How to calculate the normal force exerted by a fulcrum off-center? So I know that $F_N$ exerted by a fulcrum right under the centre of mass for, say, a long horizontal rod, is equal to $mg$. But what if the fulcrum is off the centre of mass either to the right or the left -- how would you calculate the force exerted by the fulcrum then? I guess since $\Sigma{F_y}$ stays the same, the normal force would also be the same. But doesn't some of the energy of these forces go into inducing torque about the fulcrum instead? I'm lost.
 A: Be careful - you are applying the reasoning about a static situation to a situation that is not static.  This rod will be rotationally and linearly accelerating, so you can no longer assume the net force or net torque is zero - $\Sigma F_y = m a_y$, and $a_y \ne 0$.
If the rod is instantaneously horizontal and at rest, it's not too hard to find $F_n$ at that moment:
$$\Sigma F_y = M a_y$$
$$F_N - M g = M a$$
We can get more information about the rod by using the rotational version of Newton's 2nd law:
$$\Sigma \tau = I \ddot \theta$$
Where $\tau$ is a torque, $I$ is the moment of inertia of the rod, and $\theta$ is the angle between the rod and the horizontal.  Let $l$ be the distance between the center of mass and the fulcrum:
$$M g l = I \ddot \theta$$
We can find a geometrical relationship between the angular and linear accelerations:
$$y_{CM} = - l sin(\theta), ~~x_{CM} = l cos(\theta)$$
Using the simplifying fact that we are only concerned with the instant when the rod is at rest and horizontal, we can differentiate with respect to time (what follows is only valid at that instant):
$$\ddot y_{CM} = a_y = - l cos(\theta) \ddot \theta + l sin(\theta) \dot \theta^2$$
$$\ddot y_{CM} = a_x = -l sin(\theta) \ddot \theta - l cos(\theta) \dot \theta^2$$
Since $\theta = 0$ and $\dot \theta = 0$ at this instant,
$$a_y = - l \ddot \theta, ~~a_x = 0$$
$$\ddot \theta = - \frac{a_y}{l}$$
Substituting into the rotational version of Newton's 2nd law:
$$M g l = - I \frac{a_y}{l}$$
$$a_y = - \frac{M g l^2}{I}$$
Substituting into the linear Newton's 2nd law:
$$F_n - M g = - \frac{M^2 g l^2}{I}$$
Substituting the moment of inertia about the fulcrum (which can be obtained with the parallel axis theorem) $I = \frac{1}{12} M L^2 + M l^2$$
$$F_n - M g = - \frac{M^2 g l^2}{\frac{1}{12} M L^2 + M l^2}$$
Simplifying and solving for $F_n$...
$$F_n = M g \left(1 - \frac{l^2}{\frac{1}{12} L^2 + l^2}\right)$$
To sanity-check this answer, we can return to the balanced case where the center of mass is directly over the fulcrum ($l = 0$), which gives us the familiar result $F_n = M g$.  Additionally, as $L \rightarrow \infty$, the moment of inertia increases, which should cause the rod to rotate more slowly, and indeed, $F_n$ approaches its value for the static situation in that limit.
