Question

A 75-lb child is sitting on a teeter board at (-2.5, 3.5) and the fulcrum is at (4.5, -1.5), where units are in feet. At what point should a 37.5 lb child sit to be in equilibrium? What is the distance of the second child from the fulcrum?

Answer 1

The formulae for answering this question and similar questions are:

(x-x1)/(x2-x1)=(y-y1)/(y2-y1)=M2/(M1+M2)=a/R, and a^2=(x-x1)^2+(y-y1)^2 and R^2=(x2-x1)^2+(y2-y1)^2.

Where mass M1 is at point (x1,y1), mass M2 is at point (x2,y2), the fulcrum is at (x,y); a is M1's distance from the fulcrum; R is the distance between M1 and M2.

In this question M1=75lb, M2=37.5lb, x1=-2.5', y1=3.5', x=4.5', y=-1.5'.

(4.5+2.5)/(x2+2.5)=37.5/(75+37.5)=(-1.5-3.5)/(y2-3.5).

7*112.5=37.5x2+2.5*37.5, 787.5-93.75=37.5x2, x2=18.5'.

37.5y2-3.5*37.5=112.5(-5), 37.5y2=-431.25, y2=-11.5'. So the location of the second child is (18.5',-11.5').

The distance of the second child from the fulcrum is R-a=√(21^2+15^2)-√(7^2+5^2)=17.20' approx. Note that a/R=1/3, so R=3a and M1=2M2, so the results are as expected: the second child has to sit twice as far away from the fulcrum as the first child, because the first child is twice as heavy.