Question about levers Suppose we have a lever $AB$ of length $1m$ and that the fixed point is $O$ and $AO=0,3m$. If in $B$ I have a down vertical force of $F_1=5N$, find the force $F_2$ applied in A that guarantees the equilibrium. Say $R$ is the constraint reaction. 
If I choose the down vertical $F_2=11,7N$ we have that the total moment calculated in $O$ is $0$. But to have equilibrium I have to verify that $\vec R+\vec F_1+\vec F_2=0$. How can I do? I mean do you confirm me that it is obvious by empiric way? I mean, do you confirm that the experiments (and so the observations) guarantee to me that $R=F_1+F_2$ and that there is no way to deduce it by maths?
 A: It is implicitly assumed that the lever is not accelerating i.e. it remains where it is on your worktop or wherever you left it. Since the acceleration of the lever is zero that means the net force on the lever must be zero. Assuming we are ignoring the weight of the lever, that means there are only three forces acting in it, $F_A$, $F_B$ and $R$, and therefore these three forces must sum to zero.
A: In a problems like this, it is assumed that the lever is not moving, and the moment is 0, meaning, it has no tendency to rotate. And also, $R = F_1 + F_2$ means that the lever is not accelerating because the sum of the forces is 0. Now suppose you let $R \neq F_1 + F_2$. The lever will accelerate either upward or downward, and the lever might also rotate (if it has mass, otherwise, if it has no mass [which also is usually assumed in these problems], it makes no sense to accelerate it). But it is usually assumed that $O$ is a fixed point. Therefore, whatever force on the left and right you apply on the lever, it will not move the fixed point, because at that fixed point is the point where the restraining force (which just balances those forces) is exerted on the lever.
 Now suppose you now have a balanced restraining force $R = F_1 + F_2$, and you have solved the problem for 0 moment and have known $F_1$, $F_2$, $r_1$ and $r_2$. But then, you move the position of the fixed point to somewhere (perhaps still in between, but not the same as before). You still have the balanced $R = F_1 + F_2$ with the restraining force balancing the forces at the location of the new point. But you will not anymore satisfy the 0 moment requirement, because the $r_1$ and $r_2$ will change, and now has the tendency to rotate either clockwise or counterclockwise, even spinning. But a fixed point is a fixed point, with the same restraining force at that point, regardless of how the lever spins, as long as  you apply the same $F_1$ and $F_2$.
A: You can confirm this with an experiment using a meter stick, a sharp edge, and a weighing scale, and two objects with known weight. The weighing scale is basically just measuring the restraining force that it applies on any object that you put on it. Find the center of mass of the meterstick by placing it on a sharp edge horizontally and determining the point at which it balances, and mark it.Place it on the weighing scale then check the weight. Then balance the two objects by placing them on opposite sides. The weight of two objects acts like $F_1$ and $F_2$ with distance $r_1$ and $r_2$ from center. Check the weight again. The increase in weight is due to the addition of restraining force acting on the 2 objects. Compare it with the weight with all of them just placed on the weighing scale without balancing the 2 objects (just put them on the scale). You should see the same weight as before. If so, then the restraining force when the objects' weights (forces) are balanced is equal to the weights (forces) of two objects(its weight when you just placed them there). (Note that the meterstick and sharp edge should remain the same weight because you just placed them there)
A: In response to the comment from my previous answer:
True. The Center of Mass is now Exactly at the fixed point where the moment is balanced, which was 'caused' because you balanced the moment. 
And $F_{tot}^{ext} = m_{tot}a_{CM} = R + W_1 + W_2$
So $F_{tot}^{ext} = R+m_1g + m_2g = R +(m_1+m_2)g = R + m_{tot}g = 0 = m_{tot}a_{cm}$.
But remember that balancing the moment about the center of mass does not necessarily mean that $a_{cm}$ is 0. If R is 0, then it can mean $m_{tot}a_{cm} = R + m_{tot}g = m_{tot}g$
where now, $a_{cm}=g$. The center of mass will now be accelerating downward, and now, the two masses are also accelerating downward.
But it does not change the fact that the moment is zero, since the acceleration of the two bodies is the same g, and as they fall downward, they are at the same y position as the other, so no mass is left behind means there is no rotation and the relative positions with respect to each other remain the same. In fact, by this time, they would not exert a force on the lever, cause everything is 'falling'. Balancing the moment:
 $F_1d_1-F_2d_2 = 0d_1 -0d_2 = 0$ all the same.
Now, Suppose R is not zero, but $\vec{R} +  \vec{W}_1 + \vec{W}_2 \neq 0$.
The Center of Mass will be accelerating upward or slowly downward, But the force R does not have any effect to make the moment $\neq$ 0 since you are directly applying it toward the center of mass (the point where the moment is balanced) and the distance of the force from that point is 0.
 But the center of mass is accelerating, so do the two masses ($a_{cm} = a_{masses}$), which will now be accelerating at equal rate, since there is no rotation. Taking the sum of forces on each mass:
$\sum \vec{F}_{1} = \vec{W}_1 + \vec{F}_{1L} = m_1\vec{a}$,
$\sum \vec{F}_{2} = \vec{W}_2 + \vec{F}_{2L} = m_2\vec{a}$
 where $F_{1L}$ and $F_{2L}$ are the forces due to the lever lifting the each respective mass, and actually equal to the force the objects exert on the lever (law of action and reaction)
$\vec{F}_{1L} = m_1\vec{a} - \vec{W}_1 = m_1\vec{a}-m_1\vec{g} = m_1(\vec{a}-\vec{g})$
$\vec{F}_{2L} = m_1\vec{a} - \vec{W}_2 = m_2\vec{a}-m_2\vec{g} = m_2(\vec{a}-\vec{g})$
Converting to components along y and taking $\vec{g}$ to be a downward force, so it is negative:
$F_{1L} = m_1(a-(-g)) = m_1(g+a)$
$F_{2L} = m_2(a-(-g)) = m_2(g+a)$
$a$ can be positive or negative depending on direction of acceleration.
These are now the new forces exerted on the lever by the masses.
Checking the new moment and balancing to make it 0:
$F_{1L}d_1 - F_{2L}d_2 = m_1(g+a)d_1 - m_2(g + a)d_2 = 0 $  [1]
dividing by constant (g + a):
$m_1d_1 - m_2d_2 = 0$
comparing with the original equation:
$m_1gd_1-m_2gd_2 = 0$ [2] and dividing by constant g:
$m_1d_1 - m_2d_2 = 0$
Produces the same result, which means that the equations [1] and [2] are equivalent, meaning the 0 moment is maintained regardless of the addition of acceleration. (You can actually check this in problems with one distance unknown, and increasing or decreasing the g in the weights by the same amount, and you'll get the same result for the unknown distance)
So, a balanced moment does not guarantee a balanced Net force. But in your equation:
$m_1gd_1-m_2gd_2 = 0$
You have actually put the forces on the lever as equal to the weights of the two masses $m_1g$ and $m_2g$ which means you are assuming a non-accelerating system with R just balancing the forces. Otherwise, the forces on the lever will be greater or less than the original weights (due to the addition of a) because of the acceleration due to unbalanced R. Lesser or 0 when accelerating downward (not exceeding the acceleration g), and greater when accelerating upward.
Also note (as you might have observed) that on a balanced moment, the center of mass is always located on the fixed point. This is essentially the reason why we are balancing objects, to make the position of the center of mass to be located at the fixed point, otherwise there would be an unbalanced moment about the fixed point (because the center of mass is not there, and is similar to just putting a mass equivalent to $m_{cm}$ at the that position, without any other mass to balance it. Also, putting a single point mass on the fixed point will automatically mean that there is 0 moment [This is very hard in real life, try balancing a stick standing on your hand]). 
Also note that any two masses without any restraining force will mean 0 moment about the center of mass, regardless of how they are far apart (example would be dropping a cannon ball and wood from the tower pisa).
