Dear All
I tried to pull out the real calculations to my query from different sources... But each reply I am getting leads to a new path or confusing me more...
The latest reply I got as follows:
Regarding the values you mentioned, firstly I want to remind you that the value mentioned by OSHA is a force value 5,000 pounds (22.2 kN) but not weight value (5000 pounds but not 5000 Ibs). The calculations below will be based on this fact (the force on the anchorage point is 5000 pounds = 22.2 KN = 22200 N)
The height (h) = 1.8 m
The gravity acceleration = 9.8
The mass = 140 kg
The distance that the falling object will travel after the impact (d) = X
The force of impact (F) = 22200N
Force of impact = Kinetic Energy / The distance that the falling object will travel after the impact ---> F = KE/d ---> d = KE/F (note that d is only parameter that is unknown)
F = 22200N
KE = PE (potential energy) = mxgxh = 140x9.8x1.8 = 2469.6
d = 2469.6 / 22200 = 0.111 meter
So for you to reach the vlaue of 22200 N as force on the anchrage point:
- The hieght shoudl eqaul 1.8 m
- The mass of the falling object should equal 140kg
- The distance that the rope of the harness should extend before bringing the falling object to rest should equal 0.111 m
Note that in the above calculations no safety factor included.
Can anyone come up with actual calcualtions... What the above said calculations is right or not?...
Looking forward to reach into a great coclusion...
Thanks in advance...
I tried to pull out the real calculations to my query from different sources... But each reply I am getting leads to a new path or confusing me more...
The latest reply I got as follows:
Regarding the values you mentioned, firstly I want to remind you that the value mentioned by OSHA is a force value 5,000 pounds (22.2 kN) but not weight value (5000 pounds but not 5000 Ibs). The calculations below will be based on this fact (the force on the anchorage point is 5000 pounds = 22.2 KN = 22200 N)
The height (h) = 1.8 m
The gravity acceleration = 9.8
The mass = 140 kg
The distance that the falling object will travel after the impact (d) = X
The force of impact (F) = 22200N
Force of impact = Kinetic Energy / The distance that the falling object will travel after the impact ---> F = KE/d ---> d = KE/F (note that d is only parameter that is unknown)
F = 22200N
KE = PE (potential energy) = mxgxh = 140x9.8x1.8 = 2469.6
d = 2469.6 / 22200 = 0.111 meter
So for you to reach the vlaue of 22200 N as force on the anchrage point:
- The hieght shoudl eqaul 1.8 m
- The mass of the falling object should equal 140kg
- The distance that the rope of the harness should extend before bringing the falling object to rest should equal 0.111 m
Note that in the above calculations no safety factor included.
Can anyone come up with actual calcualtions... What the above said calculations is right or not?...
Looking forward to reach into a great coclusion...
Thanks in advance...