Posted by Patrick on Friday, March 28, 2014 at 6:25pm.
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.

Physics  Steve, Friday, March 28, 2014 at 7:26pm
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
Simple, no?
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

Physics  Steve, Friday, March 28, 2014 at 7:28pm
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)((4.5d)/2) + 39L

Physics  Damon, Friday, March 28, 2014 at 7:36pm
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
so
39 x + 45 = 102
x = 1.47
so L = .07 from the pivot
In other words the little kid just about balanced the plank.
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