# How to calculate active load on a swingset? [closed]

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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.

• The tension exerted by a swinging mass at the bottom of it's swing tends to be around $3$ times it's weight. – EuYu Oct 17 '12 at 14:26
• Why is this off topic? Solving mathematical puzzles is on topic. This was a mathematical puzzle related to calculating active load. – The Evil Greebo Oct 18 '12 at 13:06

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 $3mg$.

• g being 9.81m^2 ? – The Evil Greebo Oct 17 '12 at 14:35
• @TheEvilGreebo: Right. – Ross Millikan Oct 17 '12 at 14:35
• That's right when the swinger isn't pumping her legs to speed up. When she is, that muscular energy is being driven into the swing system and gives an additional force. Rather than find a formula for that, I'd suggest just taking EuYu's formula and multiply by a safety factor of 2 or so en.wikipedia.org/wiki/Factor_of_safety – David E Speyer Oct 17 '12 at 14:36
• So 1 swing at 450 lbs will have a force of 3 x 450 x 9.81 x 9.81 = 129,918.735 lbs? That can't be right - there isn't a beam big enough for that weight. ;) – The Evil Greebo Oct 17 '12 at 14:36
• There's quite a few factors which have not been regarded here. For example, if you swing high enough then the chains will slack and I suspect that during the drop downwards forces will instantaneously surpass $3mg$ by quite a margin. I would definitely double or even triple the factor for safety. – EuYu Oct 17 '12 at 14:38

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.

• @The Evil Greebo: To get $v$, you can think about how high people swing and convert the potential energy $mgh$ into kinetic energy $\frac 12mv^2$ as EuYu has done. – Ross Millikan Oct 17 '12 at 14:33
• I get much of that but --- Angular frequency? (Bear in mind you're talking to a builder here not a mathemetician ;) ) – The Evil Greebo Oct 17 '12 at 14:34
• 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. – David E Speyer Oct 17 '12 at 14:39