It doesn't sound like your use of the word "force" is precise here. If you drop an object from 10 meters instead of 5 meters, the acceleration (and therefore the total force) will have to be higher if it stops in the same amount of time in both cases, so that's true. In the sitting case, the total force is zero, even though the force of gravity is present, nothing a. In the physical sense the mentions of "energy" here are misleading. You don't really have much conservation of energy here, because the collisions with the see-saw are inelastic and do not conserve energy. You don't have much F=ma here, because if you used some force like that you'd still have to deal with angular things. You don't even have conservation of linear momentum: the fact that the see-saw is attached to the essentially infinite mass earth ruins that, and it means that momentum is not conserved during collisions with the seesaw.
So: No energy, force (save for m*g), or linear momentum needed. Instead: Torque and angular momentum.
We need a few variables:
I0, the moment of inertia of the see-saw (it plays the same role as mass).
g, the acceleration of a free body due to gravity.
m[i], the mass of the ith particle sitting on the seesaw.
x[i], the position of the ith particle sitting on the seesaw.
q, the angle of the seesaw, with q=0 meaning it's flat.
qdt, the rate of change of the angle of the seesaw.
k, since seesaws have springs in them, "k" can be our spring constant which tries to return the seesaw to its original position.
mu, seesaws have friction and don't continue bouncing forever, so mu will be our friction variable which tries to slow the seesaw down.
So, what's the problem we want to solve. There are two things we want to do: Solve for the dynamics of the system when no one is sitting down on or jumping off the seesaw, and the instantaneous collisions, which happen when someone sits down or jumps off. The first one we can figure out using arguments about torque, and for the second we can use angular momentum.
(for a quick refresher: torque is the rate of change of angular momentum, just like how force is the rate of change of momentum. When I say "moment of inertia is like mass", that's because the equation "force=mass*acceleration" takes the form "torque=moment of inertia*angular acceleration". Likewise, "momentum=mass*velocity" is "angular momentum=moment of inertia * angular velocity")
Lets start out with the dynamics of the seesaw (that is, ignoring the people jumping on and off it). The actual moment of inertia of the seesaw with all those extra masses on it is
I=I0+m*x*x+m*x*x+.... (this is a simple consequence of the definition of moment of inertia as the sum of m*r^2 terms)
We'll denote the variable for angular acceleration as
qddt. There are only a few torques we need to take into account. As stated above, torque=moment*angular acceleration, so we know that the total torque is equal to
Torque due to the seesaw's spring: The spring constant should try to restore
q to zero, according to the spring constant k. The standard way of doing this is to add a torque of
Torque due to gravity: The torques from the
m[i]*g forces are
r x F. I won't go through the details here (because people frown on defining "cross product" for 2D vectors and I don't want to add the disclaimers), but basically this just means we tack a factor of r (=
x[i]) and sine or cosine onto the
m[i]*g force. We find the resulting torque from mass
Torque due to friction: This should act against the seesaw's velocity according to
mu. Like with the spring there are a few ways we could do it, but the easiest is to let the friction be proportional to angular velocity, so we add a torque of
The resulting equation is
I*qddt=-k*q-mu*qdt-(x*m+x*m+...)*g*cos(q), and we can divide through by
I to get the angular acceleration.
To solve for what happens during instantaneous collisions, we use conservation of angular momentum. This states that the angular momentum before a collision is the same as the angular momentum after the collision. Lets say mass
i=0 suddenly leaves the seesaw with velocity
v, at an angle
a away from the center of the seesaw. (where I take
a=0 to mean straight away from the center of the seesaw, and
a=90 degrees to be straight up off the surface of the seesaw [in the right-handed sense])
Before the collision: This is simple. With
I the same as in the last section, the angular momentum of the whole system is
After the collision: Let
I2=I-x*x*m be the moment of inertia without taking m into account. Then the angular momentum of the seesaw alone is
qdt2 is the angular velocity of the seesaw without mass 0 on it. Using the definition of angular momentum with the vector cross product
r x v, we can argue that the angular momentum of mass 0 about the center of the seesaw is
x*m*v*sin(a). So, the total angular momentum is
Putting them together, we have that
I*qdt=I2*qdt2+x*m*v*sin(a). If we solve for
qdt we can solve for what happens when a mass hits and sticks to the platform, and if we solve for
qdt2 we can figure out what happens when a mass jumps off the platform.
Summary of Equations
I=I0+m*x*x+m*x*x+... (definition of
I2=I0+m*x*x+... (definition of
I*qddt=-k*q-(x*m+x*m+...)*g*cos(q)-mu*qdt (dynamics to solve for acceleration)
I*qdt = I2*qdt2+x*m*v*sin(a) (conservation of angular momentum where mass 0 jumps off the platform and moves with velocity v away from the platform at angle a, [where a=0 implies you're heading parallel to the seesaw's platform, and a=90 degrees implies you're heading perpendicular to the seesaw's platform]. qdt is the velocity whall m is attached, qdt2 is the velocity after m is detached. Solve for qdt or qdt2 depending on which case you have.)
So, there. I always like to start from a physically-motivated model, and the terms are understandable "torque" and "angular momentum" equations, so they can be modified if you don't like the behavior! Like, if you want a different formula for pushing down when someone jumps off, you could fiddle around with that
x*m*v*sin(a) term and multiply it by some fun function, or you could apply a torque for a period of time, etc.
I wrote an uncommented application to show that the equations are valid, but there are a few issues: the velocities of the balls leaving the platform might not be right, due to factors introduced when changing between screen coordinates and