Author Topic: Possible to calculate pull force when only angles and weight are known  (Read 544 times)

Knutern

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Hi.

This is not strictly among knots, but I hope it's still within the scope on this forum.

The problem is this (theoretical):

List of what is known and what is not.
* A rope is hanging between two poles.
* The angle of the slope from the horisont at attachment point is known for both ends of the rope.
* The end points of the rope does not necessarily hang in equally distance from ground.
* The total weight on the rope and any potentially attached cargo is known.
- The centre of mass location is not known.

Then the question: Do we have enough information to calculate the pulling force of the rope at both poles.

I know that if only one of the forces are known in addition to the angles, the rest of the forces will be easy to calculate.

What do you think guys ?
I'm aiming for knots that is secure, AND that is easy to untie.

roo

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Hi.

This is not strictly among knots, but I hope it's still within the scope on this forum.

The problem is this (theoretical):

List of what is known and what is not.
* A rope is hanging between two poles.
* The angle of the slope from the horisont at attachment point is known for both ends of the rope.
* The end points of the rope does not necessarily hang in equally distance from ground.
* The total weight on the rope and any potentially attached cargo is known.
- The centre of mass location is not known.

Then the question: Do we have enough information to calculate the pulling force of the rope at both poles.

I know that if only one of the forces are known in addition to the angles, the rest of the forces will be easy to calculate.

What do you think guys ?
I think you may have enough info.  You have two unknown forces.  All you need is two equations to solve.

First equation:  Sum of forces in Y (vertical) must be zero for static equilibrium.
Second equation:  Sum of forces in X (horizontal) must be zero for static equilibrium.

Use sine & cosine of angles for vertical and horizontal components of the reactions as needed.
« Last Edit: May 18, 2020, 10:09:12 PM by roo »
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KC

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Very True.
To start, If both sides equal;
Need the support of the load thru 2 legs, so load divided by 2 to get support needed per leg/support column used
>>Then divide needed support per leg by cosine of/as efficiency to target work
>>this gives the line tension necessary for that support at that cosine efficiency per leg of support
>>as example, can logically take that same line tension and multiply it by the same cosine and get back to same support per leg
.
Just as can take that same line tension now and multiply by sine and get the side force per leg.
>>Shortcut = load divided by 2 x tangent of angle gives same number, from load down to 'tangent' side force quicker
>>because tangent itself is sine/cosine, so filters the cosine factor out, and leaves the sine/side force visible
.
If unequal angles, from rope height on poles, would not expect a center sit of load on rope
>>but if free ranging/on pulley to self adjust would expect load to sit between 2 equally opposing side forces in this float sideways
So, if not jammed against 1 end pole from that end so low
>>would expect equal line tension angle side force to either side of pulley serving it to opposite side
>> so pulley goes to center as like bubble level, between the competing forces
>> if equal line tension, then equal angles to conjure equal force from each side of pulley, inviting wander neither way/locked.
So now back to original problem of equal supports from 2 sides, just not rope center
.
If each leg of support points to 3mins after (and before) column deflections measured from 12noon pure inline position
>>18 degrees at 6 degrees per clock minute (360degrees/60mins = 6 degrees per min.)
cosine% = ~100-2-3=95% of available line tension (and length) skip -1, then use -2-3
sine%    = ~  30% of available line tension (and length) @ about 10% sine per minute in this range
  link
The sine force will pull on pole at angle to a leveraged position as measured by length from ground as pivot
>>Before 45degres 7.5mins on clock the changes in sine are most emphatic, especially at extreme from 0 sine at noon
>>in this range cosine change is most nominal change
>>this reverses after 45degrees to more nominal changes in sine and more dramatic change in cosine
>>Thus, the most dramatic change in either is coming up  from it's Zer0 value, Sine at noon towards 45degrees
>>and cosine Zer0 value at 90degrees/3o'clcock/15mins then moving towards same 45degree center
« Last Edit: May 19, 2020, 11:05:00 AM by KC »
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KC

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These numbers show how much respect (and fear) to give produced forces and their directions
Cosine/Sine TOTALLY RULE AND DICTATE, all physical displacements of space and force thru devices as mediums of transfer
(even in sound wave forms thru air as medium of transfer, electric wave form thru wire as medium, light,thermal etc. ALL WAVE FORMS)
The Ancients somehow defined that there is no other deeper pivotal concept/'common denominator' to drill to know these forces, as they span out to all.
.
Examples:  100# Load/2 support legs = 50#support needed from each support leg;
if 2 straight legs cos:100%, sine:0
>>50# needed divided by cosine 100%(1) = 50# line tension
>>50# line tension x0%sin=0 side force
>>just as 50# line tension x 100%cos = 50# support column force against load (each side)
If each support leg deflects 30? degrees (60degree spread total) cos:86%, sine:50%
>>50# needed divided by cosine 86% = ~58# line tension
>>58# line tension x50%sin=29# side force
>>just as 58# line tension x 86%cos = 50# support column force against load (each side)
This could just as easily be a trailer nose pull geometry of same forces
.
(http://mytreelessons.com/images/cosine-support-against-load-vs-sine-across-that-must-also-be-carried.png)

120? spread both legs gives 60o deflection from load force center axis, to each support axis/side
This 60? of support is very easy, memory-able benchmark: cos=.50%, sin=86%
So, 2 legs of support at 50% efficiency each, GIVES LINE TENSION = LOAD
Therefore: any more vertical angle, line tension is less than load vs an more flatter gives line tension greater than load
As more vertical increases cosine contribution from line tension etc.
NOTE: sine at 60? is greater than cosine, and the sine softens as more veritcal rope column given
.
In clock metaphor , the rope angle is past halfway point of 45?
so counting backwards towards center from 3o'clock/15mins. to 2o'clock/10mins (5 mins)
is 50% cosine, and sine=100-2-3-4-5=86%
(just as reversed at 1 o'clock cosine =100-2-3-4-5=86% and sine =50% last example before pic)
.
Just as support pull upward is a reverse of  downward direction force on load axis,
going from 1o'clock angle to 2o'clock angle is a reverse/swap of mechanical pivotal (cos/sin)forces
« Last Edit: May 24, 2020, 10:42:36 AM by KC »
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We now return you to the safety of normal thinking peoples.
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struktor

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The effect of raising point B to C.

Knutern

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Re: Calculated for actual pull force
« Reply #5 on: October 17, 2020, 12:33:49 PM »
Thank you for input's. I've figured out to solve for vertical forces (on left side).



The symbols are:
F_0 Sum of weight on the rope. Can be the rope itself or any load hang on it.
F_A and F_B : The pull forces on the rope for either side.
f_av and f_bv : Vertical forces. The total weight of rope or load must be the sum of those.
f_ha and f_hb : Horisontal forces. Those need to be of same magnitude (i.e. no wind or other external forces)
symbol 'alpha' and 'beta' : The attack angle, measured from horisontal plane.

The plan
This problem gives two equations with two unknowns (F_A and F_B, the horisontal and vertical components are used for problem solving only) and so I need to get a strategy for setting up those.

Also, for thinking sin and cos, I had to draw a help figure because the main figure is really up-side-down to what I'm used to.

At about middle , I got to the equations '1' and '2' needed to go further. From there (middle bottom) I used equation '1' and resolved F_B from '2'.


If you want to calculate F_B, simply swap side (swap all a and b's) or use simple geometry to solve for F_B (you have angle alpha and F_A and divide F_a to cos alpha to find horisontal force, then use same method in reverse for the other side).

Btw : If you find obvious errors, lease let me know.
I'm aiming for knots that is secure, AND that is easy to untie.

DDK

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Re: Possible to calculate pull force when only angles and weight are known
« Reply #6 on: October 17, 2020, 06:46:06 PM »
I came up with the same results as Knutern.  I think these results are relevant for a load secured at a fixed position on a "weightless" rope.

The equations for the sag in a rope due to its own weight can be found here  https://en.wikipedia.org/wiki/Catenary  and requires calculus to derive.

In the picture posted by struktor for a load hung by a "frictionless" pulley on a "weightless" rope, note the equal angles and how the height of the load is increased/decreased by half the distance that one of the securing points is raised/lowered (for a constant length rope), that is, a mechanical advantage of 2.

DDK