/Of course, all those movies and shows are factual, as far as the science goes.

]]>penguinchris: Years ago, I had a conversation with a technician/repairman (ok, ok, my boss), who, faced with a wobbly chain drive mechanism, proposed to increase the chain tension.

I explained, twice, that the force required would break the chain. He didn’t understand what I saying.

I ended up fixing the problem by straightening the rail/channel that the chain was travelling in, not by increasing the tension.

]]>If they put any thought into it at all, I think they probably didn’t assume it was just a pile of bricks without any additional support. It’s very intuitive that a pile of something no matter how artfully arranged is easily toppled and I’m guessing they knew that. But most column supports are not built that way these days – I might have thought the bricks were a facade around a rebar’d concrete column or even just a metal pole, for example.

]]>Most people fixate on the conversion factors, and forget that US/Imperial is a force-distance-time system where mass is measured in grains… whereas Metric is a mass-distance-time system where force is measured in newtons.

That’s the big deal with metric… whole classes of engineering calculations are easier because you don’t have to divide by 32.2 ft/s^2 to get to mass.

I started drawing a poor-quality gif of the force triangle in the case you describe. Where theta is the angle off level, the tension in the rope is an inverse of sin(theta) and for small theta the tension actually goes to infinity.

]]>Anyone remember the Pythagorean Triangles? The forces on the line are calculated just like the sides of a right-angle triangle.

If the line is bent 180 degrees (that is, straight up and down around the weight, then the force on the line is exactly the weight of the object.

Think of the upside-down triangle that the line and weight forms: the base is size zero, the height is the weight of the object suspended. The force on the line is the square root of (the square of the height plus the square of half the base) – just like right angle triangles. In other words, the force on the line is the weight of the object, or more accurately, each half of the line supports half the weight of the object.

Stretch that line taut enough to keep it almost p;erfectly straight no matter how small a weight is in the middle, you need an near-infinite amount of force.

So, let’s assume the kid on the line weighs 50kgs, the line is ten meters long, and at the midpoint, the kid on the line pulls is down half a meter.

So, the height (half a meter) is equal to 50kgs. Half the length (5 meters). Plugging these into formula, the resultant force on the line is the square root of the sum of the squares of 0.5m and 5m, or the square root of 0.25 plus 25, or 5.025.

Since the half-meter deflection, at the point where the kid is, is caused by his 50kg weight, then the force on the line is, or rather that force that is pulling on the brick column is…(cross multiplications, anyone?)…**251kg!**

Likewise for loading the column: the force required to keep the column unbent, especially given a very conservative 250kg of pull (the material does have some force-absorbing flex) would be greater than the crushing resistance of the bricks.

]]>You still need the loads and weights to work out, and a horizontal anchor to the top of the now-rigid brick column would still be a really good idea.

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