To ME and johannesbender.
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Within a long period of time you have been trying to convince all members of this forum that Sir Isaak Newton was an amateur scientist. That is why I am explaining again the basic concept. We modified STEP 1 for the second time. The second modification of STEP 1 is given below.
Read VERY, VERY CAREFULLY the text below!
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STEP 1 (modification 2)
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1) Look at the link https://www.knowledgeuniverseonline.com ... -plane.php
2) Body 1 is the block, which slides down the ramp (ONLY from the point of wiew of an observer, who is firmly attached to the ground/Earth).
3) The ramp and Earth are firmly attached one to another thus forming one united whole, which is called body 2.
4) Let us assume for a moment that gravity does not exist. Let us replace gravity with a long linear spring, whose both ends are attached to the centers of mass of bodies 1 and 2, respectively. The spring pulls one to another the centers of mass of bodies 1 and 2 and the two bodies move/slide relative to each other. FRICTION IS NEGLIGIBLE!
(Note. Look at the link https://en.wikipedia.org/wiki/Spring_(device)
Most suitable for us in this link is the spring with characteristic number 4.)
5) Under the influence of the spring body 1 exerts force F on body 2. (The absolute value |F| can be measured by a dynamometer, which is attached to the center of mass of body 1.) Force F is applied to the center of mass of body 1. Force F is directed to the center of mass of body 2.
6) Under the influence of the spring body 2 exerts force -F on body 1. (The absolute value |-F| can be measured by a dynamometer, which is attached to the center of mass of body 2.) Force -F is applied to the center of mass of body 2. Force -F is directed to the center of mass of body 1.
7) Forces F and -F have one and same line of action, which connects the center of mass of body 1 to the center of mass of body 2.
8) Forces F and -F are opposite in direction and equal in magnitude. (Equal in magnitude means that |F| = |-F|, that is, the absolute value of F is equal to the absolute value of -F. And this fact can be confirmed by the above mentioned two dynamometers, attached to the centers of mass of bodies 1 and 2, respectively.)
9) And now let us simply replace F and -F with G and -G, respectively. Nothing will change as a result of this simple operation. In one word, the effect, caused by the spring, is absolutely identical and absolutely equivalent to the effect, caused by the gravity. FRICTION IS NEGLIGIBLE!
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Do you have any objections against any of the above items 1 -- 9?
ANSWER EACH OF ITEMS 1 -- 9 SEPARATELY AND NUMBER EACH RESPONSE!
IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?
Moderator: scott
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
George,
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1) you are trying to manipulate the audience again and this fact is already obvious for anyone here in this forum (see your given red dots).
2) You are trying again and again one and same clumsy manipulation trick.
3) But yes, this new "variant" is how the contraction of a spring can be an equivalence of gravity without the incline.
4) There is an effect of the incline that's addressed in my previous reply.
5) Please address my previous reply you can find on the previous page.
-----------------------------------
1) you are trying to manipulate the audience again and this fact is already obvious for anyone here in this forum (see your given red dots).
2) You are trying again and again one and same clumsy manipulation trick.
3) But yes, this new "variant" is how the contraction of a spring can be an equivalence of gravity without the incline.
4) There is an effect of the incline that's addressed in my previous reply.
5) Please address my previous reply you can find on the previous page.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
To ME.
---------------------------------
1) You wrote:"3) But yes, this new "variant" is how the contraction of a spring can be an equivalence of gravity without the incline." Do you still support this claim of yours?
2) So you accept the fact that the effect caused by the spring is equivalent to the effect caused by the gravity, but only without the incline. In this case |F| = |-F| = |G| = |-G|. Did I understand correctly your point of view?
3) Now let us add the incline. In this case body 1 slides down the ramp (ONLY from the point of view of an observer, who is firmly attached to the ground/Earth). But though the sliding of body 1 down the ramp the two dynamometers will show the validity of the equality |F| = |-F| = |G| = |-G|, that is, the incline does not change anything. Keep in mind that most suitable for our consideration is the spring with characteristic number 4 from the link https://en.wikipedia.org/wiki/Spring_(device)
Do you agree with this item 3 and if no, then why?
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ANSWER EACH OF ITEMS 1 -- 3 SEPARATELY AND NUMBER EACH RESPONSE!
---------------------------------
1) You wrote:"3) But yes, this new "variant" is how the contraction of a spring can be an equivalence of gravity without the incline." Do you still support this claim of yours?
2) So you accept the fact that the effect caused by the spring is equivalent to the effect caused by the gravity, but only without the incline. In this case |F| = |-F| = |G| = |-G|. Did I understand correctly your point of view?
3) Now let us add the incline. In this case body 1 slides down the ramp (ONLY from the point of view of an observer, who is firmly attached to the ground/Earth). But though the sliding of body 1 down the ramp the two dynamometers will show the validity of the equality |F| = |-F| = |G| = |-G|, that is, the incline does not change anything. Keep in mind that most suitable for our consideration is the spring with characteristic number 4 from the link https://en.wikipedia.org/wiki/Spring_(device)
Do you agree with this item 3 and if no, then why?
----------------------------------
ANSWER EACH OF ITEMS 1 -- 3 SEPARATELY AND NUMBER EACH RESPONSE!
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
To ME.
-------------------------
A small addition. The equality |F| = |-F| = |G| = |-G| = const is more correct than the equality |F| = |-F| = |G| = |-G|.
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A small addition. The equality |F| = |-F| = |G| = |-G| = const is more correct than the equality |F| = |-F| = |G| = |-G|.
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
*) Please address my earlier reply you can find on the previous page.
So when the effect of that very long spring has an equivalent effect as gravity, then we don't have to make a detour by using a spring.
We can just state that there is a force acting on that box.
It could have been a magnet.
The only question we have to answer: What makes this box accelerate down that ramp?
*) Please address my earlier reply you can find on the previous page.
Let's say that the box is a magnet (box-magnet) that gets attracted by another magnet (ground-magnet) below the incline.
A force acts on that box-magnet, and two things can happen:
1) The box-magnet will slide down the ramp so it will position itself at a smaller distance.
2) The ramp (when not firmly attached to the ground) will be pushed aside, but also to cause that box-magnet to get closer to the ground-magnet.
When this magnetic force is sufficiently vertical then we can estimate the factor of acceleration of (1) this box sliding down the ramp or (2) this ramp sliding away.
a) We know, because of Newton's 3rd law of motion, that both objects (ground and box) will attract each other with equal and opposite force.
b) We know, because of Newton's 2rd law of motion, when they are free to move that the box (little mass) gains more acceleration than the ground (huge mass), so we can state that the inertial frame of reference is the ground.
c) We know, because of Newton's 1st law of motion, that the box will move because of this force acting on that box in the inertial frame.
d) The incline is a wedge between the box and the ground so the box is not completely free to move.
e) A part of the force is countered (mgcosθ), a part of this force is not countered (mgsinθ).
f) The relative acceleration between the box and the smooth incline will be gsinθ, no matter how you observe it.
I understand that point (a) is probably what you try to address, yet point(f) is what will happen.
*) Please address my earlier reply you can find on the previous page.
So when the effect of that very long spring has an equivalent effect as gravity, then we don't have to make a detour by using a spring.
We can just state that there is a force acting on that box.
It could have been a magnet.
The only question we have to answer: What makes this box accelerate down that ramp?
*) Please address my earlier reply you can find on the previous page.
Let's say that the box is a magnet (box-magnet) that gets attracted by another magnet (ground-magnet) below the incline.
A force acts on that box-magnet, and two things can happen:
1) The box-magnet will slide down the ramp so it will position itself at a smaller distance.
2) The ramp (when not firmly attached to the ground) will be pushed aside, but also to cause that box-magnet to get closer to the ground-magnet.
When this magnetic force is sufficiently vertical then we can estimate the factor of acceleration of (1) this box sliding down the ramp or (2) this ramp sliding away.
a) We know, because of Newton's 3rd law of motion, that both objects (ground and box) will attract each other with equal and opposite force.
b) We know, because of Newton's 2rd law of motion, when they are free to move that the box (little mass) gains more acceleration than the ground (huge mass), so we can state that the inertial frame of reference is the ground.
c) We know, because of Newton's 1st law of motion, that the box will move because of this force acting on that box in the inertial frame.
d) The incline is a wedge between the box and the ground so the box is not completely free to move.
e) A part of the force is countered (mgcosθ), a part of this force is not countered (mgsinθ).
f) The relative acceleration between the box and the smooth incline will be gsinθ, no matter how you observe it.
I understand that point (a) is probably what you try to address, yet point(f) is what will happen.
*) Please address my earlier reply you can find on the previous page.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
To ME.
-------------------------
1) Good answer indeed! This is already a serious and reasonable discussion! I would like to thank you for this!
2) Please consider the possibility when body 1 and body 2 have equal masses and equal inertia moments (not sizes, i.e. body 2 is still much bigger in size than body 1.) What would happen then? What would be the behaviour of bodies 1 and 2? How would they move relative one to another? What is your opinion? (The spring is in action and friction is negligible.)
Looking forward to your answer.
Regards,
George
-------------------------
1) Good answer indeed! This is already a serious and reasonable discussion! I would like to thank you for this!
2) Please consider the possibility when body 1 and body 2 have equal masses and equal inertia moments (not sizes, i.e. body 2 is still much bigger in size than body 1.) What would happen then? What would be the behaviour of bodies 1 and 2? How would they move relative one to another? What is your opinion? (The spring is in action and friction is negligible.)
Looking forward to your answer.
Regards,
George
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
Hi ME,
There is still no answer from you. What happens?
Our last post was only a transition to the final point. And here it is.
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1) Let us replace body 1 (the sliding block) with a twin of body 2. In other words, we have now two big identical balls and two identical ramps. (It is not necessary the balls to be with Earth's size. Each of the balls could have a diameter of 100 m, 10 m or even 1 m.
2) The sliding surfaces are the hypotenuses of the two identical ramps.
3) Body 1 and body 2 have one and same masses and inertia moments. (The masses of bodies 1 and 2 are not necessarily equal to Earth's mass. The mass of each of bodies 1 and 2 could be equal to, let's say, 10 kg for example.)
4) Friction is negligible. The spring is attached to the centers of mass of bodies 1 and 2 and contracts, thus causing the motion of bodies 1 and 2. The experiment is carried out under weightlessness conditions.
5) Observer A is firmly attached to body 2. In this case 1 from the point of view of observer A body 2 is "down" and is motionless and body 1 is "up" and slides "downward" at an angle θ to the horizon from right to left (as the sliding block in the initial example). Body 1 exerts force |F| = |G| on body 2 as the sliding block in the initial example.
6) Observer B is firmly attached to body 1. In this case 2 from the point of view of observer B body 1 is "down" and is motionless and body 2 is "up" and slides "downward" at an angle θ to the horizon from right to left (as the sliding block in the initial example). Body 2 exerts force |- F| = |- G| on body 1 as the sliding block in the initial example.
7) Observer C is motionless in relation to the longitudinal axis of symmetry of the spring. In this case 3 observer C sees that (a) the spring contracts, (b) the centers of mass of bodies 1 and 2 get nearer to each other and (c) each of bodies 1 and 2 rotates at some angle β. The two dynamometers will show that |F| = |G| = |- F| = |- G|, that is, body 1 exerts force |F| = |G| on body 2, and body 2 on its behalf exerts force |- F| = |- G| on body 1.
8) In one word, in the three cases 1, 2 and 3 the three observers A, B and C watch three different things but this does not change the basic fact that |F| = |G| = |- F| = |- G| = const. The last equality is a manifestation of the validity of the third Newton's law in this particular situation, that is, body 1 exerts force F (or G, respectively) on body 2 and body 2 on its behalf exerts force - F (or - G, respectively) on body 1 as forces F and - F (or G and - G, respectively) are equal in magnitude and opposite in direction. (Please keep in mind that that forces F and - F (or G and - G, respectively) are applied to the centers of mass of bodies 1 and 2, respectively.)
9) You can resolve forces F and - F (or G and - G, respectively) into as many vector components as you want. But this does not change the basic fact that |F| = |G| = |- F| = |- G| = const. Please look again at previous item 8, if you like.
10) While constructing your answer please focus on the spring-related case only. No magnets, no electrostatics, etc. Only spring!
-----------------------------------
What do you think about the above considerations? Seem to be correct, aren't they? What is your opinion? Please share it, if possible.
Looking forward to your answer.
George
There is still no answer from you. What happens?
Our last post was only a transition to the final point. And here it is.
-------------------------------------------
1) Let us replace body 1 (the sliding block) with a twin of body 2. In other words, we have now two big identical balls and two identical ramps. (It is not necessary the balls to be with Earth's size. Each of the balls could have a diameter of 100 m, 10 m or even 1 m.
2) The sliding surfaces are the hypotenuses of the two identical ramps.
3) Body 1 and body 2 have one and same masses and inertia moments. (The masses of bodies 1 and 2 are not necessarily equal to Earth's mass. The mass of each of bodies 1 and 2 could be equal to, let's say, 10 kg for example.)
4) Friction is negligible. The spring is attached to the centers of mass of bodies 1 and 2 and contracts, thus causing the motion of bodies 1 and 2. The experiment is carried out under weightlessness conditions.
5) Observer A is firmly attached to body 2. In this case 1 from the point of view of observer A body 2 is "down" and is motionless and body 1 is "up" and slides "downward" at an angle θ to the horizon from right to left (as the sliding block in the initial example). Body 1 exerts force |F| = |G| on body 2 as the sliding block in the initial example.
6) Observer B is firmly attached to body 1. In this case 2 from the point of view of observer B body 1 is "down" and is motionless and body 2 is "up" and slides "downward" at an angle θ to the horizon from right to left (as the sliding block in the initial example). Body 2 exerts force |- F| = |- G| on body 1 as the sliding block in the initial example.
7) Observer C is motionless in relation to the longitudinal axis of symmetry of the spring. In this case 3 observer C sees that (a) the spring contracts, (b) the centers of mass of bodies 1 and 2 get nearer to each other and (c) each of bodies 1 and 2 rotates at some angle β. The two dynamometers will show that |F| = |G| = |- F| = |- G|, that is, body 1 exerts force |F| = |G| on body 2, and body 2 on its behalf exerts force |- F| = |- G| on body 1.
8) In one word, in the three cases 1, 2 and 3 the three observers A, B and C watch three different things but this does not change the basic fact that |F| = |G| = |- F| = |- G| = const. The last equality is a manifestation of the validity of the third Newton's law in this particular situation, that is, body 1 exerts force F (or G, respectively) on body 2 and body 2 on its behalf exerts force - F (or - G, respectively) on body 1 as forces F and - F (or G and - G, respectively) are equal in magnitude and opposite in direction. (Please keep in mind that that forces F and - F (or G and - G, respectively) are applied to the centers of mass of bodies 1 and 2, respectively.)
9) You can resolve forces F and - F (or G and - G, respectively) into as many vector components as you want. But this does not change the basic fact that |F| = |G| = |- F| = |- G| = const. Please look again at previous item 8, if you like.
10) While constructing your answer please focus on the spring-related case only. No magnets, no electrostatics, etc. Only spring!
-----------------------------------
What do you think about the above considerations? Seem to be correct, aren't they? What is your opinion? Please share it, if possible.
Looking forward to your answer.
George
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
Perhaps you could draw a picture for the newest situation. It's hard to follow.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
1) The sliding block is replaced with a ramp which is identical to the initial ramp. The second "upper" ramp slides on the "lower" ramp which is fixed to the ground. The sliding process occurs "hypotenuse on hypotenuse".
2) After that you simply attach the second big ball to the "upper cathetus" of the "upper ramp".
3) The equatorial cross-section of each of the two identical conglomerates ball-ramp resembles a toothed wheel with only one triangular tooth. The two triangular teeth touch and slide on their "hypotenuse" parts.
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Seems to be clear? Please ask questions, if any.
2) After that you simply attach the second big ball to the "upper cathetus" of the "upper ramp".
3) The equatorial cross-section of each of the two identical conglomerates ball-ramp resembles a toothed wheel with only one triangular tooth. The two triangular teeth touch and slide on their "hypotenuse" parts.
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Seems to be clear? Please ask questions, if any.
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
Nope sorry, maybe someone else is able.
Draw a picture and then I may look at it.
Draw a picture and then I may look at it.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
I will do the picture in the very nearest future. Meanwhile let us get back to the previous variation which is given below again.
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Please consider the possibility when body 1 and body 2 have equal masses and equal inertia moments (not sizes, i.e. body 2 is still much bigger in size than body 1.) What would happen then? What would be the behaviour of bodies 1 and 2? How would they move relative one to another? What is your opinion? (The spring is in action and friction is negligible.)
Looking forward to your answer.
Regards,
George
----------------------------------
Please consider the possibility when body 1 and body 2 have equal masses and equal inertia moments (not sizes, i.e. body 2 is still much bigger in size than body 1.) What would happen then? What would be the behaviour of bodies 1 and 2? How would they move relative one to another? What is your opinion? (The spring is in action and friction is negligible.)
Looking forward to your answer.
Regards,
George
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
To ME and to all other experts in this forum who are interested in the problem.
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Let us further simplify the experiment described in our last post.
1) Let us make comparable the sizes of the bodies 1 and 2.
2) Let us assume the sliding block (body 1) has dimensions of 0.1m by 0.1m by 0.2m.
3) Let us assume that the ramp's cross-section is a rectangular triangle which has dimensions of 0.5m by 0.4m by 0.3m.
4) Let us replace Earth with a homogeneous ball, whose diameter is equal to 1m.
5) The ramp and the ball from previous items 3 and 4 are firmly attached one to another thus forming one united whole, which is called body 2.
6) The mass of body 1 is equal to the mass of body 2, that is, let us assume for example that the mass of body 1 is 10 kg and the mass of body 2 is 10 kg too.
7) The inertia moment of body 1 is equal to the inertia moment of body 2.
8) A spring is attached to the centers of mass of bodies 1 and 2. The spring contracts and causes the motion of bodies 1 and 2.
9) Friction is negligible. The experiment is carried out in a space station under weightlessness conditions.
10) Two identical dynamometers are attached to the two ends of the spring, respectively.
11) The question is what would an astronaut see while the spring contracts. The answer is that the astronaut would see 5 simultaneous things.
11-1) The distance between the centers of mass of bodies 1 and 2 decreases, that is, the center of mass of body 1 moves towards the center of mass of body 2 and the center of mass of body 2 on its behalf moves towards the center of mass of body 1.
11-2) Each of the two identical dynamometers registers a force of, let's say, 10 N for example.
11-3) Body 1 rotates at some angle α. The center of rotation is the center of mass of body 1.
11-4) Body 2 rotates at some angle β. The center of rotation is the center of mass of body 2.
11-5) The longitudinal axis of symmetry of the spring rotates at some angle γ. The center of rotation of the spring's longitudinal axis of symmetry is the common center of mass of bodies 1 and 2.
12) In one word, the above described 5 simultaneous events unambiguously show that under the influence of the spring body 1 exerts force F on body 2 and body 2 on its behalf exerts force - F on body 1 as forces F and - F are opposite in direction and equal in magnitude, that is, |F| = |-F| = 10 N = const. (Please remember that we have chosen a spring of characteristic 4 as described in the related link in one of our previous posts.) Actually this is a manifestation of the third Newton's law for this particular case.
13) The combined motion of bodies 1 and 2 does not violate and does not influence in any way the validity of item 12.
14) You can resolve forces F and - F into as many vector components as you want, but this resolving does not violate and does not influence in any way the validity of item 12.
15) You can greatly vary with (a) forms, (b) sizes, (c) masses and (d) inertia moments of bodies 1 and 2, but all these variations (which are practically infinite in number) do not violate and do not influence in any way the validity of item 12. (One of these numerous variations is our initial example case in which the ball (Earth) is much bigger and much heavier than the block.)
16) You can replace letter F with G, K, L, N, T, Q, etc., but this replacement does not violate and does not influence in any way the validity of item 12.
17) Please focus on the spring-related case only. No magnets, no electrostatics, etc. Only spring!
===========================================
What do you think about the above considerations? Seem to be correct, don't they? What is your opinion?
Looking forward to your answer.
George
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Let us further simplify the experiment described in our last post.
1) Let us make comparable the sizes of the bodies 1 and 2.
2) Let us assume the sliding block (body 1) has dimensions of 0.1m by 0.1m by 0.2m.
3) Let us assume that the ramp's cross-section is a rectangular triangle which has dimensions of 0.5m by 0.4m by 0.3m.
4) Let us replace Earth with a homogeneous ball, whose diameter is equal to 1m.
5) The ramp and the ball from previous items 3 and 4 are firmly attached one to another thus forming one united whole, which is called body 2.
6) The mass of body 1 is equal to the mass of body 2, that is, let us assume for example that the mass of body 1 is 10 kg and the mass of body 2 is 10 kg too.
7) The inertia moment of body 1 is equal to the inertia moment of body 2.
8) A spring is attached to the centers of mass of bodies 1 and 2. The spring contracts and causes the motion of bodies 1 and 2.
9) Friction is negligible. The experiment is carried out in a space station under weightlessness conditions.
10) Two identical dynamometers are attached to the two ends of the spring, respectively.
11) The question is what would an astronaut see while the spring contracts. The answer is that the astronaut would see 5 simultaneous things.
11-1) The distance between the centers of mass of bodies 1 and 2 decreases, that is, the center of mass of body 1 moves towards the center of mass of body 2 and the center of mass of body 2 on its behalf moves towards the center of mass of body 1.
11-2) Each of the two identical dynamometers registers a force of, let's say, 10 N for example.
11-3) Body 1 rotates at some angle α. The center of rotation is the center of mass of body 1.
11-4) Body 2 rotates at some angle β. The center of rotation is the center of mass of body 2.
11-5) The longitudinal axis of symmetry of the spring rotates at some angle γ. The center of rotation of the spring's longitudinal axis of symmetry is the common center of mass of bodies 1 and 2.
12) In one word, the above described 5 simultaneous events unambiguously show that under the influence of the spring body 1 exerts force F on body 2 and body 2 on its behalf exerts force - F on body 1 as forces F and - F are opposite in direction and equal in magnitude, that is, |F| = |-F| = 10 N = const. (Please remember that we have chosen a spring of characteristic 4 as described in the related link in one of our previous posts.) Actually this is a manifestation of the third Newton's law for this particular case.
13) The combined motion of bodies 1 and 2 does not violate and does not influence in any way the validity of item 12.
14) You can resolve forces F and - F into as many vector components as you want, but this resolving does not violate and does not influence in any way the validity of item 12.
15) You can greatly vary with (a) forms, (b) sizes, (c) masses and (d) inertia moments of bodies 1 and 2, but all these variations (which are practically infinite in number) do not violate and do not influence in any way the validity of item 12. (One of these numerous variations is our initial example case in which the ball (Earth) is much bigger and much heavier than the block.)
16) You can replace letter F with G, K, L, N, T, Q, etc., but this replacement does not violate and does not influence in any way the validity of item 12.
17) Please focus on the spring-related case only. No magnets, no electrostatics, etc. Only spring!
===========================================
What do you think about the above considerations? Seem to be correct, don't they? What is your opinion?
Looking forward to your answer.
George
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
George, it's all very nice that you want to run off with this new variant/contraption but i'm not really convinced that my previous answer was well received and understood.
A reply like "good answers" isn't really convincing.
Especially when the newest idea is about a "ramp upon a ramp", or "hypotenuse on hypotenuse".
This actually indicates that you still don't really grasp the principle of a Normal force, the surface pressure and the force resultant - which all form part of disagreeing with your request to agree with some statement.
Maybe you actually do understand and you try to focus on something else... but that's not really clear.
Perhaps we could solve that communication issue first.
I think the following would really help:
A reply like "good answers" isn't really convincing.
Especially when the newest idea is about a "ramp upon a ramp", or "hypotenuse on hypotenuse".
This actually indicates that you still don't really grasp the principle of a Normal force, the surface pressure and the force resultant - which all form part of disagreeing with your request to agree with some statement.
Maybe you actually do understand and you try to focus on something else... but that's not really clear.
Perhaps we could solve that communication issue first.
I think the following would really help:
And my request was made because a diagram is a useful tool. First for yourself and second for everyone else.George1 wrote:I will do the picture in the very nearest future
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
Hi ME,
Thank you for your reply.
1) About the picture. Well, actually the intention was to replace the block with а mirror reflection of the conglomerate Earth-ramp. Too sophisticated and complex.
2) About the rest of the text of your last post. Actually I can't understand what are you talking about.
2A) If the situation is "flat" (that is, θ = 0), then bodies 1 and 2 do not move and the two dynamometers register forces F and - F, respectively.
2B) If the situation is "inclined" (that is, θ > 0), then bodies 1 and 2 move relative one to another (combined motion) and the two dynamometers register AGAIN forces F and - F, respectively.
--------------------
Question:what is d the difference between 2A and 2B?
Thank you for your reply.
1) About the picture. Well, actually the intention was to replace the block with а mirror reflection of the conglomerate Earth-ramp. Too sophisticated and complex.
2) About the rest of the text of your last post. Actually I can't understand what are you talking about.
2A) If the situation is "flat" (that is, θ = 0), then bodies 1 and 2 do not move and the two dynamometers register forces F and - F, respectively.
2B) If the situation is "inclined" (that is, θ > 0), then bodies 1 and 2 move relative one to another (combined motion) and the two dynamometers register AGAIN forces F and - F, respectively.
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Question:what is d the difference between 2A and 2B?
re: IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHI
After reading a resent book. I came across the phrase 'Don't Feed the Troll'.
When a member has a bad rating then it's best to put the member into the ignore list.
That way you 'Don't Feed the Troll'.
All the Best
When a member has a bad rating then it's best to put the member into the ignore list.
That way you 'Don't Feed the Troll'.
All the Best
[MP] Mobiles that perpetuate - external energy allowed