Naked Science Forum
On the Lighter Side => New Theories => Topic started by: Yahya on 20/07/2018 23:00:58
-
The equation :
F=ma won't apply to this situation.
Both mass M and mass m are at free fall , they will have the acceleration A and a according to pulley ratio since they start from speed zero meters per second and the speed ratio is always the same according to pulleys' ratios.
I think of the same system as follows :
The mass M has acceleration say 4 m/s.s after three intervals of time 1, 2,3 seconds its speed becomes 4 , 8, 12 m/s respectively think of any other system connected in such way , let say pulley ratio is 1:4
then when pulley #1 has speed 4m/s then P2 will have 16 m/s according to pulley ratio , when P1 has 8 m/s P2 has 32 m/s , P1 has 12 m/s , P2 has 12*4=48 m/s , let's calculate acceleration on P2:
time intervals are 1,2,3 speeds are 16,32,48 respectively , a=16 m/s.s which is =A*4
At that case the equation F=ma won't satisfy for mass m, i.e force on mass m won't equal to its mass multiplied by acceleration:
Let's have mass M is 4 times mass m. M=4m, let the ratio this time be 1:6
force on mass M is its weight minus gm/6=gM-gm/6
A=F/M=(gM-gm/6)/M=(gM-gM/24)/M= (1-1/24)g
force on mass m is 6gM-gm
a=F/m=(24gm-gm)/m=23g
A is not equal to 6 times a as it should be above.
“The reason is because the third law won't apply as well , objects do not exert the same force however a force according to the pulley ratio , another law should be presented for such situations where the action and reaction are not equal, objects M and m do not touch directly so that there is equal action and reaction"
so will the second law apply? why not?
-
The equation :
F=ma won't apply to this situation.
Both mass M and mass m are at free fall , they will have the acceleration A and a according to pulley ratio since they start from speed zero meters per second and the speed ratio is always the same according to pulleys' ratios.
I think of the same system as follows :
The mass M has acceleration say 4 m/s.s after three intervals of time 1, 2,3 seconds its speed becomes 4 , 8, 12 m/s respectively think of any other system connected in such way , let say pulley ratio is 1:4
then when pulley #1 has speed 4m/s then P2 will have 16 m/s according to pulley ratio , when P1 has 8 m/s P2 has 32 m/s , P1 has 12 m/s , P2 has 12*4=48 m/s , let's calculate acceleration on P2:
time intervals are 1,2,3 speeds are 16,32,48 respectively , a=16 m/s.s which is =A*4
At that case the equation F=ma won't satisfy for mass m, i.e force on mass m won't equal to its mass multiplied by acceleration:
Let's have mass M is 4 times mass m. M=4m, let the ratio this time be 1:6
force on mass M is its weight minus gm/6=gM-gm/6
A=F/M=(gM-gm/6)/M=(gM-gM/24)/M= (1-1/24)g
force on mass m is 6gM-gm
a=F/m=(24gm-gm)/m=23g
A is not equal to 6 times a as it should be above.
“The reason is because the third law won't apply as well , objects do not exert the same force however a force according to the pulley ratio , another law should be presented for such situations where the action and reaction are not equal, objects M and m do not touch directly so that there is equal action and reaction"
so will the second law apply? why not?
Looks more like a friend type relationship to me, nice try at the drawing out of your ideas, shame they have no real weight . Perhaps you should add something electrical to your ideas like charges. Make up some charges even , why not hey if it suits . But expect the charge to repel back or escape into space , the charge may not help you at all, electrons bond protons , protons are only repulsive to protons because of negative energy between them .
M is strong where m is weaker, M holds onto m and wherever M goes , m follows because of G . You can't see an invisible link unless shown the link , a strong bond does not need contact necessarily, but relationships can be var x and incrimination of x in cyber space is meaningless and invalid .
The particle differential is fascinating hey? Disabled particles don't lie about their spatial position, under a lot of gravity and stress they are mainly fixed to a position , however in dealing with this position , the lead particle M also becomes severe stressed to a tension level where it also becomes fixed, the amount of energy levels, helping with some freedom of the particles .
Particles can't traverse without enough energy to traverse.
-
Darn't it, I gone and got myself all upset again and you know what , science is just not worth me. It is not me who should be begging science for a job , it should be them begging me to come work with them . I put more effort into my science than most, for what ?
Hope , dreams, God, not even worthy of being words, can you believe I was praying to God to win some money just so I could take my children away ......I must be insane for praying. I wish I could back in time, I would have not had children, I would of been a full bore super criminal instead.
Enough , had my moan...
-
Both mass M and mass m are at free fall
No.
They are not in free fall.
They have things tied to them.
It would be sensible for you to learn physics before trying to claim that physics is wrong.
-
will the second law apply?
What do you think "law" means?
-
They have things tied to them.
Tied , can be such a loose term to use. Subjective is subjective....
-
They have things tied to them.
Tied can be such a loose term....
Shut up, you useless idiot.
-
They have things tied to them.
Tied can be such a loose term....
Shut up, you useless idiot.
Yup that is me, useless idiot, what gives you the right to tell me to shut up?
Are you drunk again?
I love your banter, don't run off.
-
Yup that is me, useless idiot, what gives you the right to tell me to shut up?
Since you accept that your "contributions" are useless, you should stop wasting others' time + bandwidth.
There's nothing special about me telling you that- you should already know it. I'm just reminding you because I happen to be here.
-
Yup that is me, useless idiot, what gives you the right to tell me to shut up?
Since you accept that your "contributions" are useless, you should stop wasting others' time + bandwidth.
There's nothing special about me telling you that- you should already know it. I'm just reminding you because I happen to be here.
I know I am useless , but does that mean I cant have a conversation ? I have nothing else in my life but my kids and conversation.
-
Cluttering up people's threads isn't conversation, it's vandalism.
-
Cluttering up people's threads isn't conversation, it's vandalism.
You started the vandalism....not me, well maybe it was me when I got to the feeling sorry for myself. Stuff it my computer is going tomorrow , I am going have the internet turned off . Happy now ?
-
You started the vandalism....not me, well maybe it was me when I got to the feeling sorry for myself.
Nope
This "Tied , can be such a loose term to use. Subjective is subjective...." wasn't meaningful- it just cluttered the forum.
-
You started the vandalism....not me, well maybe it was me when I got to the feeling sorry for myself.
Nope
This "Tied , can be such a loose term to use. Subjective is subjective...." wasn't meaningful- it just cluttered the forum.
When objects are tied together by gravity , they can be viewed loosely, consider any object on the ground, it is loose relative to ground, do you even understand gravity ?
I think your scientific understanding is loose and not precise at all. My science is more precise, I account for all the mass. I do not think you account for all the mass, my theories is much tighter.
-
Both mass M and mass m are at free fall
No.
They are not in free fall.
They have things tied to them.
It would be sensible for you to learn physics before trying to claim that physics is wrong.
I know what free fall is I meant it in a deeper sense, the only external force affecting on the system they consist of is gravity and each other force affects on each one of them is due to gravity , and I didn't say physics is wrong here, I said the law does not apply to this situation , that does not mean it is wrong, that does not mean it should be replaced but extended as I mentioned
so your answer is Yes .Could you explain my results?
-
Objects in free fall towards a much larger mass (such as the earth) have exactly the same acceleration. If they don't, then one or other (or both) is not in free fall, by definiition.
Rather than get involved in complex arithmetic, consider Galileo's thought experiment. If rock A is heavier and falls faster than rock B, what happens if we tie them together? Does B slow down A or does A speed up B? How do they choose?
Let's suppose A speeds up B. Obviously B is reluctant to travel as fast as A so it exerts some drag, so speed (A + B) < A.
Now instead of string, we use glue. Thus since mass (AB) > mass (A), speed (AB) > A.
So the speed of a falling object depends on whether it is held together by string or glue.
Bollocks (a loose translation of Dialogo sopra i due massimi sistemi del mondo (1632)).
-
I know what free fall is I meant it in a deeper sense,
If you choose to use a phrase like "free fall"
https://en.wikipedia.org/wiki/Free_fall
which has a well defined meaning, for something else you are being foolish.
Either use "free fall" to mean what everyone understands it to mean, or use some other word or phrase.
Don't try to tell us that your use of it is somehow "better".
Try again, but this time try to communicate your meanng properly.
-
I said the law does not apply to this situation
Spoiler alert; it does.
-
I said the law does not apply to this situation
Spoiler alert; it does.
it doesn't think of a gearbox for speed reduction with very low speed and very high torque , the output gear will have acceleration relative to the input gear according to gear ratio , but the output gear will have very large torque and very large force , in that case F=ma for the output gear won't satisfy. because when I increase gear ratio mass is constant , but acceleration decreases and force increase:
F=ma m constant , acceleration decreases F should decrease as well ,but F increase according to gearbox ratio " torque " then the equation is invalid.
-
If you define F as the number of rhododendrons in my local park, M as the number of gospel saints and A as the average body-weight of a duck then there's very little chance that F=MA.
However, if you use the conventional definitions where they represent the force acting on; the mass of, and the acceleration of THE SAME OBJECT then the equation holds.
By looking at the force on one cog, but the acceleration of another you are simply not using the equation correctly.
If you don't make that stupid mistake, then F=MA.
-
However, if you use the conventional definitions where they represent the force acting on; the mass of, and the acceleration of THE SAME OBJECT then the equation holds.
By looking at the force on one cog, but the acceleration of another you are simply not using the equation correctly.
No I didn't , acceleration on the output gear is 1/(gear ratio) times acceleration on the input gear, but force is (gear ratio)*force on input gear , when increasing gear ratio acceleration on output gear decreases but on the same gear the force increases, but that makes the equation F=ma invalid.
Also the third law does not apply , because I exert force on mass on output gear by touching input gear but the output mass do not push me with the same force , however with a force related to gear ratio, inertia of the output mass would be small.
-
Your other problem is that you failed to account for the fact that F is the net force- it's the sum of the forces acting on an object.
Feel free to learn some physics + come back.
-
Your other problem is that you failed to account for the fact that F is the net force- it's the sum of the forces acting on an object.
Feel free to learn some physics + come back.
Let me fill a gap in your formal education , I meant the net force , if I have torque of 50 N.m and I put a mass at the gear edge 1 meter away from the center , then the net force is 50 newtons I can lift a 50 newton weight with this force and there are not other forces to be added or subtracted, are there ?. but this 50 newton doesn't give the actual acceleration which should be according to F=ma
another mass at the input gear will obtain the actual acceleration but inertia of first mass on it is smaller than normal.
The real problem is inertia is different than the actual value when mass m affects on mass M through gearbox , force of mass m on mass M is different than the reaction of mass M on mass m.
-
The equation in such case should be :
net force = (ratio-1)F1+F2
F2= mass times acceleration of the output gear.
F1=force on the input gear
if the ratio is 1:1 then the net force only equals F2= ma
-
At a quick glance, and your posts don't seem to deserve much more, it seems you have forgotten to take account of the forces at the axles/ bearings on which the gears turn.
-
At a quick glance, and your posts don't seem to deserve much more, it seems you have forgotten to take account of the forces at the axles/ bearings on which the gears turn.
These forces are irrelevant .
-
These forces are irrelevant .
LOL
They are the reason why the forces you look at don't seem to follow Newton's laws.
-
These forces are irrelevant .
LOL
They are the reason why the forces you look at don't seem to follow Newton's laws.
The reason is not important , but they are in fact do not follow Newton's laws.
-
Newton's law applies to the net force on an object.
If you don't consider all the forces then you are not calculating the net force and you are not using the law correctly.
It's like trying to calculate the speed at which a car rolls down a hill, but forgetting that the brakes are on.
-
Newton's law applies to the net force on an object.
If you don't consider all the forces then you are not calculating the net force and you are not using the law correctly.
It's like trying to calculate the speed at which a car rolls down a hill, but forgetting that the brakes are on.
So what is the net force ? is it just F=ma ? an output gear in gearbox has much force than F=ma .
The equation in such case should be :
Net force = (ratio-1)F1+F2
F2= mass times acceleration of the output gear.
F1=force on the input gear
if the ratio is 1:1 then the net force only equals F2= ma
-
So what is the net force
It is the vector sum of the forces acting on a body.
The fact that you needed to ask that shows (yet again) that you have no idea what you are talking about.
Go and learn some science.
While you are at it, go + reply to the thread you left dangling here
https://www.thenakedscientists.com/forum/index.php?topic=73150.msg549322#msg549322
-
Newton's law applies to the net force on an object.
If you don't consider all the forces then you are not calculating the net force and you are not using the law correctly.
It's like trying to calculate the speed at which a car rolls down a hill, but forgetting that the brakes are on.
So what is the net force for this specific situation? is it just F=ma ? an output gear in gearbox has much force than F=ma .
The equation in such case should be :
Net force = (ratio-1)F1+F2
F2= mass times acceleration of the output gear.
F1=force on the input gear
if the ratio is 1:1 then the net force only equals F2= ma
-
The equation :
F=ma won't apply to this situation.
Both mass M and mass m are at free fall , they will have the acceleration A and a according to pulley ratio since they start from speed zero meters per second and the speed ratio is always the same according to pulleys' ratios.
I think of the same system as follows :
The mass M has acceleration say 4 m/s.s after three intervals of time 1, 2,3 seconds its speed becomes 4 , 8, 12 m/s respectively think of any other system connected in such way , let say pulley ratio is 1:4
then when pulley #1 has speed 4m/s then P2 will have 16 m/s according to pulley ratio , when P1 has 8 m/s P2 has 32 m/s , P1 has 12 m/s , P2 has 12*4=48 m/s , let's calculate acceleration on P2:
time intervals are 1,2,3 speeds are 16,32,48 respectively , a=16 m/s.s which is =A*4
At that case the equation F=ma won't satisfy for mass m, i.e force on mass m won't equal to its mass multiplied by acceleration:
Let's have mass M is 4 times mass m. M=4m, let the ratio this time be 1:6
force on mass M is its weight minus gm/6=gM-gm/6
A=F/M=(gM-gm/6)/M=(gM-gM/24)/M= (1-1/24)g
force on mass m is 6gM-gm
a=F/m=(24gm-gm)/m=23g
A is not equal to 6 times a as it should be above.
“The reason is because the third law won't apply as well , objects do not exert the same force however a force according to the pulley ratio , another law should be presented for such situations where the action and reaction are not equal, objects M and m do not touch directly so that there is equal action and reaction"
so will the second law apply? why not?
Dirac. As two bodies as mass attract they .... in my paper:
The idea of negative energy associated to gravitational energy is simple: as the strength of the gravitational attraction between two objects represents the amount of gravitational energy [9] in the field which attracts them towards each other, when two objects are infinitely far apart the gravitational attraction and hence energy is close to zero. Yet when the two objects move towards each other, their motion accelerates by their mutual effect of gravity which causes an increase in the positive kinetic energy of the system. Yet, at the same time, the gravitational attraction (and thus energy) also increases in magnitude. The problem here is that the law of energy conservation [10] requires that the net energy of the system cannot change. Therefore, the change in gravitational energy must be negative to cancel out the positive change in kinetic energy. Paradoxically though, as the gravitational energy is getting stronger, this decrease can only mean that it is negative.
<<<<<<<COMMERCIAL LINK REMOVED>>>>>>>
-
So what is the net force for this specific situation?
It is the vector sum of the forces acting on a body.
... in any application.
All the forces acting on the cog- so the input force and also the output force as well as the forces acting at the bearing.
You can find it out by dividing the acceleration by the mass because (as always ) Newton's law applies.
Seriously; go and learn some science- it will help you answer questions like that for yourself and it will stop you posting silly things like you did in this thread.
https://www.thenakedscientists.com/forum/index.php?topic=73150.msg549322#msg549322
and this one
https://www.thenakedscientists.com/forum/index.php?topic=74259.0
Try this for a start
https://www.khanacademy.org/
(also, if you want to learn science, ignore opportunity- he is trying to put forward some "new theory", but he doesn't seem to know what he's on about. At best, it's irrelevant and he should know better than to clutter up mainstream threads with it.)
-
Seriously; go and learn some science
Would you stop this ? if you think I'm not educated enough then don't discuss with me.
-
if you think I'm not educated enough then don't discuss with me.
It's not that you lack the education to discuss; but you are not up to pontificating.
And it's not a matter of "I think" you should learn science, it's what the evidence says.
So what is the net force ?
-
if you think I'm not educated enough then don't discuss with me.
It's not that you lack the education to discuss; but you are not up to pontificating.
And it's not a matter of "I think" you should learn science, it's what the evidence says.
So what is the net force ?
It is just a matter of language , there is a difference between what is net force in which I want you to define it for me and what is the net force in which I want to test your thought about the exact amount for it for the specific situation and the contest shows that .
You just post without reading.
-
if you think I'm not educated enough then don't discuss with me.
It's not that you lack the education to discuss; but you are not up to pontificating.
And it's not a matter of "I think" you should learn science, it's what the evidence says.
So what is the net force ?
It is just a matter of language , there is a difference between what is net force in which I want you to define it for me and what is the net force in which I want to test your thought about the exact amount for it for the specific situation and the contest shows that you just post without reading.
No.
It's not the language.
It's the fact that you didn't know that you have to add all the forces together to get a resultant before you can use F=MA.
And it's the fact that you are pretending that you know about this stuff.
-
And it's the fact that you are pretending that you know about this stuff.
I know free fall I know net force I know all classical mechanics
-
And it's the fact that you are pretending that you know about this stuff.
I know free fall I know net force I know all classical mechanics
Then why do you keep getting it wrong?
Maybe it's this
And it's the fact that you are pretending that you know about this stuff.
-
And it's the fact that you are pretending that you know about this stuff.
I know free fall I know net force I know all classical mechanics
Then why do you keep getting it wrong?
Maybe it's this
And it's the fact that you are pretending that you know about this stuff.
I get it right and the net force is bigger than F=ma
-
I get it right and the net force is bigger than F=ma
So you think you are right and Newton got it wrong.
Do you think that's likely?
-
I get it right and the net force is bigger than F=ma
So you think you are right and Newton got it wrong.
Do you think that's likely?
He might missed it.
-
I get it right and the net force is bigger than F=ma
So you think you are right and Newton got it wrong.
Do you think that's likely?
He might missed it.
And, in spite of the fact that practically every engineer and physicist since then has used gears many times, and studied them + done exam questions about them, you are the first to notice the mistake in 400 years or so?
Do you think that's likely?
you are pretending that you know about this stuff.
-
I get it right and the net force is bigger than F=ma
So you think you are right and Newton got it wrong.
Do you think that's likely?
He might missed it.
you are the first to notice the mistake in 400 years or so?
Do you think that's likely?
I'm actually the first one to notice it
-
I'm actually the first one to notice it
Do you know that, when Newton published his stuff, he wasn't popular? Lots of the scientists of the day tried to prove him wrong.
And you- in spite of the fact that you have been told why you are wrong (and Newton is right)- still think that you are the only person in roughly 400 years, to spot that he made a mistake in one of the best known, and most widely used equations in physics.
OK,
What are you drinking/ smoking/ snorting/ whatever?
-
I'm actually the first one to notice it
Do you know that, when Newton published his stuff, he wasn't popular? Lots of the scientists of the day tried to prove him wrong.
And you- in spite of the fact that you have been told why you are wrong (and Newton is right)- still think that you are the only person in roughly 400 years, to spot that he made a mistake in one of the best known, and most widely used equations in physics.
OK,
What are you drinking/ smoking/ snorting/ whatever?
I do not understand your first two paragraphs.
-
I do not understand your first two paragraphs.
There is much you don't understand.
Have you heard of this
https://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect
-
Forget about Newton who lived 400 years ago Just tell me what is the net force:
I have a gear box of speed reduction of ratio 1:6 , I have input force of 1 N the output force would be 6 newton, the acceleration would be 1/6 of the acceleration of the input I calculated it here before :
" The mass M has acceleration say 4 m/s.s after three intervals of time 1, 2,3 seconds its speed becomes 4 , 8, 12 m/s respectively think of any other system connected in such way , let say pulley ratio is 1:4
then when pulley #1 has speed 4m/s then P2 will have 16 m/s according to pulley ratio , when P1 has 8 m/s P2 has 32 m/s , P1 has 12 m/s , P2 has 12*4=48 m/s , let's calculate acceleration on P2:
time intervals are 1,2,3 speeds are 16,32,48 respectively , a=16 m/s.s which is =A*4"
I have 1 kg mass for the input, let me summarize it:
gearbox for speed reduction ratio 1:6
input mass 1 kg force 1 newton a=1 m/s.s
output mass 2 kg force 6 N a= 1/6 m/s.s
So what is net force on mass 2 kg ? it is obvious that it is 6 Newtons because I could lift a weight of 6 N with the output gear, however the gearbox is for speed reduction so the acceleration is 1/6 m/s.s now F=ma:
6=2/6 that doesn't equal , you will notice huge force on the gear " net force " but slow speed " or acceleration " in a way that you think the gear is pushed with huge force but another thing avoid it from accelerating.
-
Mass on gearbox for speed reduction undergoes large "net force because I could lift weight with this force " but it seems there is something stopping it from accelerating in a way F=ma is not valid.
-
So what is net force on mass 2 kg ? it is obvious that it is 6 Newtons because I could lift a weight of 6 N with the output gear, however the gearbox is for speed reduction so the acceleration is 1/6 m/s.s now F=ma:
This is where you are going wrong. Although the gear could apply 6N it isn’t because the acceleration of the mass is constrained by the speed limitation.
If you connect the gear to the 2kg via a horizontal string (to remove red herring of freefall and gravity) and a spring balance then the gear will accelerate the mass at 1/6m/s2 and (ignoring friction) the force recorded by the balance will show F=ma ie 0.333=2/6
Mass on gearbox for speed reduction undergoes large "net force because I could lift weight with this force " but it seems there is something stopping it from accelerating in a way F=ma is not valid.
This is because you are calculating under no-load conditions. If you apply a 6N load at the output the reaction force will be reflected back through the gear chain to the input and reduce the net force on the input and hence the acceleration.
@Bored chemist has been trying to explain all this to you.
-
So what is net force on mass 2 kg ? it is obvious that it is 6 Newtons because I could lift a weight of 6 N with the output gear, however the gearbox is for speed reduction so the acceleration is 1/6 m/s.s now F=ma:
This is where you are going wrong. Although the gear could apply 6N it isn’t because the acceleration of the mass is constrained by the speed limitation.
If you connect the gear to the 2kg via a horizontal string (to remove red herring of freefall and gravity) and a spring balance then the gear will accelerate the mass at 1/6m/s2 and (ignoring friction) the force recorded by the balance will show F=ma ie 0.333=2/6
The force recorded by the balance won't read 2/6 N it would read 6 N.
-
So what is net force on mass 2 kg ? it is obvious that it is 6 Newtons because I could lift a weight of 6 N with the output gear, however the gearbox is for speed reduction so the acceleration is 1/6 m/s.s now F=ma:
This is where you are going wrong. Although the gear could apply 6N it isn’t because the acceleration of the mass is constrained by the speed limitation.
If you connect the gear to the 2kg via a horizontal string (to remove red herring of freefall and gravity) and a spring balance then the gear will accelerate the mass at 1/6m/s2 and (ignoring friction) the force recorded by the balance will show F=ma ie 0.333=2/6
The force recorded by the balance won't read 2/6 N it would read 6 N.
Please show your workings.
-
The force recorded by the balance won't read 2/6 N it would read 6 N.
It can’t read 6N because a 2kg mass doesn’t have enough inertial at 1/6m/s2, you would need 36kg.
As @The Spoon says “show your workings”.
I haven’t been following this but I can see why @Bored chemist is getting so frustrated with you.
-
As @The Spoon says “show your workings”.
I have two electronic scales , what I going to do is exerting force on input gearbox with one and read output gearbox force with the other.
-
As @The Spoon says “show your workings”.
I have two electronic scales , what I going to do is exerting force on input gearbox with one and read output gearbox force with the other.
No good if you fix the end of the output scale as you are only measuring static forces rather than dynamic acceleration force. I would have thought that was obvious.
-
As @The Spoon says “show your workings”.
I have two electronic scales , what I going to do is exerting force on input gearbox with one and read output gearbox force with the other.
No good if you fix the end of the output scale as you are only measuring static forces rather than dynamic acceleration force. I would have thought that was obvious.
I don't know how I would measure the force due to acceleration only , but in fact my equation :
Net force = (ratio-1)F1+F2
F2= mass times acceleration of the output gear.
F1=force on the input gear
if the ratio is 1:1 then the net force only equals F2= ma
if output mass is 0 also there would be force .
states that there are two forces one is due acceleration that you are mentioning , the other one due to gear ratio the question is :
Is the force you mentioned the net force ?
All forces are on one direction resulting in 6 N so I don't know why not the 6 N is the net force and not the 2/6 N especially the scale reads 6 N"I did an experiment moments ago "
The net force is the total useful force resulting in work and energy and I see the 6 newton will result in energy and work.
And if the 2/6 N is not the net force then the equation F=ma won't satisfy because it is about net force.
-
I have a gear box of speed reduction of ratio 1:6 , I have input force of 1 N the output force would be 6 newton
Then the gearbox will rotate, get squashed or move .
It depends on the directions + points of application of teh forces.
If you had a gearbox floating in space and pushed it, what would happen?
Well that's the problem- you need to consider the the forces on it and, to a good approximation they will also add to somewhere tween 5 and 7 N.
The gearbox is fixed to the ground.
The forces supplied by the bolts holding t in place are very nearly the same as the forces applied to teh box.
So, the net force on the gearbox is zero.
That's sensible.
It's why the box doesn't actually accelerate.
-
I have a gear box of speed reduction of ratio 1:6 , I have input force of 1 N the output force would be 6 newton
Then the gearbox will rotate, get squashed or move .
It depends on the directions + points of application of teh forces.
If you had a gearbox floating in space and pushed it, what would happen?
Well that's the problem- you need to consider the the forces on it and, to a good approximation they will also add to somewhere tween 5 and 7 N.
The gearbox is fixed to the ground.
The forces supplied by the bolts holding t in place are very nearly the same as the forces applied to teh box.
So, the net force on the gearbox is zero.
That's sensible.
It's why the box doesn't actually accelerate.
I don't know what this has to do with the main subject .
-
I get it right and the net force is bigger than F=ma
So you think you are right and Newton got it wrong.
Do you think that's likely?
He might missed it.
And, in spite of the fact that practically every engineer and physicist since then has used gears many times, and studied them + done exam questions about them, you are the first to notice the mistake in 400 years or so?
Do you think that's likely?
And it is unlikely they would believe me even if I'm right.
-
I don't know what this has to do with the main subject .
So, once again, you proclaim that you don't know what you are talking about.
-
And it is unlikely they would believe me even if I'm right.
They would if you had evidence.
-
I don't know what this has to do with the main subject .
So, once again, you proclaim that you don't know what you are talking about.
I repeated this hundred times: what is the net force ? this is the subject
-
I repeated this hundred times: what is the net force ? this is the subject
And I answered a hundred times that it is the vector sum of all the forces acting on the object- including the forces at the bearings. Unfortunately, you are too lazy to understand that- so you keep asking the same pointless question and pretending that it's somehow my fault.
There is nothing like enough information to calculate it to any better precision that I already did.
Well that's the problem- you need to consider the the forces on it and, to a good approximation they will also add to somewhere between 5 and 7 N.
-
I repeated this hundred times: what is the net force ? this is the subject
And I answered a hundred times that it is the vector sum of all the forces acting on the object- including the forces at the bearings. Unfortunately, you are too lazy to understand that- so you keep asking the same pointless question and pretending that it's somehow my fault.
There is nothing like enough information to calculate it to any better precision that I already did.
Well that's the problem- you need to consider the the forces on it and, to a good approximation they will also add to somewhere between 5 and 7 N.
Then it is not F=ma according to Colin2B a=1/6m/s.s and F=2/6 not 5,6 or 7
The equation F=ma is about net force F does not equal ma
-
Colin has made the assumption that the box is fixed to the floor. I didn't.
You have not specified the problem properly- so there are many possible answers.
Your failure to ask a proper question is not evidence of a problem with physics, it is evidence that you need to learn some science before you carry on.
-
Colin has made the assumption that the box is fixed to the floor. I didn't.
My gearbox is fixed to the floor
-
Colin has made the assumption that the box is fixed to the floor. I didn't.
My gearbox is fixed to the floor
Good start.
And I answered a hundred times that it is the vector sum of all the forces acting on the object- including the forces at the bearings. Unfortunately, you are too lazy to understand that- so you keep asking the same pointless question and pretending that it's somehow my fault.
There is nothing like enough information to calculate it to any better precision that I already did.
-
you say net force is 5 to 7 right ?
my response was:
Then it is not F=ma , a=1/6m/s.s and F=2/6 not 5,6 or 7
The equation F=ma is about net force F does not equal ma
-
you say net force is 5 to 7 right ?
my response was:
Then it is not F=ma , a=1/6m/s.s and F=2/6 not 5,6 or 7
The equation F=ma is about net force F does not equal ma
In reality it will obey the rules of physics.
You should learn what they are, and how to apply them.
You have not considered all the forces in spite of my repeated explanation that you need to do so.
Are you an idiot?
-
Colin has made the assumption that the box is fixed to the floor. I didn't.
You have not specified the problem properly- so there are many possible answers.
Your failure to ask a proper question is not evidence of a problem with physics, it is evidence that you need to learn some science before you carry on.
Yes, I did assume that.
As you say, there are many failures in this topic, not least of which is the OP’s failure to understand basic forces.
I don't know how I would measure the force due to acceleration only
I explained that here https://www.thenakedscientists.com/forum/index.php?topic=74254.msg549757#msg549757
If you can’t figure it out I suggest you could take a child's toy with free running wheels (low friction) to take a 2kg weight, attach one end of a spring balance to it and other end of balance to string connected to gear output. Measure 1m and get a stopwatch. When you release the input weight start the stopwatch and time the 1m journey while recording the reading on the balance. Now you have the F and can work out a.
If the wheels are running free you will have a good approximation of the accelerating force. Remember to allow for the weight of the toy and balance when putting on the mass, total should be used for m.
-
So what is net force on mass 2 kg ? it is obvious that it is 6 Newtons because I could lift a weight of 6 N with the output gear, however the gearbox is for speed reduction so the acceleration is 1/6 m/s.s now F=ma:
This is where you are going wrong. Although the gear could apply 6N it isn’t because the acceleration of the mass is constrained by the speed limitation.
If you connect the gear to the 2kg via a horizontal string (to remove red herring of freefall and gravity) and a spring balance then the gear will accelerate the mass at 1/6m/s2 and (ignoring friction) the force recorded by the balance will show F=ma ie 0.333=2/6
Mass on gearbox for speed reduction undergoes large "net force because I could lift weight with this force " but it seems there is something stopping it from accelerating in a way F=ma is not valid.
This is because you are calculating under no-load conditions. If you apply a 6N load at the output the reaction force will be reflected back through the gear chain to the input and reduce the net force on the input and hence the acceleration.
@Bored chemist has been trying to explain all this to you.
The spring balance will read 2/6 N .The large force is something like static force , static force doesn't do work it is not considered as a net force but here we have a static force affecting on a moving object in the same direction without causing it to accelerate that what tricked me.
-
The spring balance will read 2/6 N .
You got it.
The only reason you got 6N in the static situation is because the output was being resisted by an equal & opposite force - your arm muscles holding the balance still. So net accelerating force is zero.
-
Yahya,
You might want to look at the example I gave earlier here
https://www.thenakedscientists.com/forum/index.php?topic=74334.msg549969#msg549969