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From: raylopez99 on 12 Aug 2010 03:55
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.

RL

cc: sci.physics.
From: Androcles on 12 Aug 2010 07:27

"raylopez99" <raylopez88(a)gmail.com> wrote in message
news:6a61bd50-9f7f-41d7-955b-03af7a3e96c6(a)y11g2000yqm.googlegroups.com...
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.

========================================
This is a common fallacy.
http://en.wikipedia.org/wiki/File:Motorcycle_power_wheelie.jpg
In fact, you are totally clueless.





From: None on 12 Aug 2010 13:29
On Aug 12, 7:27 am, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote:

>
> This is a common fallacy.  In fact, weight of a car on an incline is
> evenly distributed on all wheels.

Yet it is easier to lift the higher end then the lower end of such a
car.
From: Excognito on 12 Aug 2010 13:46
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?
From: raylopez99 on 12 Aug 2010 16:49
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
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