Calculate the force upon the slings

dvsDave

Benevolent Dictator
Administrator
Senior Team
CB Mods
Fight Leukemia
If you lift a 1,000 pound load with two slings bridled at 30 degrees from horizontal (See Image), what is the resultant force imposed upon each of the two slings when the load is lifted? (For purposes of our QOTD ignore any additional stresses due to initially overcoming inertia and / or wind loading, etcetera)

QOTD-sling-angle-30-degree.jpg



for bonus points, show your work and formulas using the handy math function on CB.

1580275703818.png


Many thanks to @RonHebbard for coming up with the question and @egilson1 for the graphic!
 
A note for the new members of CB, new questions are reserved for students for 1 week, after that anyone can answer.
 
Hint for students: the *angles* matter.
 
And riggers will know the math for 90-degrees and 120-degree bridles off the top of their head. This is a question you would find on the ETCP Rigging exams.

Good Question of the day. I can't wait for the next one in 2025.
@DavidJones 2025?? Now, now David, we can't rush into these things; if you're becoming impatient, feel free to contact @dvsDave and / or any of the mod's with all of your proposed suggestions for CB's QOTD. (Like I've done for the last two or three.)
Toodleoo!
Ron Hebbard
 
@DavidJones 2025?? Now, now David, we can't rush into these things; if you're becoming impatient, feel free to contact @dvsDave and / or any of the mod's with all of your proposed suggestions for CB's QOTD. (Like I've done for the last two or three.)
Toodleoo!
Ron Hebbard
Let's give David the benefit of the doubt, perhaps he just meant Feb 25th? ;)
 
Think "trigonometry".
 
That's the 120 degree bridle in this illustration (because adding twice 30 degrees gets you to 180), and hence the entire 1000lb load weight bears on each line/strap.

It's indeed 1000 lb tension on each line, but to be pedantic about it only 500 lbs of the load is supported by each one; the other "extra" force ends up as compression across the top of the load. Put another way, the vector sum of all forces on the load (gravity and the tension from the two lift lines) is zero, rather by definition since this is a problem in statics where the load is not undergoing acceleration. The horizontal components of the tension forces from the two lines cancel each other out in a free body diagram/analysis of the load.
 
This is something I was never taught and haven't really needed. But now I want to know what the math is so that I could get info on something that isn't on that very handy chart Jay posted.
 
This is something I was never taught and haven't really needed. But now I want to know what the math is so that I could get info on something that isn't on that very handy chart Jay posted.
Trig. Seriously. There is a vector math approach that @DrewE alludes to and using either is acceptable when taking the ESTA rigging certification test.

I suggest a copy of Delbert Hall's "Rigging Math Made Simple." Amazon or your favorite bookmonger...
 
Rigging Handbook by Jerry Klinke is also a decent, easy to carry text full of info, math and graphics. The caveat being it is for/from the crane rigging side of things.

 
I've worked out like that for the simple beams, but when it comes to Cascaded one, I'm stuck with finding load at the upper beam.. I've to choose between either same lifting load or slightly larger one due to the beams&slings underneath the upper beam.. but hey, thank you for your input! :)
 

Users who are viewing this thread

Back