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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]]

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]]

Nov 22, 2022, 3:23:14 AM11/22/22

to

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.
> 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.

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?

Nov 23, 2022, 7:39:50 AM11/23/22

to

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)
> 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.

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.

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.

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.

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?

(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

Nov 23, 2022, 7:44:26 AM11/23/22

to

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.

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.

Nov 24, 2022, 4:32:06 PM11/24/22

to

Jonathan Thornburg [remove -color to reply] mercoled=EC 23/11/2022 alle=20

ore 13:39:47 ha scritto:

I made the new animation

https://www.geogebra.org/m/t3zezftf

where there are no points but only bodies.

It's right?

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?

Nov 26, 2022, 4:05:57 PM11/26/22

to

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

> 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.
>> 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?

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