On 5/29/19 8:15 AM, charlie@thegallos.com wrote:
Hey Gang,
This isn't a snark answer - this is me trying to learn, I'm NOT an
engineer (although some call me a 'software engineer' - bah)
Where do those 2x and 4x numbers come from? I _ASSUME_ it is standard
engineering "stuff". Now I have a Machinery's Handbook, and a Marks
Manual sitting here - is there a section where I can look this up, so I
can understand it?
You analyze this in terms of the moments which are Force times distance.
The tower is 160 lb (4* 40lb/sec, and you can assume it's uniformly
distributed along the length, so it's the same as a point mass at half
the length.
The load is 50lb and at the end of the 40 ft tower.
So the moment is
160 *20 = 3200 ft lb
50 * 40 = 2000 ft lb
or 5200 ft lb.
In order to just lift it, you need to figure out what force is applied
at the 10 ft mark.
5200/10 = 520 lb (pulling straight up).
You're not pulling straight up, though, you're using a cable from a
winch attached to the garage wall.
For simplicity, let's assume the winch is at 10 ft also, so the cable is
at 45 degrees when the tower is on the ground.
The tension in the cable is 520 lb/cos(45) = 735 lb.
Now, what's the horizontal force on the garage wall? Since it's 45
degrees, it's exactly equal to the vertical force on the tower. 520 lb.
I'm assuming it is the simple ftlb model, with the ratio of the 10ft and
40 ft, but that doesn't seem to take into account any vectors on the
load too (of course that gives you a nice safety factor, always a GOOD
thing)
73 de KG2V
<snip>
if the 'hinge' is at ground level and your winch is at the 10'
level, the total force of those two loads will be:
2x the tower load PLUS
4x the 'top' load.
that's quite a bit for the position of the winch .. AND .. if the
winch is connected to the garage, it just might pull the garage
wall down.
73
Don
N8DE
<snip>
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