Force on the Hitch during a recovery pull

[XJ_ranger]

after reading this thread - i wondered if a winch truely does put more tention on a tow point than a strap.
http://www.naxja.org/forum/showthread.php?t=44725
here is what i found...

High school physics will tell you that force (F) = Mass (M) x Acceleration (a)
therefore f=ma

the mass of my XJ is around 3500lbs which is a weight and not a true mass measurement. Therefore we have to convert that into a mass measurement - Kg or Kilograms happens to be one and 3500lbs = 1590kg.
we are calculating the line connecting the system, so we have to double the weight because there are 2 XJ’s that will be accelerating and we get 3180kg.
My acceleration from a dead stop can be formulated (assuming that acceleration is constant - which its not, but ill consider the change in acceleration from when the strap has tension till the vehicle is either moving or your JC Whitney hitch pin bends negligible) via the 1/4 mile time for a stock Cherokee (which with my 28.5" tires and stock gears it is) from a search on this page Dragtimes
ok so we have the equation from physics class of:
final Position (X) = Initial Position (X.) + Initial Velocity (V.) x final Time (T) + 1/2 Acceleration (A) x final Time (T)^2 or time squared
looks like this:
X = X. + V.T + 1/2AT^2
since I will be using the 1/4 mile time to find the acceleration, I’m going to set up my plane to use Start as 0 therefore X. = 0 and since we are starting from a stop, initial velocity = 0 therefore V. = 0
and since I’m using metric units above, I need to convert 1/4 mile into meters which is about 402 meters.
so our equation now looks like:
402 = 0 + 0(15.2) + 1/2A(15.2)^2
simplification that we learned in 6th grade gives me an acceleration of:
3.5 m/s^2

now we plug this back into our F=ma equation and get the force on your "Pin cure all cheap tow point"
F = 3180 x 3.5 then the force on that pin is 11310 kg x m/s^2 or 11310 Newton’s.
now we convert back into pounds, (1 Newton ~ ¼ lbs of force)

[font=&quot]2782 lbs of force.

[/font][font=&quot]ok - where did i mess up? or is all that fairly correct?
[/font]

 

[bocaj]

i asked some one about this with the strap on the pin compared to the hitch on the pin. i cant remember much about it but... they said something about the shear strength compared to bending strength. with the hitch in, the pin cant bend and can only break. and thats what its rated for. but the strap spreads its weight out unevenly easily bending the pin. then its broken easily after that. some better words were used when they told me and a better explination was given about the pin being easier to break with the strap. i didnt read all the other posts and as for the math i dont know what it means but good job.

 

[jeepguy97]

All I can say is LET IT GO. Use the proper recovery gear, physics has nothing to do with lif and death.

 

[bocaj]

The hitch pin with a hook or D-ring receiver tube is loaded in double-shear. The load on the pin is isolated to two shear loads.

The hitch pin with a strap loads the pin in bending and double-shear.

Will the hitch pin fail? The pin should not fail in bending if the strap is used properly, if shock loads are not applied to the pin. As long as the pin does not bend the shear forces on the pin are identical, with or without a tube in the receiver

went back and read the rest. this guy already had it.

 

[Beej]

Hey Ranger, interesting stuff. You're not just showing us how clever you are, are you? We already think you're super! I wonder if your particular method of mathematical exploration will answer your initial question, which is: "i wondered if a winch tru[]ly does put more tenion on a tow point than a strap". Although your calculations yield an accurate answer arithmetically, I'm not sure you can answer your initial question from the numbers you generated. Just a thought....

 

[Dirk Pitt]

Those numbers would be great if you were stuck on a flat, asphalt surface similar to the one you pulled your drag acceleration numbers from.

Figure in the reaction forces of the mud/rocks you are stuck in and get back to us.

 

[Beej]

Those numbers would be great if you were stuck on a flat, asphalt surface similar to the one you pulled your drag acceleration numbers from. Figure in the reaction forces of the mud/rocks you are stuck in and get back to us.

I think he's right Ranger, however, you could create a guesstimation of this by say, adding 50% mass to both vehicles to account for mud and reduce your acceleration time by about a third or so to get some numbers that might be closer to real world forces (not too accurate though, as I just pulled those figures out of my ass...). However, even with this calculation, I don't know if you'll be any closer to answering your original question.

PS did I mention that I loved your gauge-swap writeup?

B.