Blog

Wait a moment...

I’ve just had a very interesting discussion with Hollie. It got me thinking so much that despite my initial aim of going to sleep at midnight, I just couldn’t resist crunching the numbers for it and blogging. 

It came from a challenge presented to the Bianco riggers. The challenge was to weigh one of the towers so it could be hung from the roof like so. They have a giant version of the newton meter you drag shoes around with in science lessons. There is however, still the task of generating enough force to get it off the ground using just four people. 

This lead me to the thought of how you could do it using levers to reduce the work which I explained to Hollie. It looked something like this. image

So the tower would be counterweight from one end, lifting it just an inch off the floor allowing it to pivot around its opposite end. This gives, by balancing moments;

mg(L/2) = FL      

where L is the length of the tower, m is the mass of the tower and F is the force shown on the scales in Newtons. The L’s cancel and shuffling thing around gives;

m=(2*F)g

This means just a single 82kg me could get the end of the tower off the floor to be measured as long as it was less then 164kg. 

From there, Hollie pointed out that the one inch off the floor would create error in the calculation. At the time I agreed, but I wasn’t worried (as sinθ is a good approximation of θ for small angles). Meanwhile, Hollie suggested lifting the tower to a pre arranged angle (30 degrees for example) and then adjusting the calculation accordingly and working from there. This however sounded like more algebra then necessarily and involves lifting the heavy tower quite far. 

My solution was to just lift the fulcrum an inch to make the tower level and then you know everything is hunky dorey. 

The thing that started to worry me was when I went back to have a closer look at my original approach. I wanted to see how big the error caused by the 1 inch lift would be. It was then I noticed that in theory no error at all would occur, the L’s canceled completely and so the horizontal distance between the lifting point and the fulcrum did not occur in the final calculation. Because moments are just perpendicular-distance multiplied by mass, this means that you should get the same value on the scales if the tower is held at say 45 degrees as when you hold it level to the horizon. This felt instinctively wrong. I mean, if you were to lift a long pole up from one end and gradually stand it up it gets easier to lift as it get more upright yeah? Here I was making a mistake and mixing up forces acting in different directions. There are two ways of lifting a pivoting object and usually we use a combination of the two. The two ways are marking belowimage

We usually use the upwards arrow early on and then gradually shift to the sideways one and while we could in theory use the sideways force very early on there is one thing that real life throws at us…Slipping, if you just push sideways the bottom will generally move away from you, not good. 

The essence here is that a solid object inclined and supported in the vertical direction only by two surfaces of different heights will apply the same force to both surfaces (even if one is the floor and one is the end of a rope.)

So, theory says that F will always read the same no mater what angle the tower is at, as long as the pulley is always directly above the point you are picking it up from. This however still doesn’t sit with me quite right, imagine the tower if it lifting slowly in this way, eventually the the tower could end up balanced on it’s end, and then all the scales will be weighing is the weight of the person attached to the other end of the rope…wait a moment, when did that happen?

I guess I need to sleep on it, night.

ps I’ll let you know how heavy they turned out to be.