
A 40lb child sits on a seesa, 3 ft from the fulcrum.
What torque is exerted when the child is 2 ft above the horizontal?
What is the maximum torque exerted by the child?.

Let t be the angle between the horizontal and the seesaw. The torque
T of the weight of the child at Q about the fulcrum P is,

> PQ := 3*cos(t)*i+2*j;
PQ := 3 cos(t) i + 2 j> F := 2*j;
F := 2 j
and the torque T is, 
> T := crossprod(PQ,F);
T := [0, 0, 6 cos(t)]
From the picture we have that, 
> t := arcsin(2/3);
t := arcsin(2/3)
thus, 
> T := T[3]*k;
1/2 T := 2 5 k
Clearly the maximum value of the magnitud of the torque is 6 when cos(t)=1 i.e. when t=0 (i.e. horizontal seesaw). 