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Q: Swing cable breaking force ( Answered 5 out of 5 stars,   7 Comments )
Subject: Swing cable breaking force
Category: Miscellaneous
Asked by: itsjustmatt-ga
List Price: $4.00
Posted: 16 Jul 2005 20:25 PDT
Expires: 15 Aug 2005 20:25 PDT
Question ID: 544391
I'm thinking of building a large swing between 2 huge pine trees on my
land. Here's the skinny: I have a large tree, I hang a 15m steel cable
from a branch. A 114kg (approx) man swings from it starting at 15m
above the ground (from another tree). What is the max tension will be
exerted the cable? I need equations too so that I can modify cable
length, etc.
Subject: Re: Swing cable breaking force
Answered By: hedgie-ga on 14 Aug 2005 23:11 PDT
Rated:5 out of 5 stars
The tensioning force is M * g ( 1 + cos (deflection))

where deflection is angle from the vertical.

This is the sum of gravity and centripetal force 

 They add or subtract, so that

 Max force is at the vertical position (cos(0) =1) as
described here in detail:

Search terms: Forces Pendulum

As myoarin-ga commented, fairly thin steel cable (1.2 mm diameter) can
support that.

BONUS : use  calculator build into google seacrh engine to get:

you enter:  cos(180 degrees) to get   
cos(180 degrees) = -1  (this is rope straight up, total tension zero 

g is acceleration of gravity (9.81 m  /s /s in 'metric') 
you enter 

1 kg * (9.81 m/s/s)   to get weight of 1 kilogram

(140 pounds) * (9.81 m/s/s)
to get
(140 pounds) * (9.81 ((m / s) / s)) = 622.963761 newtons

weight of the man

to get tension at deflection 90 (horizontal) on a swing you enter:

(140 pounds) * (9.81 m/s/s) * (1+cos( 90))
and get
(140 pounds) * (9.81 ((m / s) / s)) * (1 + cos(90)) = 343.830136 newtons

Request for Answer Clarification by itsjustmatt-ga on 15 Aug 2005 06:05 PDT
Your answer totally ignores tension on the cable due to centriptal
force. I could have figure out that max "static" tension would just be
the weight of the person. That is a no brainer. But because the person
is swinging, there is an angular velocity, therefore, there will be an
added tension due to centriptal force which can be many G's. This
force will depend on weight, initial drop height and cable length. All
that data is provided in the question.

Clarification of Answer by hedgie-ga on 16 Aug 2005 01:22 PDT

 I had said quite clearly: 
"This is the sum of gravity and centripetal force"

   In the vertical position, formula gives double of the weight.
  (gravity alone produces just the weight).

 As you correctly pointed out, one has to use his/her brain to figure
 this out. Please reread my answer AND elementary explanation
provided, in the  in such a mode, this time.
Formula is simple, giving vector sum of gravity and centripetal force, 
becouse it is a simple problem.

  If you really need clarification, pleast post another RFC
 however, considering the price, I cannot to engage in an
a prolonged argument.

Rating is always appreciated.

Clarification of Answer by hedgie-ga on 16 Aug 2005 22:59 PDT

 Thanks for the rating.
 We all sometime act on impuls, before thinking things through. 

And so here are few notes of CAUTION if you actually build it:

 1)You need to use a safety factor, not theoretical limit.
   I would guess cable should support at least 6 time the weight of the load. 
2) The formula leading to 2 g makes an assumption about the energy of
the pendulum.
 Clearly, if energy get's higher and higher,  the bob swings over the
top and
      Centrifugal force=  M * v *v /L  
 will be  large compared to
 contribution  yo the tension due to gravity= M * g *cos (deflection)  

  Of, course that is not a swing aby more.  Energy of a swing is less
then E0=M * g *L = is energy of the bob at horizontal position at

If energy of the swing is more then that, e.g. if you hold the rope
and jump of the point above the horizonal level of the pivot, you are
not restrained by
the rope at first. You will be in the free fall until your vertical
(or parabolic) trajectory intersect the circle of pendulum's path. At
that moment,
there is a jerk, which can exceed the forces we are considering here and which
depends on elasticity of the rope. It can far exceed the 3g we can get at E0
Finally, a note on terminology: We consider three forces here:
Centripetal is the force 'towards the center' - a reaction of the rope 
Centrifugal is the force of inertia, (away from the center)
  [fuga is same root as in word refugee = to run away from]
  and then, there is gravity.
We are really adding all three - and so, it is more complicated 
 then I said on the beginning, after all.

     Happy swinging
itsjustmatt-ga rated this answer:5 out of 5 stars

Subject: Re: Swing cable breaking force
From: myoarin-ga on 17 Jul 2005 04:43 PDT
Hi Matt,

Just look at this site:

I don't think you have to worry about any steel cable breaking under
the weight of a 114 kg person, regardless what he does.  I would be
more worried about the limb of the pine giving way.

Subject: Re: Swing cable breaking force
From: hedgie-ga on 16 Aug 2005 02:19 PDT
 in the  in such a mode, this time

I mean:  You have to think about it a bit to see how it works.

The formula is deceptively simple, since some terms cancel out.
(Perhaps this is what is confusing you.)
This is due to the fact that velocity and deflection angle are
related because energy (potential+kinetic) is constant.

The article explains the situation at different angles of deflection.
It says that at the vertical you get 2g exactly
 (1 + cos (angle)) =2

and you get 0g at the top
 (a+ cos (angle)) =0

Please do read the article before posting another RFC.
It says at the end:

 For a 180-pound person, the seat might have to supply 360 pounds of
upward pull. This is twice the usual amount experienced by our
180-pound rider. For this reason, we would say the rider experiences 2
g's of force (a seat force that is 2 times the gravity force).

Subject: Re: Swing cable breaking force
From: itsjustmatt-ga on 16 Aug 2005 06:34 PDT
Wow, what an idiot I am. I first missed the statment "This is the sum
of gravity and centripetal force..." then saw the formula and wrote it
off as wrong. I thought "this can't be right, it's way too simple".
I'm sorry for that. Thank you.
Subject: Re: Swing cable breaking force
From: racecar-ga on 29 Aug 2005 15:29 PDT
There are a lot of words here in the Answer and the comments, but the
correct answer is nowhere to be found.  Here it is:

Maximum Tension (in Newtons) = M * g * (1 + 2*h/L)

M = mass of swinger
g = acceleration of gravity
h = height above bottom of swing from which you start
L = length of swing

Here is how to calculate the answer:

Conservation of energy: .5 M v^2 = M g h     [v is velocity at the bottom]

Tension at bottom  =  M v^2 / L  +  M g  =  2 M g h / L  +  M g  =  Mg(1+2h/L)

Putting this formula in terms of X, the angle from vertical at which
the swing starts, we have

Tension at bottom = M g (3 - 2 cos(X))   

(which is quite different from the formula posted in the Answer)

In any case, if you start out with the cable horizontal (h = L, or X =
90), as described in the question, the maximum tension is three times
the weight of the swinger.  For a 114 kg swinger, that's 432 kg or
4236 N.

It is rediculous to suggest (as someone did) attempting this with 1.2
mm diameter cable.  It WILL break.  In order to have a reasonable
safety factor, you need cable at least 6 mm (1/4 inch) in diameter.

The cable can become fatigued at the point where it flexes.  It would
be best to use a piece of chain or something for the very top where
the flexing occurs, and then attach the chain to your cable.
Subject: Re: Swing cable breaking force
From: myoarin-ga on 29 Aug 2005 16:44 PDT

Someone miscalculated 1/8 in to 1.2mm; it is 3.175mm :

The info from the site I linked:

SS1  	Stainless Steel Cable, 1/8" dia., 1,700 lbs. breaking strength 
	  	 In Stock  	$114.00
SS3 	Stainless Steel Cable, 1/4" dia., 6,400 lbs. breaking strength
		In Stock 	$237.00

1700 lbs breaking strength should be adequate, though I will admit
that I would be a bit queezy swinging on such a small cable.
But you are very right, 1.2mm would never do.
Subject: Re: Swing cable breaking force
From: racecar-ga on 30 Aug 2005 10:40 PDT
Well, the tension on the cable is very nearly 1000 lbs (see above). 
In an application where cable failure means injury or death, you want
a comfortable safety factor.  You're right that the 1/8 cable probably
wouldn't break (1000 lb load on a cable that breaks at 1700 lb), but
that's a safety factor of less than 2.  I would definitely go for the
1/4 inch.
Subject: Re: Swing cable breaking force
From: racecar-ga on 30 Aug 2005 10:59 PDT
Oops, I just noticed I made a dyslexic mistake in my comment: I wrote
that 3*114 is 432 instead of 342.  So the maximum tension is 342
kg(force) or 3355 N or 754 lb.  So the 1/8 cable has a safety factor
of more than two.  Still pretty iffy though...

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