F=ma

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Luigi Fortunati

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Nov 20, 2022, 12:36:26 AM11/20/22
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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.

By clicking on the "Braked car" check box, the car becomes free to move
and we can accelerate it with the appropriate button.

The question is this: By applying the same force F=3 to the car as
before, does the red reaction force which I assumed to be less than 3
really exist or not at all?

[[Mod. note -- To understand what's going on, let's say for purposes
of exposition that the thing that's applying the force F is a person's
hand. And let's ignore any vertical forces and dynamics, and focus on
the horizontal forces and dynamics. Then:

1. We are given that the hand applies a force of 3 units (rightward)
on the car. By Newton's 3rd law, the reaction force has the same
magnitude (i.e., 3 units), opposite direction (leftward), and is
*applied by the car* *to the hand*. This is a crucial point: the
reaction force is applied *to the hand*, not to the car.

[More generally, if two (or more) objects are touching
at a point, and you have a force applied at that point,
it's essential to always be clear as to which object the
force is being applied to. In this case, the reaction
force is being applied to the *hand*, not the car.]

Notice that the reaction force is the same (3 units leftward)
whether or not the car is accelerating. Luigi's animation is wrong
in showing the reaction force as being smaller for the acceleration
case.

Because the reaction force isn't applied to the car, the reaction
force makes no difference to the car's motion.

2. Now let's consider the *braked* car, i.e., the case where the car's
brakes prevent the car's wheels from rotating, so the car is held in
place by the friction of the tires on the ground. As is often the
case in Newtonian mechanics, a free-body diagram is useful. (See
https://en.wikipedia.org/wiki/Free_body_diagram
for more information on free-body diagrams.) Here's a free-body
diagram for the car:

<----------*---------->

The rightward-pointing arrow is the force (of magnitude 3 units)
applied by the hand to the car. (This is shown in blue in the diagram.)
The leftward-pointing arrow is the friction force, also of magnitude
3 units, applied by the ground to the car (more precisely, this force
is applied to the car's tires, but since the wheels aren't rotating
we can treat the car+tires as a single body). This friction force
is missing in Luigi's diagram.

Notice that the reaction force doesn't appear at all here: this
free-body diagram only shows forces applied *to the car*.

The net force applied to the car is zero, so by a = F_net/m the
car is unaccelerated. Since its initial condition is stationary
(zero velocity with respect to the ground), it will thus stay
stationary.

3. Now let's consider the car with wheels free to rotate. For this case
there's (by assumption) no friction force of the ground on the car,
so the free-body diagram looks like this:

*---------->

The rightward-pointing arrow is again the force, of magnitude 3 units,
applied by the hand to the car. This is the only (horizontal) force
applied *to the car*. (To repeat, the reaction force isn't relevant
here, because it's not applied *to the car*.) Because there's now a
net force (to the right, of magnitude 3 units) applied to the car,
the car will accelerate (to the right). Luigi's animation is wrong
in showing the net force as only 1 unit for the accelerating case.
-- jt]]

Luigi Fortunati

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Nov 22, 2022, 3:23:14 AM11/22/22
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Luigi Fortunati sabato 19/11/2022 alle ore 14:36:21 ha scritto:
> 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.
>
> By clicking on the "Braked car" check box, the car becomes free to move and we can accelerate it with the appropriate button.
>
> The question is this: By applying the same force F=3 to the car as before, does the red reaction force which I assumed to be less than 3 really exist or not at all?
>
> [[Mod. note -- To understand what's going on, let's say for purposes
> of exposition that the thing that's applying the force F is a person's
> hand. And let's ignore any vertical forces and dynamics, and focus on
> the horizontal forces and dynamics.

Ok.

> Then:
>
> 1. We are given that the hand applies a force of 3 units (rightward)
> on the car.

If we want to be rigorous, this is not the case.

The hand force is applied to point P and not to the whole car.

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.

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.

What is the necessary condition for the point P to accelerate, if not
that of the inequality between the opposing blue and red forces?

Jonathan Thornburg [remove -color to reply]

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Nov 23, 2022, 7:39:50 AM11/23/22
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Luigi Fortunati <fortuna...@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.th...@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

Luigi Fortunati

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Nov 23, 2022, 7:44:26 AM11/23/22
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Maybe, with this animation
https://www.geogebra.org/m/gzuwcueh
the forces in action are better identified.

There are two spaceships A and B in far space, nose to nose, facing
each other with their engines running.

Spaceship A exerts the blue force F on spaceship B, and spaceship B
exerts the red force R on spaceship B.

Since the forces F and R are equal and opposite, the two spaceships
remain stationary and do not accelerate.

If we click on the button that increases the force F, the equilibrium
is broken and the point P (together with the two spaceships) moves to
the right, accelerating from zero speed to v other than zero speed.

If we click on the button that increases the force R, the reverse
occurs and the two spaceships accelerate to the left.

Is the animation correct?

Are the forces F and R the action and the reaction?

I would like to point out a fundamental difference between the
acceleration of my animation and that resulting from a collision
between two spaceships with their engines off.

In my animation, when the force F is different from R, the common
center of mass accelerates and the difference between F and R seems to
contradict the third principle.

Instead, in the acceleration resulting from the collision between two
spaceships with the engine off, the common center of mass remains in
place and does not accelerate, so that the third principle is certainly
respected.

Luigi Fortunati

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Nov 24, 2022, 4:32:06 PM11/24/22
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Jonathan Thornburg [remove -color to reply] mercoled=EC 23/11/2022 alle=20
ore 13:39:47 ha scritto:
> Forces aren't applied to points, they're applied to objects.

You are right.

I made the new animation
https://www.geogebra.org/m/t3zezftf
where there are no points but only bodies.

It's right?

Luigi Fortunati

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Nov 26, 2022, 4:05:57 PM11/26/22
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Luigi Fortunati gioved=EC 24/11/2022 alle ore 22:32:03 ha scritto:
> Jonathan Thornburg [remove -color to reply] mercoled=3DEC 23/11/2022 al=
le=3D20
> ore 13:39:47 ha scritto:
>> Forces aren't applied to points, they're applied to objects.
>
> You are right.
>
> I made the new animation
> https://www.geogebra.org/m/t3zezftf
> where there are no points but only bodies.
>
> It's right?

I've added hand movement to the animation.
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