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
Best Answer
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 $.