 metre end as the "moment" (engineering term) will be greater acting about the centre of thats sides mass i.e. 6kn x 3 metres = 18knm, where as the 4 metre side would be 4kn x 2 metres giving 8knm. To balance this out we need an extra 10knm on the 4 m side so if you know the weight of the kid say 4 kn then we can find the position for equilibrium (balance) by dividing the required additional moment 10knm, by the kid i.e. 10knm divided by 4kn giving a distance of 2.5 metres from the fulcrum on the 4 metre side of the seesaw.