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If gravity would disappear, would Newton's third law make everything that was pressed to the ground by gravity get pushed upwards?
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If gravity would disappear, would Newton's third law make everything that was pressed to the ground by gravity get pushed upwards?
newtonian-mechanics forces newtonian-gravity free-body-diagram planets
edited 14 mins ago
AccidentalFourierTransform
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Casimir Rönnlöf
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If gravity would disappear, would Newton's third law make everything that was pressed to the ground by gravity get pushed upwards?
newtonian-mechanics forces newtonian-gravity free-body-diagram planets
edited 14 mins ago
AccidentalFourierTransform
24.2k1366121
asked 2 hours ago
Casimir Rönnlöf
203
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If gravity would disappear, would Newton's third law make everything that was pressed to the ground by gravity get pushed upwards?
newtonian-mechanics forces newtonian-gravity free-body-diagram planets
edited 14 mins ago
AccidentalFourierTransform
24.2k1366121
asked 2 hours ago
Casimir Rönnlöf
203
If gravity would disappear, would Newton's third law make everything that was pressed to the ground by gravity get pushed upwards?
newtonian-mechanics forces newtonian-gravity free-body-diagram planets
newtonian-mechanics forces newtonian-gravity free-body-diagram planets
edited 14 mins ago
AccidentalFourierTransform
24.2k1366121
asked 2 hours ago
Casimir Rönnlöf
203
edited 14 mins ago
AccidentalFourierTransform
24.2k1366121
asked 2 hours ago
Casimir Rönnlöf
203
edited 14 mins ago
AccidentalFourierTransform
24.2k1366121
edited 14 mins ago
AccidentalFourierTransform
24.2k1366121
edited 14 mins ago
AccidentalFourierTransform
24.2k1366121
24.2k1366121
asked 2 hours ago
Casimir Rönnlöf
203
asked 2 hours ago
Casimir Rönnlöf
203
asked 2 hours ago
Casimir Rönnlöf
203
203
add a comment |Â
add a comment |Â
3 Answers
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No. The only reason a reaction force exists is because you are pushing down on the floor as a result of gravity pulling you down.
Perhaps one way to visualise this is to imagine a block on a slope at an angle $theta$ to the horizontal.
When $theta=0$ (i.e. the slope is flat), the block has force $mg$ down and so the reaction upwards is $R=mg$.
As $theta$ grows steadily, the downwards force is still $mg$, but now the reaction force (which is the force at a right angle to the slope) becomes $R=mgcostheta$.
Imagine this surface has very large friction, so that you can get quite a large $theta$ without the block slipping down. When you finally do reach a large enough $theta$, the block will slide down parallel to the slope. Note that if $R$ had retained its value of $mg$, the block would have by now accelerated away from the slope, which doesn't happen.
answered 2 hours ago
Garf
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You can simulate this experiment in real life with an electromagnet.
For instance, you can hold a vertically oriented steel plate by a horizontally oriented electromagnet, so that the plate is free to fall down, when the electromagnet is deenergized.
If the falling plate has a horizontal velocity component and describes a parabola, you can conclude that the normal force has pushed it. If the plate falls straight down, you can conclude that the there was no push.
Even without performing such experiment, you, probably, can predict that the plate will fall straight down. This is because the normal force is a reaction force and it never exceeds the applied force causing it, be that electromagnet attraction or gravity. So, as the applied force disappears, gradually or suddenly, the normal force will disappear with it and, therefore, there won't be any push.
answered 2 hours ago
V.F.
9,90021024
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Yes, but in almost all cases the push would be imperceptible.
Reaction forces from surfaces occur when the molecules in the wall are displaced from their equilibrium position. The harder they are pushed the more they are displaced, and the more they are displaced the harder they pushed back. When you stand on a surface without falling it is because you have displaced the surface enough for the reaction force to match your weight.
As an extreme example of this imagine standing on a trampoline. Heavier people make the trampoline surface sink lower than lighter people. The same is true on hard surfaces, but the displacement is basically imperceptible.
If you suddenly removed gravity, the reaction force from the displaced surface would still be there, and it would push you away until the equilibrium state of the surface is restored.
Again, imagine standing on a trampoline holding heavy weights. When you throw the weights away the trampoline will begin to push you up until you reach a new equilibrium. If the weights were heavy enough it could even launch you into the air.
The same thing would happen harder surfaces, but the amount of time the remaining reaction force would act on you would be tiny, and you would notices hardly any effect.
answered 1 hour ago
Luke Pritchett
2,147610
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
No. The only reason a reaction force exists is because you are pushing down on the floor as a result of gravity pulling you down.
Perhaps one way to visualise this is to imagine a block on a slope at an angle $theta$ to the horizontal.
When $theta=0$ (i.e. the slope is flat), the block has force $mg$ down and so the reaction upwards is $R=mg$.
As $theta$ grows steadily, the downwards force is still $mg$, but now the reaction force (which is the force at a right angle to the slope) becomes $R=mgcostheta$.
Imagine this surface has very large friction, so that you can get quite a large $theta$ without the block slipping down. When you finally do reach a large enough $theta$, the block will slide down parallel to the slope. Note that if $R$ had retained its value of $mg$, the block would have by now accelerated away from the slope, which doesn't happen.
answered 2 hours ago
Garf
979117
add a comment |Â
up vote
3
down vote
accepted
No. The only reason a reaction force exists is because you are pushing down on the floor as a result of gravity pulling you down.
Perhaps one way to visualise this is to imagine a block on a slope at an angle $theta$ to the horizontal.
When $theta=0$ (i.e. the slope is flat), the block has force $mg$ down and so the reaction upwards is $R=mg$.
As $theta$ grows steadily, the downwards force is still $mg$, but now the reaction force (which is the force at a right angle to the slope) becomes $R=mgcostheta$.
Imagine this surface has very large friction, so that you can get quite a large $theta$ without the block slipping down. When you finally do reach a large enough $theta$, the block will slide down parallel to the slope. Note that if $R$ had retained its value of $mg$, the block would have by now accelerated away from the slope, which doesn't happen.
answered 2 hours ago
Garf
979117
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
No. The only reason a reaction force exists is because you are pushing down on the floor as a result of gravity pulling you down.
Perhaps one way to visualise this is to imagine a block on a slope at an angle $theta$ to the horizontal.
When $theta=0$ (i.e. the slope is flat), the block has force $mg$ down and so the reaction upwards is $R=mg$.
As $theta$ grows steadily, the downwards force is still $mg$, but now the reaction force (which is the force at a right angle to the slope) becomes $R=mgcostheta$.
Imagine this surface has very large friction, so that you can get quite a large $theta$ without the block slipping down. When you finally do reach a large enough $theta$, the block will slide down parallel to the slope. Note that if $R$ had retained its value of $mg$, the block would have by now accelerated away from the slope, which doesn't happen.
answered 2 hours ago
Garf
979117
No. The only reason a reaction force exists is because you are pushing down on the floor as a result of gravity pulling you down.
Perhaps one way to visualise this is to imagine a block on a slope at an angle $theta$ to the horizontal.
When $theta=0$ (i.e. the slope is flat), the block has force $mg$ down and so the reaction upwards is $R=mg$.
As $theta$ grows steadily, the downwards force is still $mg$, but now the reaction force (which is the force at a right angle to the slope) becomes $R=mgcostheta$.
Imagine this surface has very large friction, so that you can get quite a large $theta$ without the block slipping down. When you finally do reach a large enough $theta$, the block will slide down parallel to the slope. Note that if $R$ had retained its value of $mg$, the block would have by now accelerated away from the slope, which doesn't happen.
answered 2 hours ago
Garf
979117
answered 2 hours ago
Garf
979117
answered 2 hours ago
Garf
979117
answered 2 hours ago
Garf
979117
979117
add a comment |Â
add a comment |Â
up vote
2
down vote
You can simulate this experiment in real life with an electromagnet.
For instance, you can hold a vertically oriented steel plate by a horizontally oriented electromagnet, so that the plate is free to fall down, when the electromagnet is deenergized.
If the falling plate has a horizontal velocity component and describes a parabola, you can conclude that the normal force has pushed it. If the plate falls straight down, you can conclude that the there was no push.
Even without performing such experiment, you, probably, can predict that the plate will fall straight down. This is because the normal force is a reaction force and it never exceeds the applied force causing it, be that electromagnet attraction or gravity. So, as the applied force disappears, gradually or suddenly, the normal force will disappear with it and, therefore, there won't be any push.
answered 2 hours ago
V.F.
9,90021024
add a comment |Â
up vote
2
down vote
You can simulate this experiment in real life with an electromagnet.
For instance, you can hold a vertically oriented steel plate by a horizontally oriented electromagnet, so that the plate is free to fall down, when the electromagnet is deenergized.
If the falling plate has a horizontal velocity component and describes a parabola, you can conclude that the normal force has pushed it. If the plate falls straight down, you can conclude that the there was no push.
Even without performing such experiment, you, probably, can predict that the plate will fall straight down. This is because the normal force is a reaction force and it never exceeds the applied force causing it, be that electromagnet attraction or gravity. So, as the applied force disappears, gradually or suddenly, the normal force will disappear with it and, therefore, there won't be any push.
answered 2 hours ago
V.F.
9,90021024
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You can simulate this experiment in real life with an electromagnet.
For instance, you can hold a vertically oriented steel plate by a horizontally oriented electromagnet, so that the plate is free to fall down, when the electromagnet is deenergized.
If the falling plate has a horizontal velocity component and describes a parabola, you can conclude that the normal force has pushed it. If the plate falls straight down, you can conclude that the there was no push.
Even without performing such experiment, you, probably, can predict that the plate will fall straight down. This is because the normal force is a reaction force and it never exceeds the applied force causing it, be that electromagnet attraction or gravity. So, as the applied force disappears, gradually or suddenly, the normal force will disappear with it and, therefore, there won't be any push.
answered 2 hours ago
V.F.
9,90021024
You can simulate this experiment in real life with an electromagnet.
For instance, you can hold a vertically oriented steel plate by a horizontally oriented electromagnet, so that the plate is free to fall down, when the electromagnet is deenergized.
If the falling plate has a horizontal velocity component and describes a parabola, you can conclude that the normal force has pushed it. If the plate falls straight down, you can conclude that the there was no push.
Even without performing such experiment, you, probably, can predict that the plate will fall straight down. This is because the normal force is a reaction force and it never exceeds the applied force causing it, be that electromagnet attraction or gravity. So, as the applied force disappears, gradually or suddenly, the normal force will disappear with it and, therefore, there won't be any push.
answered 2 hours ago
V.F.
9,90021024
answered 2 hours ago
V.F.
9,90021024
answered 2 hours ago
V.F.
9,90021024
answered 2 hours ago
V.F.
9,90021024
9,90021024
add a comment |Â
add a comment |Â
up vote
1
down vote
Yes, but in almost all cases the push would be imperceptible.
Reaction forces from surfaces occur when the molecules in the wall are displaced from their equilibrium position. The harder they are pushed the more they are displaced, and the more they are displaced the harder they pushed back. When you stand on a surface without falling it is because you have displaced the surface enough for the reaction force to match your weight.
As an extreme example of this imagine standing on a trampoline. Heavier people make the trampoline surface sink lower than lighter people. The same is true on hard surfaces, but the displacement is basically imperceptible.
If you suddenly removed gravity, the reaction force from the displaced surface would still be there, and it would push you away until the equilibrium state of the surface is restored.
Again, imagine standing on a trampoline holding heavy weights. When you throw the weights away the trampoline will begin to push you up until you reach a new equilibrium. If the weights were heavy enough it could even launch you into the air.
The same thing would happen harder surfaces, but the amount of time the remaining reaction force would act on you would be tiny, and you would notices hardly any effect.
answered 1 hour ago
Luke Pritchett
2,147610
add a comment |Â
up vote
1
down vote
Yes, but in almost all cases the push would be imperceptible.
Reaction forces from surfaces occur when the molecules in the wall are displaced from their equilibrium position. The harder they are pushed the more they are displaced, and the more they are displaced the harder they pushed back. When you stand on a surface without falling it is because you have displaced the surface enough for the reaction force to match your weight.
As an extreme example of this imagine standing on a trampoline. Heavier people make the trampoline surface sink lower than lighter people. The same is true on hard surfaces, but the displacement is basically imperceptible.
If you suddenly removed gravity, the reaction force from the displaced surface would still be there, and it would push you away until the equilibrium state of the surface is restored.
Again, imagine standing on a trampoline holding heavy weights. When you throw the weights away the trampoline will begin to push you up until you reach a new equilibrium. If the weights were heavy enough it could even launch you into the air.
The same thing would happen harder surfaces, but the amount of time the remaining reaction force would act on you would be tiny, and you would notices hardly any effect.
answered 1 hour ago
Luke Pritchett
2,147610
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, but in almost all cases the push would be imperceptible.
Reaction forces from surfaces occur when the molecules in the wall are displaced from their equilibrium position. The harder they are pushed the more they are displaced, and the more they are displaced the harder they pushed back. When you stand on a surface without falling it is because you have displaced the surface enough for the reaction force to match your weight.
As an extreme example of this imagine standing on a trampoline. Heavier people make the trampoline surface sink lower than lighter people. The same is true on hard surfaces, but the displacement is basically imperceptible.
If you suddenly removed gravity, the reaction force from the displaced surface would still be there, and it would push you away until the equilibrium state of the surface is restored.
Again, imagine standing on a trampoline holding heavy weights. When you throw the weights away the trampoline will begin to push you up until you reach a new equilibrium. If the weights were heavy enough it could even launch you into the air.
The same thing would happen harder surfaces, but the amount of time the remaining reaction force would act on you would be tiny, and you would notices hardly any effect.
answered 1 hour ago
Luke Pritchett
2,147610
Yes, but in almost all cases the push would be imperceptible.
Reaction forces from surfaces occur when the molecules in the wall are displaced from their equilibrium position. The harder they are pushed the more they are displaced, and the more they are displaced the harder they pushed back. When you stand on a surface without falling it is because you have displaced the surface enough for the reaction force to match your weight.
As an extreme example of this imagine standing on a trampoline. Heavier people make the trampoline surface sink lower than lighter people. The same is true on hard surfaces, but the displacement is basically imperceptible.
If you suddenly removed gravity, the reaction force from the displaced surface would still be there, and it would push you away until the equilibrium state of the surface is restored.
Again, imagine standing on a trampoline holding heavy weights. When you throw the weights away the trampoline will begin to push you up until you reach a new equilibrium. If the weights were heavy enough it could even launch you into the air.
The same thing would happen harder surfaces, but the amount of time the remaining reaction force would act on you would be tiny, and you would notices hardly any effect.
answered 1 hour ago
Luke Pritchett
2,147610
answered 1 hour ago
Luke Pritchett
2,147610
answered 1 hour ago
Luke Pritchett
2,147610
answered 1 hour ago
Luke Pritchett
2,147610
2,147610
add a comment |Â
add a comment |Â
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