Given that the mass of the boy is Mb= 36 Kg
mass of the girl is Mg = 27 Kg
The distance between the pivot and the boy is d = 2.1 m
---------------------------------------------------------------------------------
The free body diagram as follows
if the system is in equlibrium then the summation of all forces along is y direction is zero
ΣFy = 0
Normal force - M*g - Mb*g - Mg*g = 0
Fn = M*g + Mb*g + Mg*g ------------------------ (1)
(a) if the seesaw is in equlibirum then the torque acting on it is zero .
Σ τ = 0
(M*g)*(0) + Mb*g(2.1m) - Mg*g (x) = 0
Mb*g(2.1m) = Mg*g (x)
x = Mb(2.1m) / Mg
= --------- m
--------------------------------------------------------------------
(b) if the mass of the seesaw is M = 20 Kg
then from the equation (1) we get Fn = --------------------- N