IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[ME]

Ey George, you did not understand anything from my previous post, nor from that link you provided yourself.

Because this post happened to fall on a new page I'll add your text as a courtesy for avoiding unnecessary page swaps.

Hi ME,
Thank you for your reply.
But you are not reading carefully my posts.
1) Firstly, we are not talking for the present for V', V", Vi and Vr at all! Please pay special attention to this fact!
2) Secondly, please have a look again at the link https://www.knowledgeuniverseonline.com ... -plane.php
3) Thirdly, R is not equal to mg, this is clear. We are not argueing about this fact.
4) Fourthly, G = mg = weight of the body.
4) Fifthly, if you accept the validity the considerations in the above link, then you have to accept the validity of the equality G = - G. The body exerts a vertical downward force G on the inclined plane. At the same time however the inclined plane on its behalf exerts a vertical upward force -G on the body. Besides G=-G, that is, the two forces G and -G are opposite in direction and equal in magnitude.Do you accept the validity of the considerations in this item 4?
Looking forward to your answer.
George

1. Firstly: Please read again. 'V, Vi etc' was a side note... To specifically repeat: Letters do not tell the story, they indicate.
2. Secondly: I did. I did not invent this 'R'. The "R", of Reaction force, may be the cause of your confusion (I don't know).
3. Thirdly: This is related to the first.... This letter R equates to the Normal force. Which is indeed not mg, there is no argument other than you making it a third item.
4. Fourthly: yes
5. Fiftly: Nonsense!

• 5.1. Fiftly-dot-firstly: Mathematically G = -G is only valid when G=0. This is clearly not the case!!
  5.2. Fiftly-dot-Secondly: G is non-zero and directed straight down, so yes I agree that "The body exerts a vertical downward force G on the inclined plane."
  5.3. Fiftly-dot-Thirdly: The plane does not, I repeat DOES NOT, "exerts a vertical upward force -G on the body".
  • 5.3.1. Fiftly-dot-Thirdly-dot-Firstly: That's where this 'R' comes in.
    5.3.2. Fiftly-dot-Thirdly-dot-Secondly: We better name it 'N'.
    5.3.3. Fiftly-dot-Thirdly-dot-Thirdly: It is the N of NORMAL-force of the plane and is provided by the plane because that plane is a sturdy one and not, for example, made out of foam.
    5.3.4. Fiftly-dot-Thirdly-dot-Fourthly: This Normal force is perpendicular to the plane. Thus at an angle θ
    5.3.5. Fiftly-dot-Thirdly-dot-Fiftly: This Normal force has a magnitude of mg·Cos(θ).
    5.3.6. Fiftly-dot-Thirdly-dot-Sixtly: Please have a look again at the link https://www.knowledgeuniverseonline.com ... -plane.php
    5.3.7. Fiftly-dot-Thirdly-dot-Sevently: ... it tells you the same value when you look at the vector indicated with 'R', which is actually the Normal-force.
  5.4. Fiftly-dot-Fourthly: "Do you accept the validity of the considerations in this item 4". You mixed up 4 and 5. I Accept item 4. But NO I DO NOT accept your additions.
  5.5. Fiftly-dot-Fiftly: To give a counter-advice for us not being "experts in theoretical and applied mechanic": Please learn some basic physics.

Something to consider:
- When G would be countered in total ("G = -G" as you like to put it), then which force would be responsible for accelerating an object down a slope?

Lesson to learn, or a hint (whichever you like):
- When θ=0 then the plane is a flat plane, like a TABLE. On this FLAT HORIZONTAL TABLE the Gravitational force is countered IN TOTAL by the Normal force of the table. Now that object will not go anywhere, and we may only now put is as "G = -G".

[George1]

Hi ME,
Thank you for your reply.
--------------------------------------
Yes, I mixed 4 and 5. Accept the second "4" (actually 5). Please excuse me for my incorrectness.
--------------------------------------
But I cannot understand what are you talking about in your last post. Because:
1) G = mg is the vector sum of X = mgsin(angle) and N = mgcos(angle);
2) -G = -mg is the vector sum of - X = - mgsin(angle) and - N = R = - mgcos(angle).
Where (a) G is the force exerted vertically downward by the body on the inclined plane and (b) - G is the force exerted vertically upward by the inclined plane on the body.
-------------------------------------
Do you accept that G = - G? (Because this is the third Newton's law.)
Looking forward to your answer.

[ME]

Please note that your link is about able a "smooth inclined plane"

1) Agreed
2) It is mathematically correct, yet... Not applicable here.


a) The Normal-force "N = mgcos(angle)" is provided by the structure to counter the force of that box (G=mg) - or else it would sink in that plane.
Right?
b) The Normal force is perpendicular to the surface, because [-it's complicated-] I try to explain: Because the plane-material inside that plane is in the perpendicular direction, all that material pushes back perpendicularly on any dent it might encounter to keep that molecular shape intact. When that box would (try to) dig itself straight down into the plane, that box would (try to) compress the plane-material at an angle on one side and (try to) extend plane-material on the other side. This potential-energy build-up is in the parallel direction down the slope. The molecular shape pushes and pulls it back up the ramp with Newton's third-law.

Right?
c) ... That's it basically.


The force of gravity on that box (mg) is only partially countered by the Normal-force.
The situation is UNBALANCED.
There is a residual force of gravity, namely: X = mgsin(angle)
This residual force is responsible for that box to accelerate down that smooth inclined plane.

Do you accept that G = - G? (Because this is the third Newton's law.)

I accept Newton's law. Yet in this situation the two objects don''t encounter each other straight on.
Because it is not a heads-on collision (maybe hard to notice it is a collision) Newton's third-law acts at an angle.
Hence, not all that force is countered in full.

To repeat:
That's why there's a residual force that's still working on that box.
It causes that box to accelerate down that smooth ramp.

But I cannot understand what are you talking about in your last post.

Didn't you have, as you claimed, "extremely gifted engineers" at your disposal?

But don't worry. We have the same issues with your words.
I hope you see how difficult it is to understand someone with just mere words and thoughts in their heads, like your initial handwritten pdf.
We need more pictures.

[George1]

Hi ME,
Thank you for your reply.
---------------------------------
But you are not reading carefully my posts and even worse:), you are trying to violate directly and severely the third Newton's law. But it's OK. I will explain the things in another manner.
---------------------------------
1) The third Newton's law clearly states that for every action, there is an equal and opposite reaction. In other words, no force F in this Newtonian world can exist in in isolation. Forces always exist in couples. That is, if there is a certain force F (action), then there is always another certain force - F (reaction) as F and - F are always equal in magnitude and opposite in direction. In other words, if body 1 exerts force F on body 2, then body 2 exerts force - F on body 1. And, as mentioned above, the two forces F and - F are always equal in magnitude and opposite in direction.
----------------------------------
2) Let us get back to our inclined plane example. We simply replace F and - F with G and - G, respectively. In other words, body 1 (this is the body, which is mentioned in the link https://www.knowledgeuniverseonline.com ... -plane.php) exerts force G on body 2 (the inclined plane and Earth are firmly attached one to another thus forming one united whole, which is called body 2). On the other hand body 2 exerts force - G on body 1. The two forces G and - G are equal in magnitude and opposite in direction. And this is a clear manifestation of the third Newton's law.
----------------------------------
3) In one word, if F and G exist and if - F and - G do not exist, then this is a direct and severe violation of the third Newton's law. (Which seems to be an as if excessively brave hypothesis.)
----------------------------------
4) And one more important thing, which seems to be evident, but let us focus on it anyway. Any possible vector resolution of forces F, - F, G and -G (or of any Newtonian couple of forces, which are related to the third Newton's law and which are always opposite in direction and equal in magnitude) into their vector components (if any and if necessary) do not violate in any way the validity of the above considerations.
------------------------------------
It seems to me that now everything is crystally clear and understandable.
Looking forward to your answer.
George

[johannesbender]

Newtons law of gravitation states.

F = G(m1 m2 )/R 2.

... ?

[ME]

you are trying to violate directly and severely the third Newton's law. But it's OK.

There's no violation.
Johannesbender seems to know it too.

the inclined plane and Earth are firmly attached one to another thus forming one united whole, which is called body 2).

Yes. The inclined plane and the Earth are.
Yet that object is not glued to the ramp!

The title itself tells you: "Motion of a body on a smooth inclined plane"
This alone reveals that you can expect this body to move on a smooth inclined plane.
This follows:

• a body is placed on a smooth inclined plane...

When you look at the graph there is no force-vector straight up. Thus, as clearly visually shown by force decomposition, gravitational force is not opposed in full... by the inclined-plane. That we can take gravity for granted you can read below.
I don't seem to get your linked page up and running, so I reconstructed the picture in the attachment.

Forces always exist in couples.

Yes!!
But see, while the Earth may exert a force onto a box-object so it causes an acceleration, so does the object itself ALSO exert a force on the Earth.
Those forces are equal and opposite. Determined by this well know gem: F=G·M·m / r²

With F=ma, here it comes:
While the Earth has an extremely huge mass 'M' then, relative to both their common center of mass, it will accelerate extremely little: a=F/M
While that box has a relative small box mass 'm' then, relative to both their common center of mass, it will accelerate a lot more: a=F/m

We, as observers standing on the ground, are attracted to the Earth as well.
We will only notice the attraction of the box down towards the Earth: 9.81 m/s²

It only looks like a one sided force.
Therefore, without anything between that box-object and the ground we see this box-object accelerated down to the ground,
Agreed?

[George1]

To johannesbender and ME.
--------------------------------------
Hi guys,
Thank you for your replies.
You both are not reading carefully my posts. But it's OK. I will explain the situation again.
-----------------------------------------
1) Please look at the link https://physicsabout.com/newtons-3rd-law/. Please read very, very carefully the text in this link. For your covenience i am giving below the most important part of the text of this link.
==============================
"Newton’s Third law of motion definition
Newton's third law of motion
Newton’s 3rd law of motion deals with the reaction of a body when a force acts on it. Let a body A exerts a force on another body B , body B reacts against this force and exerts a force on body A.The force exerted by body A on B is the action force whereas the force exerted by body B on A is called the reaction force. Newton’s 3rd law of motion states that:
“To every action, there is always an equal but opposite reaction.�
According to this law, the action is always accompanied by a reaction force and the two forces must always be equal and opposite. Note that action and reaction forces act on different bodies.
Note:
Newton’s 3rd law of motion tells us four characteristics of forces.

Forces always occur in pairs (action and reaction forces).
Action and reaction are equal in magnitude.
Action and reaction are opposite in direction.
Action and reaction act on different bodies."
=============================
2) According to Galileo's principle of relativity of motion let us assume that body 1 (this is the body from our initial inclined plane example(or body A from the link above)) moves and body 2 (the inclined plane + Earth firmly attached one to another thus forming one united whole(or body B from the link above)) is motionless. In this case body 1 exerts force G on body 2 as G is resolved into mgsinθ and mgcosθ.
=============================
3) According to Galileo's principle of relativity of motion let us assume now that body is motionless and body 2 moves. In this case body 2 exerts force -G on body 1 as -G is resolved into -mgsinθ and -mgcosθ.
=============================
4) Actually it doesn't matter which of the bodies 1 and 2 (A and B) is assumed to be motionless or to be in motion. In any case body 1 (body A) exerts force G on body 2 (body B) and body 2 on its behalf exerts force -G on body 1 as G and -G are always opposite in direction and equal in magnitude. This is the third Newton's law.
=============================
5) I am doing the above considerations because you both are obviously confused by the related resolutions of forces in our initial inclined plane example and do not have in mind the principle of relativity of motion.
=============================
Dear ME and johannesbender, you are obviously smart men. Why must I explain to you basic postulates and concepts of theoretical and applied mechanics elementary course?

[George1]

To johannesbender and ME.
--------------------------------------
Hi guys,
Thank you for your replies.
You both are not reading carefully my posts. But it's OK. I will explain the situation again.
-----------------------------------------
1) Please look at the link https://physicsabout.com/newtons-3rd-law/. Please read very, very carefully the text in this link. For your covenience i am giving below the most important part of the text of this link.
==============================
"Newton’s Third law of motion definition
Newton's third law of motion
Newton’s 3rd law of motion deals with the reaction of a body when a force acts on it. Let a body A exerts a force on another body B , body B reacts against this force and exerts a force on body A.The force exerted by body A on B is the action force whereas the force exerted by body B on A is called the reaction force. Newton’s 3rd law of motion states that:
“To every action, there is always an equal but opposite reaction.�
According to this law, the action is always accompanied by a reaction force and the two forces must always be equal and opposite. Note that action and reaction forces act on different bodies.
Note:
Newton’s 3rd law of motion tells us four characteristics of forces.

Forces always occur in pairs (action and reaction forces).
Action and reaction are equal in magnitude.
Action and reaction are opposite in direction.
Action and reaction act on different bodies."
=============================
2) According to Galileo's principle of relativity of motion let us assume that body 1 (this is the body from our initial inclined plane example(or body A from the link above)) moves and body 2 (the inclined plane + Earth firmly attached one to another thus forming one united whole(or body B from the link above)) is motionless. In this case body 1 exerts force G on body 2 as G is resolved into mgsinθ and mgcosθ.
=============================
3) According to Galileo's principle of relativity of motion let us assume now that body 1 is motionless and body 2 moves. In this case body 2 exerts force -G on body 1 as -G is resolved into -mgsinθ and -mgcosθ.
=============================
4) Actually it doesn't matter which of the bodies 1 and 2 (A and B) is assumed to be motionless or to be in motion. In any case body 1 (body A) exerts force G on body 2 (body B) and body 2 on its behalf exerts force -G on body 1 as G and -G are always opposite in direction and equal in magnitude. This is the third Newton's law.
=============================
5) I am doing the above considerations because you both are obviously confused by the related resolutions of forces in our initial inclined plane example and do not have in mind the principle of relativity of motion.
=============================
Dear ME and johannesbender, you are obviously smart men. Why must I explain to you basic postulates and concepts of theoretical and applied mechanics elementary course?

[johannesbender]

"Dear ME and johannesbender, you are obviously smart men. Why must I explain to you basic postulates and concepts of theoretical and applied mechanics elementary course?"

i simply posted an equation from newton , stating that F FORCE is the result of G and the masses .

that means that F FORCE is not G gravity ,so explain to us ignorent how G = -G because G is not F in an equation according to newton.

look at those "laws" and read the word "force" in your lessons your teaching us apparently, or establish that gravity is the force in your newton examples and his equations.

[George1]

To johannesbender.
-------------------------------
Your phrase "F FORCE is the result of G and the masses" unambiguously shows that you are an absolute ignorant in the field of theoretical and applied mechanics. I will not discuss any more the recent topic with you. Firstly educate yourself in theoretical and applied mechanics and just then take part in this discussion.

[George1]

To ME.
-----------------------
Looking forward to your answer related to our post sent before our last post.

[johannesbender]

To johannesbender.
-------------------------------
Your phrase "F FORCE is the result of G and the masses" unambiguously shows that you are an absolute ignorant in the field of theoretical and applied mechanics. I will not discuss any more the recent topic with you. Firstly educate yourself in theoretical and applied mechanics and just then take part in this discussion.

newtons law is true about action and reaction but your statement "G=-G" is utter bs.

and yes it would be my pleasure to avoid you.

[ME]

George,

Your point 1:
Let's say we both understand Newtons laws

Your point 2:

In this case body 1 exerts force G on body 2 as G is resolved into mgsinθ and mgcosθ.

That's all I talked about...
So you simply agree with my earlier reply, and I don't understand why you complain and sputter.
Do we disagree to agree?

Now your point 3:

According to Galileo's principle of relativity of motion let us assume now that body is motionless and body 2 moves.

Why are we suddenly assuming a motion? Ok, motion it is.

Please don't call it "body1"and "body2"... when they already have a perfect designation. I'm not a computer.
We simply have a BOX on an smooth INCLINED PLANE and one of them moves. Let's say the box moves

Questions:
a. In which direction is the box (relatively) moving?
b. Is it colliding with a stationary inclined plane?
c. Are you implying that this affects the earlier force decomposition?
d. Let's say the box initially glides up the ramp with a certain velocity. Just parallel. Do you think it will eventually return and glide down again?

Let's leave it to this for now. And let us all await your answers.

[George1]

I am explaining one and same simple things for the fourth time.
------------------------------------------------------------
1) Please look again at the link https://knowledgeuniverseonline.com/iit ... -plane.php
There are three bodies there -- (a) a body, sliding downward from right to left, (b) an inclined plane and (c) Earth as the inclined plane and Earth are firmly attached one to another thus forming one united whole.
-------------------------------------------------------------
2) An observer A is firmly attached to Earth and/or to the inclined plane. From the point of view of observer A the inclined plane and Earth are motionless and the the body is in motion. Observer A detects that the body exerts force G on the inclined plane. Besides obsever A resolves G into mgsinθ and mgcosθ.
-------------------------------------------------------------
3) An observer B is firmly attached to the body. From the point of view of observer B the body is motionless and the inclined plane (together with Earth) is in motion. Observer B detects that the inclined plane (together with Earth) exerts force - G on the body. Besides obsever B resolves - G into -mgsinθ and -mgcosθ.
-------------------------------------------------------------
4) According to the third Newton's law it is evident that:
a) G = - G, that is, G and - G are opposite in direction and equal in magnitude;
b) mgsinθ = -mgsinθ, that is, mgsinθ and -mgsinθ are opposite in direction and equal in magnitude;
c) mgcosθ = -mgcosθ, that is, mgcosθ and -mgcosθ are opposite in direction and equal in magnitude.
-------------------------------------------------------------
5) If angle θ is equal to 1° (or even smaller or a little bigger) for example, then G is practically equal to R (in absolute values) and then practically G = - G = R.
-------------------------------------------------------------
How to explain these simple things in a simpler and easier manner? Please stop imitate ignorance in such a clumsy and unskillfull manner! Your tricking method does not work!

[ME]

Sorry to say, but the only thing "evident" is that you are the ignorant and unskilled one here.

What's the first sign: You don't even attempt to explain what those force-factors are other than you following some formula.
Also an indicator: See all the red dots.

Besides obsever A resolves G into mgsinθ and mgcosθ.

If we have to point to something that causes that division, then it is simply that inclined plane that resolves it in such manner. The vector-sum remains "mg"

Know a tiny bit of physics and know when to add "-mgsinθ".
Because it's a smooth inclined plane we know it's not there.

"mgcosθ" is Newton's-3rd's reaction from a very sturdy incline.
That incline is simply unable to provide a full reaction force.
Hence there is a residual of "mgsinθ".

Because there is NO reaction possible for this residual force, that factor causes the box to be still in FREEFALL* by gravity.
Not in full, but partially.
We see that as an accelerated slide down the slope.
*Technically not "Free"-falling, but called it as such anyway to underline the reactionlessness of this vector "mgsinθ".

Some extreme examples that should ring a bell:
-----------------
Extreme situation 1:
-----------------
Let's say the plane is a horizontal table: θ=0°
This table is able to provide a reaction to the block's force "mg" IN FULL
Thus: mgcos0° = mg

Where does this block go?
mgsin 0° = 0
In other words: Absolutely nowhere!!

-----------------
Extreme situation 2:
-----------------
Let's say the plane is a vertical wall: θ=90°
This wall is unable to provide a reaction to the block's force "mg".
Thus: mg cos 90° = 0

Where does this block go?
mgsin 90° = mg
In other words: This block remains in total FREEFALL, despite that wall!!
---




What does this even mean:

If angle θ is equal to 1° (or even smaller or a little bigger) for example, then G is practically equal to R (in absolute values) and then practically G = - G = R.

The formula you try to construct is possibly:
mg - mgcos 1° ≈ 0.000152 mg

Yet this is a totally fabricated construct with no validity regarding the incline.