How do you account for the weight of the lever? I have a 6 foot lever where the fulcrum is 2 feet from the input force. In an ideal situation, this gives a mechanical (dis)advantage of 0.5. However, let's just assume there is no weighted object being lifted, but the lever itself weighs 50lbs. How do you calculate how much downward force is required to lift the other side of the lever?
 A: Find the weight of only the side you want, and apply that force halfway between the fulcrum and the edge (ie the center of gravity of that side).
A: This is how I would try to approximate it. I don't know if this is the correct method or how far it deviates from such.
Imagine that the lever has its weight evenly distributed into buckets between each foot mark.
$|-D-|-C-\uparrow=C=|=D=|=E=|=F=|$
Apply the principle of The Law of the Lever formula to each bucket $\frac{F_b}{F_a}=\frac{a}{b}$.
This is the balanced lever.
$|\small{D+\frac{5}{3}E+\frac{7}{3}F}|-C-\uparrow=C=|=D=|=E=|=F=|$
$C$ and $D$ cancel.
If $F=E=\frac{50}{6} $
Then
$\frac{5F}{3}+\frac{7F}{3}=12F=100 $
A: If lever fulcrum is displaced, i.e. it's position not in the COM of lever, then from the balance of torque between both lever sides you can calculate how much mass you need to put on the shorter lever side for balancing-out longer empty lever side. Consider such displaced lever scheme :

Then added mass $m$ and left lever side mass $\mu$ ratio is :
$$ \frac m\mu = \frac 12 \left(\frac {L^2}{\ell^2} -1 \right) $$
Where $L, \ell$ is right and left lever side lengths respectively.
A: Assuming you're considering a static, condition, you can solve the equilibrium of the moments (sum of external moments equals zero) around the fulcrum without considering inertia.
Assuming:

*

*the lever is orthogonal w.r.t. the gravity and the force applied

*mass distribution is uniform along the lever

you can:

*

*evaluate the resultant of the mass force, i.e. the weight of the whole lever (that's the point previous answers were missing), $Mg$

*apply it to its application point, i.e. the centre of mass of the lever, i.e. its middle point of the mass distribution is uniform, far $L/6$ from the fulcrum

*write the out-of-plane equilibrium of moments, that gives you $F \frac{L}{3} = Mg \frac{L}{6}$
and thus the solution
$F = \frac{1}{2}Mg = 0.5 \cdot 50 lbs \cdot 0.454 \frac{kg}{lbs} \cdot 9.8 \frac{m}{s^2} = 111.2 \, N$.
