posted by Patrick .
The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is uniformly distributed. The child on the left end has a mass of 14 kg and is a distance of 1.4 m from the pivot point while a second child of mass 39 kg stands a distance L from the pivot point, keeping the seesaw at rest. Find L the distance of the child on the right.
so, the mass of the seesaw can be ignored, since it is uniformly distributed. Now for the loads and their moments. They must balance on both sides, so
14*1.4 = 39*L
Actually, since I have no diagram, I can't say that the seesaw mass may be ignored. If the pivot point is not at the center, but rather at some arbitrary distance d from the left), then we have to allow for the different mass of the seesaw needing to be balanced. In that case,
20(d/4.5)+14*1.4 = 20(1 - d/4.5) + 39L
Oops. I just gave the mass of the seesaw on each side. That must be multiplied by the distance of its center of mass:
20(d/4.5)(d/2)+14*1.4 = 20(1 - d/4.5)((4.5-d)/2) + 39L
Take moment about the left end
total mass = 20 + 14 + 39 = 73 kg
That is force up from pivot
(I am going to do force in kilograms because g will cancel everywhere.)
pivot distance from left = 1.4
Distance left to 39 kg = x
(our L is (x -1.4) )
moments about left end
14 * 0
- 73 * 1.4 up from piviot
+ 20 * 2.25 mass of plank halfway along
+ 39 * x
39 x + 45 = 102
x = 1.47
so L = .07 from the pivot
In other words the little kid just about balanced the plank.