Weight distribution of a Hammock - quick question.

Factual Questions
1

I have a hammock that I’d like to string under my porch. I have two large screws I am planning on holding it up with. My question is this:

If I weigh 250 pounds how much weight is being applied on each individual screw when I stretch out on the hammock?

Is it 125 pounds on each screw or 250 on both? I know this seems simple but for those of us teemings who are mathematically retarded it is a legitimate question.

2

Assuming equal weight distibution, half on each screw. It’s just the same as standing on two scales, one foot on each. If you lean slightly to one side or another, the weight distribution will shift, but the total of the scales’ readings will always sum to your weight.

3

There may be some complicated geometrical considerations, but I can say that it you put a plank of wood between two nails and lay on it, the force on each would be 125 pounds.

That’s assuming an equal weight distribution - if you’re a top-heavy guy, it may be more like a 150-100 split.

Also, keep in mind that the force applied will be greater when you jump into the hammock, hop out, twist around, etc.

4

Argh, submitted too soon.

Given that the peak force applied will be much greater than your weight, I would say that you’d want to design each mounting point to withstand at least 250 pounds of force, and probably more.

5

It depends on the angel of the string to hook. If they are both straight down then it will be 125 each, if they are straight out (impossiable) the force approaches infinity. anywhere inbetween it is some trig function, which I would WAG would be a Tan since infinity is involved.

6

I agree. The weight (downward force) on each end is 125 lb, but the tension of the string (i.e. the force on the peg) can be much greater. (125lb)/sin(a) to be exact, where a is the angle of the string measured from horizontal. At 45 degrees it’s 177 lb. As angle approaches a=0, tension approaches infinity. That is, it takes infinite tension to pull the string perfectly taut.

7

Figure out how far apart your hooks are going to be, and one can solve for the worst case (hammock doesn’t stretch, load = 250lbs) and apply a safety factor. Note that because of the sin function, 15 degrees is worse than 30 degrees by a much greater margin than 30 is worse than 45. Measure carefully!

8

More to the point, an un-reinforced hammock can be considered mathematically as a simple suspension bridge. As such, you have to contend with both the shear force on the screw, which will be 125lb each, and the tensile force, which will be much larger. How much larger? A simple rule of thumb is to divide the downward (shearing) force on each screw by the sine of the angle the tangent of the hammock (where it connects to the screw) makes with respect to the horizontal, as scr4 states above. This isn’t quite right because the weight is distributed along the functional length of the hammock, and thus the tension is somewhat less, but this offers a nice, conservative number.

And you thought a hammock was going to be relaxing? Ah, heck, just grab some big honkin’ hooks down at the hardware store that are grossly oversized and recline with your lemonade and a good book.

Stranger

9 


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\  o                                    /
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      \                           /
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downward_force = 250 lbs.

downward_force / sin o

10

Also another factor is that the wight distrubution won’t be 50% 50% but will varry depending where you are in the hammock, and might have a dynamic force as well (say you are having sex in the hammock).