# How is weight dispersed among multiple points?

I was recently viewing remembrance photos from the 1995 bombing of the Alfred P. Murrah Federal Building on MSN news where slide nine shows a photo of firemen teaming up to carry a $$40'$$ steel bar weighing $$1,000lbs$$.

This got me thinking; I've participated in log runs that simulate such feats with telephone poles, though these are usually limited to $$5$$ to $$10$$ people per log.

I have a few questions related to this topic, and I believe (probably horribly wrong though) the math would be simple for the weight felt by each person while carrying the object:

$$w = t / n$$

Where,

• $$w$$ is the weight each person feels.
• $$t$$ is the total weight of the object.
• $$n$$ is the number of people carrying the object.

For the calculations, we can assume all persons participating meet the criteria of a perfect test:

• They are all $$5'11"$$.
• They all use the same amount of strength to lift and carry.
• No one is slacking off.
• They all move in unison.

It can be assumed that all conditions that provide any type of inconsistency will never occur ($$+50$$) bounty if you can provide explanations for how impactful these inconsistencies are.

## My Questions

Out of curiosity, how do you calculate how much weight each person involved will feel when:

• Initially lifting the object.
• While carrying the object.
• What if all persons decided they should switch shoulders at the same time?
• When placing the object back to the ground.

The most useful theoretical model is probably a "Winkler Medium": imagine a rigid rod resting on row of springs. Each spring obeys Hooke's law: $$F=kx$$, where $$F$$ is the force applied by the spring, $$x$$ is the distance the spring is compressed downward, and $$k$$ is a constant for each spring, sometimes called the "stiffness". If each spring has the same constant $$k$$, then the force in each spring will be the same, as you noted above. But now let's consider the case where they aren't all the same.

A weaker (technically, less stiff) spring will have a lower value of $$k$$, so it applies less force for a given amount of displacement. To equilibrate the total weight, this force must be taken up by other springs. Similarly, a stronger (more stiff) spring will have a higher value of $$k$$, taking more force, so the others take less. But now we can calculate how much. Because if the weight is rigid, all the springs have the same value of $$x$$.

$$W = \sum_{i=1}^N F_i = \sum_{i=1}^N k_i x$$

Where $$W$$ is the total weight, we have $$N$$ springs, and we've given each spring its own force $$F_i$$ and constant $$k_i$$. Solving for $$x$$:

$$x = \frac{W}{\sum_{i=1}^N k_i}$$

And so the force in any given spring is

$$F_i = \frac{k_i W}{\sum_{i=1}^N k_i}$$

Now back to reality: people aren't springs. So what determines how much force an individual must exert? The answer is, the individual chooses it. If they push harder with their hands, or try to stand up straighter, they take more of the load, and others take less.