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Given a beam (a swingset) with three swings suspended from it. Each swing holds 450 lbs of stationary load for a total of 1,350 lbs of stationary weight suspended from the beam.

All three swings begin swinging in synch so that the load hits the bottom of the arc simultaneously.

How does one calculate the effective weight of the moving mass on the three swings combined?

To clarify - the need is to determine the maximum load that will be put on the beam, so that the proper sized beam can be selected to ensure it can be used without breaking. The charts that indicate maximum active load are here. I think we're looking for total uniformly distributed load (w from the tables) that will be applied.

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    Why is this o$f$f topi$c$? Solving mathematical puzzles is on topic. This was a mathematical puzzle related to calculating active load.2012-10-18

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Let $m$ be the mass of the load. I will make the assumption that the swing will never pass $90^\circ$ from the vertical. Letting $\ell$ be the length of the swing, the height difference will never pass $\ell$. From conservation of energy, we have $\frac{1}{2}mv^2 = mgh \implies v^2 = 2gh$ At the bottom of the swing, we will have $h = \ell$. From the centripetal force, we have $F_c = \frac{mv^2}{\ell} = 2mg$ This is totaled with the actual weight of the masses to give an effective force of $3mg$.

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    Read the last line - it's totaled with the weight o$f$ the masses - that means add the original weight for a total of 3mg.2012-10-18
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By "effective weight", presumably you mean the force exerted by the swings and the beam on each other? The swings experience a gravitational force $mg$ (where $m$ is the mass of the swings and $g\approx9.81\text{ms}^{-2}$ is the gravitational acceleration), which they pass on to the beam. The beam also exerts a centripetal force on the swings, which is $m\omega^2r=mv^2/r$, where $\omega$ is the swings' angular frequency and $v$ is their veocity. The two forces add constructively, so the total force exerted by the swings and the beam on each other is their sum.

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    Someone making practical measurements might prefer $\omega = 2 \pi / T$, where $T$ is the time it takes the swing to through one forward-and-back cycle.2012-10-17