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From: Androcles on 12 Aug 2010 17:51

"raylopez99" <raylopez88(a)gmail.com> wrote in message
news:ffe19ddc-6167-4f81-b530-0d7abd905e0f(a)v8g2000yqe.googlegroups.com...
On Aug 12, 8:46 pm, Excognito <stuartbr...(a)gmail.com> wrote:
> On 12 Aug, 08:55, raylopez99 <raylope...(a)gmail.com> wrote:
>
> > On Aug 12, 2:34 am, The Master <colossalblun...(a)gmail.com> wrote:
>
> > > At one time, I lifted the back end of a car off the ground with my
> > > bare hands! Of course, that 'car' was a Ford Fiesta, and it was
> > > parked on a steep grade which shifted much of the weight forward.
> > > But still, not just anyone could do it. My grandmother for instance,
> > > would never have even attempted such a daring feat of strength.
> > This is a common fallacy. In fact, weight of a car on an incline is
> > evenly distributed on all wheels. It may be harder to lift a car from
> > one end or another depending on where its center of gravity is, but
> > that has nothing to do with the car being on an incline.
>
> Consider a car on an extreme slope of, say, 80 degrees from the
> horizontal (or at whatever angle keeps the cg 6 inches upslope of the
> wheel). How much force, applied at the rear, would it take to tip
> the car over? Is this the same force required to lift the front of
> the car by 6 inches?

This is a common fallacy. You are changing the problem, essentially
saying that a moment arm exists between the front wheels, piveted to
the ground, and the back end. Sure, in that case you get leverage.
But in fact, for most inclines, the front and back wheels are both
resisting the road via static friction, so your problem conditions is
in fact are never achieved.

Fact is: takes as much force to raise the front end of a car on an
incline as the back end, if the center of gravity is in the middle of
the car. Basic physics.

RL
=============================================
Fact is: the centre of gravity is higher than the wheels, not on the ground.
Fact is: the centre of gravity is over one axle when the car is at a steep
enough angle and no weight is on the other axis.
Fact is: you are a fallacious idiot. Basic stupidity.


From: The Master on 12 Aug 2010 17:54
On Aug 12, 3:55 am, raylopez99 <raylope...(a)gmail.com> wrote:

> On Aug 12, 2:34 am, The Master <colossalblun...(a)gmail.com> wrote:
>
> >   At one time, I lifted the back end of a car off the ground with my
> > bare hands!   Of course, that 'car' was a Ford Fiesta, and it was
> > parked on a steep grade which shifted much of the weight forward.
> > But still, not just anyone could do it.  My grandmother for instance,
> > would never have even attempted such a daring feat of strength.
>
> This is a common fallacy.  In fact, weight of a car on an incline is
> evenly distributed on all wheels.  It may be harder to lift a car from
> one end or another depending on where its center of gravity is, but
> that has nothing to do with the car being on an incline.


Phil, if you had a brain, you'd be dangerous. Now go 'round to the
front of my current vehicle and try to pick it up from that end -- the
end
in which lies its massive engine. What's the matter? Too heavy, or
is
the center of gravity 'out to get you'? Can't do nuthin' but grunt
and
strain, then collapse? Maybe it's just that you're out of shape, old
boy.
Time to get off your computer and start losing some weight -- do a
few
laps in the pool, play some tennis, lift some weights. But first,
let's
see where you are... weight: about 90 pounds... muscle mass: DNA...
Hmm, looks like we've got our work cut out for us.


From: The Master on 12 Aug 2010 18:10
On Aug 12, 4:49 pm, raylopez99 <raylope...(a)gmail.com> wrote:

> Fact is:  takes as much force to raise the front end of a car on an
> incline as the back end, if the center of gravity is in the middle of
> the car.  Basic physics.


Who is 'changing the problem' now?

In my story, I the car was a Ford Fiesta: a car whose main source
of 'mass' was the engine, way up front. A car whose weight rested
primarily on its front wheels, and which altogether was (yes, *was*:
I
smacked into an Oldsmobile Delta 88) far lighter than your average
Chevy or Dodge gas guzzler.

After bouncing that pitiful Olds into a ditch, I drove home, wishing
I
had not been *quite* so brutal in displaying my car's superiority
over
his. NOT! However, I did take some small satisfaction in the fact
that his radiator was put out of service and his chrome bumper got
dinged up just a bit. Let that be a lesson to him -- next time,
he'll
think twice before crushing a Fiesta (or a Yugo) with his behemoth
Oldsmobile.
From: raylopez99 on 13 Aug 2010 04:47
On Aug 13, 12:54 am, The Master <colossalblun...(a)gmail.com> wrote:
> On Aug 12, 3:55 am, raylopez99 <raylope...(a)gmail.com> wrote:
>
> > On Aug 12, 2:34 am, The Master <colossalblun...(a)gmail.com> wrote:
>
> > >   At one time, I lifted the back end of a car off the ground with my
> > > bare hands!   Of course, that 'car' was a Ford Fiesta, and it was
> > > parked on a steep grade which shifted much of the weight forward.
> > > But still, not just anyone could do it.  My grandmother for instance,
> > > would never have even attempted such a daring feat of strength.
>
> > This is a common fallacy.  In fact, weight of a car on an incline is
> > evenly distributed on all wheels.  It may be harder to lift a car from
> > one end or another depending on where its center of gravity is, but
> > that has nothing to do with the car being on an incline.
>
>   Phil, if you had a brain, you'd be dangerous.  Now go 'round to the
> front of my current vehicle and try to pick it up from that end -- the
> end
> in which lies its massive engine.  What's the matter?  Too heavy, or
> is
> the center of gravity 'out to get you'?  

You are changing the problem, a common fallacy.

The problem is this: given a car with a center of gravity at the
center of the car, parked on a steep incline, which end is heavier to
lift, the front or the back (car facing downhill)? The force on each
wheel is the same, as it touches the ground. But the problem is this:
gravity. In a gravity free environment, which was my original intent,
the force to lift each end is the same. But on earth, the back end
will be easier to lift since you are not working against the gravity
vector.

Got it now? I hope I put to rest your common fallacies. At least one
of them.

RL
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