Calculating the acceleration of an object on the end of a lever. I have a lever that is 16 feat long and pivots on a fixed point 4 ft from the left/heavy end, and 12ft from the right/light end.  I wanted to calculate the acceleration of the right end due to gravity.  The right end ways 10 lbs.  The left end weighs 1000 lbs.  
I calculated the acceleration, but I am skeptical of my accuracy, and was wondering if I did it correctly.  What I did was this:
I calculated the force on each side(unit: lb*ft/s^2):
 - F = 1000*32ft/s^2 = 32000
 - (force) = (mass, 1000lbs) * acceleration due to gravity)
 - F = 70*32ft/s^2 = 2240
 - (70 because I am accounting for the weight of the beam, which overall is about 90lbs)
Because of the mechanical advantage of the right side, I multiplied the force of the left side by .3 (3.5ft/11.5ft) (the weight would really be attached approximately 6 inches from the end of the board), which gave me 9600lb*ft/s^2.
Then, since these forces are on opposite ends of the pivot point, they counter each other, so, I subtracted the answers, and got 9600-2240 = 7360lb*ft/s^2. So since I want acceleration, which is measured in ft/s^2, I divided by the mass of the light end. 7360/70 = 105.14286 ft/s^2.  Did I reach the answer correctly?  It's been a while since I took physics :). 
Note: I am aware I did have a few inaccuracies in rounding, and not accounting for the mass of the beam on the left side, I am more concerned that in principal, I used valid math.
Also, once I have the acceleration, if I wanted to figure out how fast it would be going in say... 3 seconds, I would just do: 105*3^2 = 105*9 = 945?(Which obvoiusly doesn't acc
 A: First of all, you were given the weights, which already are forces; you do not multiply by acc due to gravity.  Anyway, the following is what I think is a standard method of attack in physics:
The net torque $\tau$ on the lever is $(1000) (4) - (10) (12) = 3880 \,\text{ft} \,\text{lb}$.  We find the angular acceleration by dividing the torque by the moment of inertia $I$, which is split into two components: $I_w$ (due to the weights on the ends), and $I_b$ (the uniform load)
$$I_w = [(1000) (4^2) + (10) (12^2)]/32.2 \approx 541.6 \, \text{ft}\, \text{lb} \, \text{sec}^2 $$
$$I_b = \left[\frac{1}{12} (97) (16^2) + (97) (4^2)\right]/32.2 \approx 112.5 \, \text{ft}\, \text{lb} \, \text{sec}^2$$
(The second term is an addition due to the parallel axis theorem; the lever has an off-center fulcrum.  Also, the factor of $32.2$ is acc due to gravity.)  The angular acceleration of the lever is then
$$\alpha = \frac{\tau}{I} \approx 5.93/\text{sec}^2 $$
The linear acceleration of the right end is then the length of the right arm times the angular acceleration, or about $(5.93)(12) \approx 71.2 \,\text{ft}/\text{sec}^2 $.
