# 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 figure will help to solve the problem.So if you can upload a fig. – iostream007 Jul 19 '13 at 5:54
• Are the 1000 and 10 pound weights distributed along the length of the bar, or concentrated at the ends? – Ross Millikan Jul 19 '13 at 16:49
• @RossMillikan concentrated. The evenly distributed weight is 97 – csga5000 Jul 31 '13 at 18:24

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 $(5.93)(12) \approx 71.2 \,\text{ft}/\text{sec}^2$.

• Thanks, I had a feeling what I was doing was totally off. – csga5000 Jul 19 '13 at 16:42
• You are assuming the 10 pound and 1000 pound forces act at the ends of the bar. As they are specified as weights of the bar, maybe they should be distributed. But that makes the bar vary greatly in mass per unit length. The original problem didn't make it clear. – Ross Millikan Jul 19 '13 at 16:50
• @RossMillikan: the picture shows a uniformly distributed weight on the bar with two additional weights on the end. That is what I assume in my solution. – Ron Gordon Jul 19 '13 at 16:55
• @RonGordon You are correct. Also, if I wanted to calculate the acceleration for the other end, it would just be 5.93*4?(And would be in the other direction? – csga5000 Jul 19 '13 at 16:59
• @csga5000: yes, and yes. – Ron Gordon Jul 19 '13 at 17:00