Angles and load in Rope rescue

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Greetings; I am a firefighter so physics is pretty much Greek to me. I work primarily as a rope rescue/USAR tech. We work a lot with rope, pulley's, and anchor points. All of these systems have angles that affect the loading in our system. Now I been taught rules of thumb when constructing a rescue system, for example; I am told when using a change of direction pulley, the greater the angle the less force required to pull the load, so fine I just follow this rule and all is golden, but I always wondered why.
Anyways, I recently attended a training session where we built a Kootenay high line patient retrieval system (See image). This system is basically a long rope that extends from one anchor point, the line can span over a ravine, river, etc. The patient is placed in a litter and transferred across the line using tag lines.
One critical aspect is the slack in the mainline, too tight and the load on the anchors can cause a anchor failure. Too loose and the patient will not be retrievable once approaching the ravine edge or similar.
Now my question;
I have a rope rigging software called Vrigger that will do the force calculations. It is apparent from my simulations that the greater the angle on the line, the less force on each side of rope where the pulley is. (hope that made sense). The numbers are quite drastic, can someone explain to a firefighter why this is so, I just don't get why the angles make such a difference. I wanted to find a formula for this but did not know where to look in my old trig book.

http://img14.imageshack.us/img14/9094/ti4.gif [Broken]
 
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  • #2
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I would consider Newton's second law to analyze this situation. F=ma, but to be more technical, [tex]\sum\vec{F} = m\vec{a}[/tex]

This sum of forces equals mass times acceleration. If your load is staying put then its acceleration is zero. Thus the sum of forces should equal zero. Now its time to recall (or learn) that forces are "vectors", they have a value and a direction. Direction matters. You mentioned you have a trig book, thats good. What you need to do is use trig to "break" the force value of your rope into "components". That is, instead of considering the 613lb of force at an angle in your first picture, we can consider that as a small piece of force in the up/down direction and a large piece of force in the left/right direction. A key to analyzing this situation is being able to break a force in some angle up into its "components".

At this point I would put forth the exercise to you, lets break the rope with 613lbs up into components using trig. Note the angle. Its 161.2 total. We want to draw a right triangle on it so we can use trig. Note my picture. The new angle in my right triangle there is half the old angle. Using trig (SohCahToh) you can find the values of the up/down and left/right components. Try it.
 

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  • #3
Thanks for the quick reply. We use the term vectors a lot. Especially when using rescue tripods, our force vectors must stay within the tripod footprint or risk the tripod will topple. I am grabbing my trig book and will see if I can do this thanks.
 
  • #4
Well I thought the first thing I should do for this problem is set up my right triangle and label it properly. Not sure I did. I took the 161.2 deg angle and divided by 2 for 80.6 then subtracted (90+80.6) from 180 for my other angle of 9.4 deg.
Now where I am unclear is my angles. Is the 80.6 degree the opposite and 9.4 deg adjacent? I'm really not clear on that.

http://img822.imageshack.us/img822/9408/e9o.gif [Broken]

Thanks
 
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  • #5
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You flipped it upside down! I think we should leave it oriented the way it was, so we dont get confused. The angles look right to me. To determine which one is the opposite and which one is the adjacent we need to consider which angle we are using. We are using the 80.6 degree angle (thats our choice). The lables you have for opposite and adjacent would be flipped in this case. The short side is adjacent to the angle, so its Adj and the long side is opposite of the angle so its Opp. See my picture.

Now the big error I see you your picture is that you put 200lb on the short side. I suspect you did this because you intuitively think its the case, but its not! This is what we are going to use trig for. At this point you dont know what the force is on that small side. With my picture in mind lets consider the sine equation. Sin(angle) = opposite/hypotenuse (thats the Soh in SohCahToa) Using algebra we can multiply both sides of that equation by the hypotenuse and get the equation Hyp*Sin(angle)=Opp. I did this on my calculator, 613*Sin(81.6)=606.4. So we know the long side has 606.4lb of force, the hypotenuse has 613lb and now we need to know the short side's lb of force. Use Cosine for this. Cos(angle)=Adj/Hyp. Multiply both sides by Hyp and get the equation Hyp*Cos(angle)=Adj. Plug this into your calculator and see what you get for the force on the small Adj side. Remember that you are using degrees so your calculator should be in degree mode not radian mode.
 

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  • #6
I really appreciate the help. I had to go and watch a few you tubes on right triangles to understand how to name the adjacent and opposites. I thought they were fixed but now understand it depends on what angle I am looking at.
So I did what you suggested, sin(80.6)* opposite/613 = sin(80.6)*613= 604.7
Now when I do cos(80.6)*613= 100
Should I not get 200?

Thanks again∞)
 
  • #7
SteamKing
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I really appreciate the help. I had to go and watch a few you tubes on right triangles to understand how to name the adjacent and opposites. I thought they were fixed but now understand it depends on what angle I am looking at.
So I did what you suggested, sin(80.6)* opposite/613 = sin(80.6)*613= 604.7
Now when I do cos(80.6)*613= 100
Should I not get 200?

Thanks again∞)
Remember, each of the lines supports half of the 200-lb load.
 
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  • #8
Well heck, that makes sense, I think I get it. Did I just learn some Trigonometry? Very cool. This math stuff is kinda cool and to think even a guy like me can do it :)
Thanks all, I think I have some more basic questions if that is okay.

I would really like to take a trig class, but my schedule is pretty full. I could do a online class but learning math from books is hard for me.
 
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  • #9
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Now that you have done the steps of breaking up a vector into components using trig I would suggest another exercise. Look again at the equation for newton's second law in my first post. F=ma, but the "F" have the big symbol in front of it which means you have to do a sum. You have to add up all the forces for F, you cant just consider one. The other thing to remember is that "F" is a vector (because of the arrow on top of it). Since its a vector we need to add the forces that are going in the same (or opposite) direction. That is, in our case we want to add the forces going up/down and then in a separate calculation we want to add the forces going right/left.

Lets consider the left/right direction and lets consider the blue attachment as the body we are looking at. We need to call one direction positive and one direction negative, lets call left negative and right positive. On the blue attachment I see a red rope pulling to the left and a blue rope pulling to the right. Apparently they each contribute zero but I want to include them for completeness and to illustrate that we have to sum all forces. Then you have the left/right component of the rope we "decomposed" into components together. You calculated that it is pulling to the right with 604.7 lbs. We can do the trig again for the left rope, but we can also consider symmetry and conclude that it is pulling to the left with 604.7lbs, or more precisly -604.7lbs since we consider going to the left to be negative. Now we add all four pieces up, BlueRope+RedRope+RightSupport+LeftSupport = 0 - 0 + 604.7 - 604.7 = 0. It equals zero meaning that we have no acceleration in the left/right direction. We do the same thing for the up/down direction. This time we call down negative and up positive. The weight is 200lbs down, so -200lbs. The up/down component of our black ropes is 100lbs each, as you calculated. They are each pulling up. We add these three pieces up; Load+RightSupport+LeftSupport = -200 + 100 + 100 = 0. It also equals zero meaning we have no acceleration in the up/down direction.

To help solidify this I would run through the calculations again for your bottom picture. Decompose the two black ropes into "components" using trig. Then sum all the forces in the up/down direction and see that they do equal zero. Do the same thing for the left/right directions and see that they equal zero. Since we have a different angles we have different forces on the rope.
 
  • #10
Hi, I have another rescue system math question pertaining to angles again.

Here is the background: NFPA defines 1 and 2 person loads as being 300 lbs for 1 and 600 lbs for 2.
In rescue we have this setup called a ladder Gin, a method of creating a High point anchor point as shown in figure.

http://www.firehouse.com/forums/attachment.php?attachmentid=22773&d=1361317283

Fire ladders are rated at a working load of 750 lbs.

What I am trying to figure out is if I have a 2 person load on the ladder gin, am I exceeding the rating of the ladder.

I was gonna use my SOH CAH TOA, but the Guy line is throwing me, also I have only 1 load vector to work with which leaves 2 unknown. See image.

http://img713.imageshack.us/img713/9666/nqi.gif [Broken]

Am I correct that the short side is opposite? Will the guy line affect the calculation? The guy line is rated for 40 kilonewtons or around 8000 lbf.

Thanks
 
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  • #11
jbriggs444
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The angle between the guy wires and ladder is 70 minus 45 = 25 degrees.
Then angle between the load rope and the ladder is 90 minus 70 = 20 degrees.

If the setup is to be stable, there must be no net torque on the ladder -- the torque from the guy lines must cancel the torque from the load rope. To a back-of-the-envelope approximation the force from the load and the guy wires must be in a 5 to 4 ratio. [Bigger angle = less force for the same torque]. A better approximation would be solve for 600 sin(20 degrees) = x sin(25 degrees).

Back of the envelope says 480 pounds total from the guy lines. The better approximation says 485 pounds.

The vector sum of those two forces acts to compress the ladder. To a back of the envelope approximation, the angle is sharp so the forces simply add. A better approximation would be that the sum is 600 cos(20 degrees) + 485 cos(25 degrees).

Back of the envelope says 1080 pounds. The better approximation says 1003 pounds.
 
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  • #12
Thank you very much. From your math, 600 lbs. would exceed the ladder rating which supports what I have been told that the ladder gin does not support 2 person load.
 

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