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B Real World Gravity on an incline question

  1. Jul 13, 2016 #1
    Hi, I'm brushing up on my physics, and I have a real world problem I've chosen to solve!

    First the Word Problem part of it, then the Physics after that...

    I have a 3000 lb. Recreational Trailer that I wish to put on my driveway on the side. The driveway is at a 10 degree angle from level. Above the driveway (on the side where the trailer will go) is a Palm tree that's just huge, that I wish to attach it to.

    I have had welded onto the back of the trailer two D rings that swivel, welded to the frame (one on each side, left and right), and have backed the trailer up the driveway. I have attached two tow straps, one has a breaking strength of 20,000 lbs., and it's 8 ft. fixed length, and I've simply gone from D ring around the tree to D ring. As a backup, I have an adjustable-length tow strap that has a WLL of 3300 lbs. and a breaking strength of 10,000 lbs., and it attaches to the trailer at the 1 1/4" hitch welded at the back/middle of the rear of the trailer, goes around the tree also, and returns to the same hitch. It's tightened to match the first tow strap, so that if either strap breaks, the other will hold it with no shock loading. Essentially the trailer is held with two separate loop systems. It seems to be holding fine, and the tree seems fine as well.

    What I want to calculate, is the actual load I'm putting on the straps/tree. So I'm hoping someone can validate my math. This *should be* a gravity on an incline problem. So the value I want should be F parallel, rather than F perpendicular, which is the same as F normal. I am assuming that the trailer is at rest and held in place and friction of the wheels in the axle in layers of grease are minimal, and so I'm not bothering subtracting friction for this problem.

    Seems like the equation should be:

    F parallel = m * g * sin (theta)

    where m = mass (3000 lb. trailer = 1360 kg), gravity is 9.80665 (technically it's 9.79593 as I've calculated local gravity, but I'll go with 9.8 for this explanation).

    F parallel = 13,328N * sin(10)
    F parallel = 13,328 * 0.1736481777
    F parallel = 2,314.3829 N
    so the load (foot pounds of force?) = F parallel / g
    load = 236.161 kg or 520.646 lbs.

    I assume I screwed at least one thing up royally. It just seems too easy and I must have botched something. Any help would be appreciated!!!!

    Note: Please don't go over my head without explanation. The more detailed the answer the more likely I am to understand (especially with math steps taken and none skipped). THANK YOU!!!! It's been many years since I took physics classes...

    Cheers,
    jeff
     
  2. jcsd
  3. Jul 13, 2016 #2

    PhanthomJay

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    You don't need to convert to metric, the gravity force parallel to incline is ( 3000 )(sin 10 ) pounds or 520 pounds.
     
  4. Jul 13, 2016 #3
    I see. That makes sense. I also see I didn't need to convert to mass by dividing by g then multiplying by g.... I assume that I am correct that the gravity force parallel to incline is in fact what I am looking for, the load on the tow straps/tree?

    Thank you!!!

    Cheers,
    Jeff
     
  5. Jul 13, 2016 #4

    PhanthomJay

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    Yes correct. The D eyes need similar strength. You don't want the eyes to be the weak spot. Protect the straps from cutting into the tree. The loading is rather small I would say.
     
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