1.A see-saw has a 40kg boy sitting at a distance of 80cm from one end and his little brother of 30kg is counter balanced on the other side. The fulcrum is in the middle of the uniform plank of length 4.0 .How far is the little brother from the other end of the see-saw ?

2.In the above problem on see-saw the mass of the plamk is 10kg. (a) What is the force exerted(magnitude and direction) on the support system by the see-saw and the children? (b.)What is the normal reaction on the see-saw? Do both calculations for the see-saw remaining horizontal.

1- Total length of plank=4 m=400 cm
Distance of 40 kg boy (Suppose boy A) from fulcrum=(400/2)-80=120 cm
Distance of 30 kg boy (Suppose boy B) from fulcrum=x (suppose)
Using
Weight of A x distance from fulcrum=Weight of B x distance from fulcrum
400 x 120=300 x  (since 1 kg mass means mg=1x 10=10 N)
x=160 cm
Thus its distance from the other end will be (400/2)-160=40 cm
2-
The easiest way to find the force is to use the first condition for equilibrium, which is
Net F=0.
The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as
net Fy=0
Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that
Fy-400-300-100=0
or F=800N
Which is equal to normal reaction i.e.

$$\begin{gathered} \mathrm{Normal} \mathrm{reaction}={\mathrm{m}}_1\mathrm{g}+{\mathrm{m}}_2\mathrm{g}+{\mathrm{m}}_{\mathrm{plank}}\mathrm{g} \\ =400+300+100 \\ =800 \mathrm{N} \end{gathered}$$