Can the law of the lever be derived from only Newton's Three Laws of Motion? Can the law of the lever be derived from only Newton's Three Laws of Motion?    
 A: The lever law is just a result from newton's laws in the angular form.
Let $\vec{F} = F_\perp + F_\parallel$ be the force acting upon a point-like particle and $\vec{R}$ the position of its rotation axis, $F_\perp$ being the force component perpendicular to the axis and $F_\parallel$ being the parallel one.
Taking the vector product on both sides we got
\begin{equation}
\vec{R} \times \vec{F} = \vec{R} \times (F_\perp + F_\parallel) = \vec{R} \times \frac{d\vec{p}_\perp}{dt} = \frac{d}{dt}(\vec{R} \times \vec{p}) \quad \Rightarrow \quad \tau = \frac{dL}{dt}
\end{equation}
Where $\tau$ is the resulting torque and $L$ the angular momentum. The Law of levers arises when the total torque is zero. Then 
\begin{equation}
\tau = 0 \quad \Rightarrow \quad \frac{dL_1}{dt} + \frac{dL_2}{dt} = 0 \quad \Rightarrow \quad \vec{R_1} \times \vec{F_1} = \vec{R_2} \times \vec{F_2}
\end{equation}
or, as usually 
\begin{equation}
R_1F_1\sin{\alpha_1}=R_2F_2\sin{\alpha_2}
\end{equation}
A: The statement "If the lever's at equilibrium, there must be zero net torque on it" is the Law of the Lever.
Write down the equation for the torque of masses pushing on a lever and it's algebraically equivalent to the "law of the lever". So that's a proof.
Suppose. I have two masses $A$, and $B$,  so 
masses $A$ and $B$ balance if
$$
m_A x_A = m_B x_B
\tag{1}
$$
where $m$ is the mass and $x$ the distance to the pivot.
You could prove it via conservation of angular momentum about the pivot of the lever. Multiplying $(1)$ by $g$, we see that the torque about the pivot is $0$, hence there is no change in angular momentum; the balance beam remains at rest.
A: You can derive the lever law just by using a force diagram and Newton first law, there's a proof of this in David Morin's book Introduction to classical mechanics. P.27. :)
