Luigi Fortunati <
fortunati.luigi@gmail.com> wrote:
In my animation
https://www.geogebra.org/m/g32ywuep
there is a braked car to which I apply a force 3 (action) and it responds with a force R=-3 (reaction).
The net force is zero, the car does not accelerate.
In Newtonian mechanics, we almost never want to compute the net (vector)
sum of a 3rd-law action-reaction pair, for two reasons:
(1) That sum is always zero (by Newton's 3rd law), so it's not very
interesting.
(2) And, these two forces *act on different objects*. In this case,
the reaction force acts on the hand, not on the car.
Instead, we usually want to compute the net (vector) sum of all the
forces *acting on a given object*. In this case, what we want is
the net force *acting on the car*. As Luigi notes, the reaction force
is applied at the point P... but the reaction force isn't acting on
the car, rather it's acting on the hand, so we can ignore this force
when trying to analyze the car's movement or non-movement.
The horizontal forces *acting on the car* are
(a) the applied force (3 units pointing to the right), and
(b) a friction force (pointing to the left, of a magnitude which
we're going to figure out in the next paragraph) exerted by the
road on the car's tires.
In this scenario the words "tires don't skid on road" mean that the
friction force (b) is sufficiently large that there's no relative
motion of the tire bottoms with respect to the road. Since we are
told that the wheels are locked (don't rotate), this means that
(after an initial displacement as the tires stretch a bit) the
car's center of mass doesn't accelerate horizontally. By F = ma
this means that the net horizontal force acting on the car must be
zero. Since that net horizontal force is precisely (a) + (b), we
conclude that (b) = -(a), i.e., (b) must be of magnitude 3 units
pointing to the left.
[It's also instructive to consider a hypothetical situation
where the car's tires *do* skid on the road. For example,
suppose the car is on an icy surface (still with wheels locked)
and the tires start skidding, with only a small frictional
force of (let's say) (b) = 0.2 units pointing to the left.
Then the sum of (a) and (b) would be 2.8 units pointing to
the right, and the car would accelerate to the right under
the influence of that net force.
What distinguishes the "tires don't skid on road" and the
"tires skid on road" scenarios is how much friction there is
between the tires and the road. If the frictional force is
less than the applied force, then the car will accelerate to
the right. If not, then the car won't accelerate (horizontally).]
Luigi also wrote:
The hand force is applied to point P and not to the whole car.
That's true, but irrelevant. Since the car is (apart from the minor
stretching of the tires noted above) a rigid body, where on the car
we apply the force doesn't matter for determining the motion of the
car's center-of-mass. (It *does* matter for determining if the car
is going to start rotating about a vertical axis, but that's a
different question.)
And the force of the car is applied to the tip P of the hand and not to
the whole arm or body of the man.
The protagonist of my animation is point P.
It seems to me undeniable that the blue force and the opposite red force
are both applied to the point P.
They're applied *at* the point P... but they are applied to *different
objects* at that point. The blue (applied) force is applied to the car;
the red (reaction) force is applied to the hand.
If the two blue and red forces are equal and opposite, the point P will
not accelerate, because there is no "net" force on it.
Forces aren't applied to points, they're applied to objects.
In this context it's instructive to think about some variant scenarios
in which we have a *non-contact* applied force. For example, cars
generally contain a fair bit of iron, so what if we bring a large magnet
near to, but not touching, the right side of the car? In this case
there's still a net force (say of magnitude 3 units) to the right
applied to the car, and a corresponding reaction force to the left
applied to the magnet (as per Newton's 3rd law). The forces acting
on the car are the same as in Luigi's original scenario, so the car's
motion is also the same. But now there's no contact point P! We can
still paint a dot on the car and call that dot "point P", but it's
clear that the reaction force has nothing to do with point P, but
rather is applied to the magnet (which isn't touching the car).
What is the necessary condition for the point P to accelerate, if not
that of the inequality between the opposing blue and red forces?
The necessary condition for *the car* to accelerate" is that the net (horizontal) force *on the car* be nonzero. As described above, this
net force is the (vector) sum of (a) and (b). (a) is the applied force
shown in blue in your diagram. (b) is not shown in your diagram; the
red reaction force shown in your diagram is not involved in whether or
not the car moves because it's applied to the hand, not the car.
If you want to ask about the point P accelerating, then you need to
be a bit more precise, and say whether you want to consider the point
P as attached to the car (in which case obviously P moves if and only
if the car moves), or (say) attached to the magnet mentioned above
(in which case P may stay stationary even while the car moves).
--
-- "Jonathan Thornburg [remove -color to reply]" <
dr.j.thornburg@gmail-pink.com>
currently on the west coast of Canada
"Open source code is not guaranteed nor does it come with a warranty."
-- the Alexis de Tocqueville Institute
"I guess that's in contrast to proprietary software, which always
comes with a full money-back guarantee." -- anon
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