# Thread: Suspension Force Calculation - Need Mathletes

1. Originally Posted by OneEye
The only difference between this eyebolt and an ideal pulley is the friction force between the suspension line and the eye.
The friction force between eyebolt and line will slightly reduce the tension in the line from the wall eyebolt to the ceiling. The magnitude of this reduced tension will have a small effect on the idealized calculations presented above, reducing the net vertical force on the eyebolt.

2. Originally Posted by OneEye
Each ceiling bolt will be subjected to T, the tension in the cord. Assuming 30 degree hang angle that T is W, the weight of the occupant. In your example, 200 lbs. The wall bolts happen to be subjected to a force of T, but it is based on the free body diagram of the wall bolt. Sum the forces of the lines coming into the bolt, and the bolt must handle any unbalanced forces to remain at rest. There are two forces acting on the bolt, one from the line to the ceiling, and one from the line to the hammock. The force in each is T or W (200 lbs). One is directed straight up, so Fy=+200lbs. One is directed towards the hammock, so Fy=-100lbs and Fx=+173lbs. Summing the forces we have Fy= +200-100 = +100 lbs. Fx = +173 lbs. The bolt must oppose these net forces, so 100 lbs down and 173 lbs towards the wall. Vector sum of these forces is sqrt(100^2+173^2) = 200 lbs.

The net result being W is a coincidence of the angles chosen...
yep. That's the solution I came up with (free body), except that I interpreted the vertical incorrectly as somehow floating to the ceiling, rather than being applied to the bolt. I do better with transient Markov chains....

Now to distribute the force on the wall bolt perhaps one could install a whole column of bolts on the walls, and thread the suspension line through them like lacing a shoe...hammock to bolt, across the room to an opposing bolt, across the room again to a higher bolt still.... now there's a modeling problem...

thanks for chiming in One-Eye. Important to get this stuff right.

I do better with transient Markov chains....
What is this? I don't even...

20 minutes of wikipedia later, this is the only part I understand:

Originally Posted by Wikipedia
Games
Markov chains can be used to model many games of chance. The children's games Snakes and Ladders and "Hi Ho! Cherry-O", for example, are represented exactly by Markov chains. At each turn, the player starts in a given state (on a given square) and from there has fixed odds of moving to certain other states (squares).
Back to the statics problem...

Now to distribute the force on the wall bolt perhaps one could install a whole column of bolts on the walls, and thread the suspension line through them like lacing a shoe...hammock to bolt, across the room to an opposing bolt, across the room again to a higher bolt still.... now there's a modeling problem...
That's one way to take the negligible friction in the original problem and crank it up so it can't be ignored...I think I will add the Grizzly Adams shoelace configuration to the Markov Chains article, and cite it to this website. I think it fits the definition... The solution presented is random, and doesn't depend on any of the previous solutions offered except the one immediately preceding it.

Am I doing it right?

Back to my previous suggestion: Place eye bolts directly above where you want the ends of the hammock to be. Use a separate line for the vertical component of suspension and the horizontal component. Forces are now purely withdrawal forces (no pesky moments on eye bolts), and the nominal capacity can be simply calculated.