Why does a teeter-totter (see-saw) tend to incline towards the heavier end? Why does a teeter-totter (see-saw) tend to incline towards the heavier end? Since objects of different mass tend to fall at the same speed (assuming a vacuum), why do then heavier objects push harder on a scale? Why are they heavier? Even though they have different masses, their "resistance" against Earth's pull is proportional. Right?
This seems like a very basic question.
 A: The reason that objects tend to fall at the same rate is because on earth we can approximate the force of gravity to be $F=mg$.  In this case, the force is really $$|F|=G\frac{Mm}{r^2}$$ where $r$ is the distance to the center of the earth, $M$ is the mass of the earth, and $G$ is the gravitational constant.  Since these things tend to be the same near the earths surface, we approximate $g = \frac{GM}{r^2}$.  This we can relate to $F=ma$ (Newton's Laws), and we can infer that for objects of any mass, the acceleration is the same, at $a=g$.
Now, to the point about the see-saw.  The accelerations are the same, but the forces are not the same, since they are also proportional to the mass, $m$.  Where a see-saw balances at is dependent on the torque, which is given by $\tau = \vec{r}\times\vec{F}$, where $\vec{r}$ is the distance from the pivot point.  Now, for masses the same distance away from the pivot point, the torque is greater for the greater force.  For objects under the influence of gravity, the greater force belongs to the object with greater mass.  Therefore, the see-saw tends to tip toward the side with greater mass.
A: When you say "push harder" you are talking about force, which makes objects move. The manner in which they move is governed by Newton's second law,
$$F_{net}=ma,$$
where $m$ is the mass of the object and $a$ is the acceleration. Now, in a gravitational field, the weight of an object is equal to it's mass times the gravitational field of the earth, $g$. So the net force is simply $mg$. But if we put that into Newton's second law,
$$F_{net}=mg=ma\rightarrow a=g$$
So, the acceleration is constant, exactly as you expect.
The see-saw works a bit differently. A see-saw is fixed to a point in the center (I'll call that the pivot point), so it will never move (unless you put something way too heavy, and then it will break!). However, the see-saw can still rotate, and rotations are caused by torques. The torque $\tau$ due to a force $F$ acting a distance $r$ from the point of rotation is
$$\tau=rF$$
(I am assuming this see-saw is perfectly horizontal, see details below). The further the force is from the pivot, the greater the torque. 
In this case, our force is caused by gravity, so each mass is applying a force $mg$ to each end of the see-saw. If you have two masses on either side of a see-saw, they will be balanced if the two torques are equal to each other,
$$r_1m_1g=r_2m_2g$$
So if one mass is smaller $m_1<m_2$, it must be placed a farther distance away $r_1>r_2$ in order for the system to remain balanced.
Details: Actually, torque is a bit more complicated. It depends on the angle $\phi$ between the application of the force and the vector from the pivot to the place the force is acting:
$$\tau=rF\sin\phi.$$
That's why I made sure the see-saw was horizontal, because the force of gravity will be straight downwards and I don't have to worry about that angle.
EDIT: More details, so everyone understands that these answers are all the same. The actual definition of torque is indeed a cross-product, $\vec{\tau}=\vec{r}\times\vec{F}$, I'm just talking about the magnitude $\tau=rF\sin\phi$.
A: I'm not sure exactly what you are asking, but this may help: Let's say gravity causes a blob of clay to push downward on a weight scale with a force of 1 Newton. Now let's say we get another blob of clay identical to the first and add it to the scale along with the first blob of clay, so there are two blobs of clay on the scale. The downward force will be 2 Newtons because force is additive--the total force is the sum of the forces from each blob (as long as the force is in the same direction, of course). Now smoosh those two blobs of clay together so that its now one big blob and keep the big blob on the scale. You haven't changed anything about the two blobs of clay except they are stuck together. So the two blobs, now in the form of one blob, still push with 2 Newtons. Also, consider that two identical blobs released at the same time will fall at the same rate that a single blob, by itself, will fall. Now connect the blobs together with a light string and again drop them at the same time. Surely that string won't cause them to suddenly fall faster, so totally smooshing them together should also not change how fast they fall, because you can still think of them as two blobs that just happen to be connected together. 
