Safety Consultant
Raghuvaran Chakkaravarthy
Environmental, Health And Safety
Kesava Pillai
Occupational Safety & Health
Safety & Production
+2 Others

Dear All
Please help me in clear the concept about the following:
In lot of write up I have read the below snetnace:
Anchor points for fall arresting systems (including lifelines) must be capable of withstanding a 5000 lb. (2300 kg) load per person attached.
But how we are coming to this conclusion?
Is it really required to ensure a lifeline or anchorage point having a strenght to withstand 2300 Kg?
When a normal person (Weight of 60 or 70 Kg) falls how much force is generating? How we can calculate the same?
Hope to get a great participation from all of you in the thread...
Dipil Kumar V

From India
Dear dipil, Sorry to say i dont have idea regarding above mentioned query Give me some time i will check with my seniors then i will catch u . . . Thanks & Besafe Raghu
From United States, Fpo
Dear Sir,
As we know Physics Formula F=ma
Where F= Force
m= Mass
a = Acceleration due to gravity (9.8 m/s)
Normal length of Safety belt is 2-3 meter.
If person fall from the height he require 2-3 meter to get support from anchor point. Lets keep the 3 meter length
Now calculate the acceleration for 3 meter.......... it is 9.8*3= 29.4
F= 70 (Normal weight of person) * 29.4 = 2058.
So it is recommended to to have the capacity more than this.

From India, Gurgaon
Dear vinoddigwal
Thanks a lot for your quickest repsonse and easiest explanation...
Keep on participating and share your expertise with us...
@ Raghu
Great positive attitude... Keep on participating in each and every thread and try to deliver what we have or our concerns...
Dipil Kumar V

From India
Dear Vinoddigwal, Thank you for explaining the subject in great detail. It was a very useful piece of information that I lost track almost 25 years ago. M.V.KANNAN
From India, Madras
Dear Vinod,

Excellent reply thanks a lot.

Dear Dipil,

I check with my seniors finallly i got the solution please read below mentioned points this one i got from my seniors but i didnt get the calculation:::

References in the OSHA and EM-385 are based upon ANSI standards for Fall Protection.

All standards allow for engineered fall protection systems (designed by "qualified" person, usually a Registered Profession Engineer - licensed by a State or Federal Agency) that meets safety factors. All engineered fall protection systems must be tested (drop test of a standard weight). The employer who designs and uses such a system assumes all product and regulatory liability for use of the system.

However, if the emplorer choses not to use an engineered system, the the anchors and fall protection system components must meet the minimum ANSI standard.

Fall protections systems are designed with several goals:


Arrest the fall before the person reaches grade -


Arrest the fall in the shortest fall distance reasonable - maximum 2m (6ft) and preferred .75m (2ft.)


Arrest the fall with minimum or no injury to the person, no shock force greather than 900 ft pounds to person -


Keep the system as light as possible without sacrificing durability and integrity.

The different components of the system are designed to different strengths - 1800lbs (e825Kg) strength webbing and fittings on harness, 3600 lbs (e1500Kg) attachments and lanyard, and 5000 lbs (2300Kg) on anchors and attachments. These numbers are based upon calculated loads of a person weighing up to 310 lbs (e150 Kg) free falling a distance of 6 ft (2m).

These numbers are not flexible. Doing anything less is most often considered negligence on the part of the employer.

Hope this helps.

Thanks & Besafe


From United States, Fpo
Dear Friends,

I really appreciate the efforts taken by Mr.Reghuvaran Raghavan in answerering this typical question.

I also feel disappointed with the post from Mr.Vinoddigwal. This calculation is just confusing for me. If further clarifications can be provided it will be of much help for me in understanding the calculations.

It is basics in fall protection that the lanyard length is strictly 1.8 meters to be effective. Now shock absorbing lanyards are only used. With this a fall will be arrested at about 1.07 meters and as such even the 1.8meter length can not be considered before deceleration starts. We at any cost can not use 3 meters length for (safety belt) lanyard. I would appreciate if any one can come out with a calculation which is not even available with OSHA.

Much time has passed since this value was originally defined by OSHA way back in 1970. Unfortunately, there exists no clear explanation as to the origin of the 5,000 lb value.

In general it is required by OSHA regulations to ensure a lifeline or anchorage point capable of supporting at least 5000 lb (2300 Kg) per employee attached. However it need not be so if it is designed, installed, and used as part of a complete personal fall arrest system which maintains a safety factor of at least two, under the supervision of a qualified person.

This clearly establishes that the lifeline or anchorage need not be strictly of 5000 lb (2300 Kg) capacity. It can be substantially reduced by designing a suitable fall arrest system by qualified personnel.

Nowadays fall arrest equipment are labeled to produce a 900 pound maximum arrest force if the equipment is used in accordance with the manufacturer's recommendations. OSHA also mandates that the maximum arrest force that may be delivered at one's full body harness D-Ring during fall arrest to be 1,800 pounds. Under these circumstances a 2 to 1 safety factor against anticipated peak dynamic load during a fall arrest event would yield 1,800 pounds (2 times 900 pounds) or 3,600 pounds (2 times 1,800 pounds) respectively. Both these figures are clearly well below the 5,000 pound requirement outlined above. It again establishes that it is not 5000 lb capacity that is required of anchorage all the time.


Kesava Pillai

From India, Kollam
@ Raghu
Thanks for your answer...
Your answer states as follows:
"These numbers are based upon calculated loads of a person weighing up to 310 lbs (e150 Kg) free falling a distance of 6 ft (2m)."
If we apply the formula which brings by Mr. Vinoddigwal into this situation:
a= 2*9.8=19.6
then F=19.6 * 150 = 2940 kg which is higher than 2300kg.
So question arises that can we use this formula for calculating this scenario?
@ Keshav Sir
Thanks a lot to put light into the wrong direction were the discussion was heading...
Your post stating that "Now shock absorbing lanyards are only used"... Is this mandatory? We are still using safety harness with double lanyard only...
Hope to get more participation from all of you...
Dipil Kumar V

From India
"Now calculate the acceleration for 3 meter.......... it is 9.8*3= 29.4"
Dear Sir,
I would appreciate a clear understanding of why 9.8 m/s is multiplied with the distance travelled (3m) and not the time taken to cover that 3 m?
Thanks in advance

From India, Pune
Dear Friends,
My humble request is not to waste your time over this if you consider it valuable. Even OSHA is not giving an explanation as to how they arrived at the 5000 lb capacity for anchorage though it is their regulation.
It is true as Mr.Raghu has mentioned about the weight of a person in this calculation with OSHA as 310 lb which is equal to 140.9 Kg (140 as rounded off) and not ro assume as 100 kg.
A fall is arrested at 1.8 meters with the safety harness. In reaching that distance it will never attain the terminal speed of 9.8m/s. The calculation part provided by Mr.Vinoddigwal needs authentication.
He may be able to advise us as to where he could arrive at that calculation. Let us wait and see for clarification.
Kesava Pillai

From India, Kollam

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