Weight and force of gravity

  • #1
adjurovich
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A relatively simple question has been bothering me for some time: Why is the weight of an object (the force the object exerts on a surface/rope) equal to the force of gravity that the body experiences in Earth’s gravitational field?

Earth’s gravitational field accelerates all bodies in its atmosphere, meaning that there has to be a force acting on a body: ##F_g=mg##
According to Newton’s third law, the body also exerts the same force on Earth but it’s negligible considering earth’s mass.

The only reasonable idea I had is that weight is just a different name of the same force (gravity). The bodies come in contact but they are still some distance ##r## apart if observed from their centers and bodies exert force of gravity on each other regardless of being in contact or not.
 
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  • #2
adjurovich said:
The only reasonable idea I had is that weight is just a different name of the same force (gravity).
Weight is the force of gravity on the body, yes. Or sometimes the magnitude of the force - depends on context.
adjurovich said:
According to Newton’s third law, the body also exerts the same force on Earth but it’s negligible considering earth’s mass.
The force isn't negligible - it's the same as the weight. The acceleration is, though, yes.
 
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  • #3
Ibix said:
Weight is the force of gravity on the body, yes. Or sometimes the magnitude of the force - depends on context.

The force isn't negligible - it's the same as the weight. The acceleration is, though, yes.
Force of 50 newton’s can’t accelerate 5.972•10^24 kilograms that much, so why isn’t it negligible?
 
  • #4
adjurovich said:
Force of 50 newton’s can’t accelerate 5.972•10^24 kilograms that much, so why isn’t it negligible?
It's the same magnitude as the force on the smaller weight and if you neglect it you get obvious violations of momentum conservation. For example, if you drop a ball from rest the initial momentum is zero. When it reaches the floor the momentum of the ball is non-zero - which is a violation of momentum conservation if you don't account for the momentum of the Earth.

You can treat the Earth as outside the system you are studying and the force of gravity as an external force. Or you can choose to ignore the momentum change of the Earth and live with the fact that you can't conserve momentum (it may not matter for your calculation). But you cannot neglect the momentum change, nor the force, because it's not small. The resuting acceleration certainly is small, though.
 
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  • #5
adjurovich said:
Why is the weight of an object (the force the object exerts on a surface/rope) equal to the force of gravity that the body experiences in Earth’s gravitational field?
Newton's second law. And Newton's third law.

One assumes that the object in question is at rest in the chosen laboratory frame. According to Newton's second law, this means that the net external force on the object is zero. The only two forces on the object are the force from the surface/rope on the object and the force of gravity on the object. Those two forces must therefore be equal and opposite.

According to Newton's third law, the force that the object exerts on the surface/rope is equal and opposite to the force that the surface/rope exerts on the object.

Two forces equal and opposite to the same force are equal to each other. So the force of gravity on the object is the same as the force the object exerts on the surface/rope.

To put it in a pithy way, it is a second law daisy-chain. The force in on one side of an unaccelerating body is the same as the force out the other side.

If the object is accelerating, the two forces need not match. For instance, if you jump up and down on your bathroom scale, the weight reading on the scale will fluctuate wildly while the force of gravity on your body remains constant.
 
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  • #6
jbriggs444 said:
One assumes that the object in question is at rest in the chosen laboratory frame. According to Newton's second law, this means that the net external force on the object is zero. The only two forces on the object are the force from the surface/rope on the object and the force of gravity on the object. Those two forces must therefore be equal and opposite.
Just a point to be highlighted. The chosen laboratory frame is assumed to be inertial, otherwise one cannot apply Newton's second law without introducing inertial forces in that (non-inertial) frame.
 
  • #7
cianfa72 said:
Just a point to be highlighted. The chosen laboratory frame is assumed to be inertial, otherwise one cannot apply Newton's second law without introducing inertial forces in that (non-inertial) frame.
I look at it differently. The laboratory frame need not be inertial. As you point out, one then needs to introduce inertial forces to obtain the second law force balance.

In a rotating space station, one can then confidently step on a bathroom scale and judge that the apparent centrifugal downforce from the station's rotation matches the weight reading on the scale.

On the surface of the rotating Earth in the context of Newtonian mechanics, one can confidently step on the bathroom scale and judge that the resultant of gravity and the centrifugal repulsion from the Earth's rotation axis matches the weight reading on the scale.

Or on the surface of the Earth in the context of general relativity, one can confidently step on the bathroom scale and judge that the downward inertial force from the upward proper acceleration of the Earth's surface matches the weight reading on the scale.
 
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  • #8
jbriggs444 said:
In a rotating space station, one can then confidently step on a bathroom scale and judge that the apparent centrifugal downforce from the station's rotation matches the weight reading on the scale.
You mean the apparent centrifugal force like in the Rotor.

jbriggs444 said:
Or on the surface of the Earth in the context of general relativity, one can confidently step on the bathroom scale and judge that the downward inertial force from the upward proper acceleration of the Earth's surface matches the weight reading on the scale.
In this context, the person on the bathroom scale is "pushed away" from its natural geodesic path through spacetime by the upward proper acceleration of the Earth's surface. Is the bathroom scale indication due to the force from the Newton 3rd law pair ?
 
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  • #9
cianfa72 said:
Is the bathroom scale indication due to the force from the Newton 3rd law pair ?
The contact forces between scale and person form a Newton 3rd law pair in any of the above scenarios. But trying to bring causation into this ("due to") just leads to confusion.
 
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  • #10
A.T. said:
The contact forces between scale and person form a Newton 3rd law pair in any of the above scenarios. But trying to bring causation into this ("due to") just leads to confusion.
Ok, so one can simply claims that a force from the Newton 3rd law pair acts on the bathroom scale.
 
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  • #11
cianfa72 said:
Ok, so one can simply claims that a force from the Newton 3rd law pair acts on the bathroom scale.
Well yes. The force of the object on the scale is an interaction force. Every interaction force is part of a third law pair.
 
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  • #12
jbriggs444 said:
Newton's second law. And Newton's third law.

One assumes that the object in question is at rest in the chosen laboratory frame. According to Newton's second law, this means that the net external force on the object is zero. The only two forces on the object are the force from the surface/rope on the object and the force of gravity on the object. Those two forces must therefore be equal and opposite.

According to Newton's third law, the force that the object exerts on the surface/rope is equal and opposite to the force that the surface/rope exerts on the object.

Two forces equal and opposite to the same force are equal to each other. So the force of gravity on the object is the same as the force the object exerts on the surface/rope.

To put it in a pithy way, it is a second law daisy-chain. The force in on one side of an unaccelerating body is the same as the force out the other side.

If the object is accelerating, the two forces need not match. For instance, if you jump up and down on your bathroom scale, the weight reading on the scale will fluctuate wildly while the force of gravity on your body remains constant.
I’ll reformulate the question to make it more precise.

Between any two bodies with mass (according to Newton), there arises attractive force (gravity).

The bodies in Earth’s atmosphere “fall” due to gravity.

The bodies that are standing on surface experience the force of gravity (Earth pulls them), and they also pull Earth (let’s neglect this for it has no important effects on this observation). The body exerts force on surface called weight. Why is weight equal to the force of gravity between Earth and body when body is at rest on flat surface.
 
  • #13
adjurovich said:
The bodies that are standing on surface experience the force of gravity (Earth pulls them), and they also pull Earth (let’s neglect this for it has no important effects on this observation). The body exerts force on surface called weight. Why is weight equal to the force of gravity between Earth and body when body is at rest on flat surface.
It's not accelerating, so the forces must sum to zero. A free body diagram would help here: a body in free fall has one for acting on it, and the one sitting on the scale has two.
 
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  • #14
adjurovich said:
The body exerts force on surface called weight. Why is weight equal to the force of gravity between Earth and body when body is at rest on flat surface.
As said before, the body weight measured by the scale (the body standing still on it) is one of the Newton 3d pair.
 
  • #15
adjurovich said:
Why is weight equal to the force of gravity between Earth and body when body is at rest on flat surface.
Try to be precise. Exactly what do you mean by "weight"?

A) The force of Newtonian gravity from the Earth on the body? ##W=G\frac{mM_e}{r^2}##
B) The support force supplied by the scale on which the body rests? ##W=F_n##
C) The down-force from the body on the surface? ##W=-F_n##
D) The resultant of all such non-gravitational support forces, including atmospheric buoyancy? ##W=\sum F - F_g##
E) The force of apparent gravity? ##W=mg##
F) Other

Personally, I consider A to be approximately correct, B to be approximately correct and C to be approximately correct. I take D and E to be exactly correct. I take E to be definitional, but am not opposed to other choices.
 
  • #16
jbriggs444 said:
Try to be precise. Exactly what do you mean by "weight"?

A) The force of Newtonian gravity from the Earth on the body? ##W=G\frac{mM_e}{r^2}##
B) The support force supplied by the scale on which the body rests? ##W=F_n##
C) The down-force from the body on the surface? ##W=-F_n##
D) The resultant of all such non-gravitational support forces, including atmospheric buoyancy? ##W=\sum F - F_g##
E) The force of apparent gravity? ##W=mg##
F) Other

Personally, I consider A to be approximately correct, B to be approximately correct, C to be approximately correct. I take D and E to be exactly correct. I take E to be definitional, but am not opposed to other choices.
It would be B). Why is it equal to the force gravity (you mentioned as A) in the case of body resting on a flat surface?
 
  • #17
adjurovich said:
It would be B). Why is it equal to the force gravity (you mentioned as A) in the case of body resting on a flat surface?
As has been suggested, create a free body diagram and apply Newton's second law.
 
  • #18
jbriggs444 said:
As has been suggested, create a free body diagram and apply Newton's second law.
Fine but I wanted to take a more theoretical approach. It would only be reasonable to me that weight has to be equal to the gravity in that case in order to provide normal reaction of surface that is equal to and opposite of gravity and balances it? But it it were true, it would look like we are just trying to make sense of what we see but not understand it?

Tension force has a pretty logical description. Molecules “don’t want to be pulled from their comfort zone” so attractive force arises between them that tries to get them back into the previous position
 
  • #19
I would says C) since ##F_n## and ##-F_n## are a Newton 3rd pair.
 
  • #20
adjurovich said:
weight has to be equal to the gravity in that case in order to provide normal reaction of surface that is equal to and opposite of gravity and balances it?
I want to be very sure that I understand what you are saying. You have tossed out three terms here:

"weight"
"gravity"
"normal reaction"

You've already indicated that "weight", for you, is a synonym for the normal up-force of surface on body. The "normal reaction" would then be the third law partner of that: the down-force of body on surface.

But then you say that this down-force of body on surface is equal and opposite to gravity and balances it?!

So "gravity", for you, must be the gravitational up-force of body on earth?! That hardly seems right.

Please provide the free body diagram.

adjurovich said:
But it it were true, it would look like we are just trying to make sense of what we see but not understand it?
I do not see the distinction between "make sense of" and "understand".

Science is about experiments and predictive models. If you have a model that works, that's all there is.
 
  • #21
adjurovich said:
Fine but I wanted to take a more theoretical approach. It would only be reasonable to me that weight has to be equal to the gravity in that case in order to provide normal reaction of surface that is equal to and opposite of gravity and balances it? But it it were true, it would look like we are just trying to make sense of what we see but not understand it?
I don't see what's hard to understand about why I'm not accelerating down through the floor where I stand. The floor is solid, capable of resisting/providing a wide range of forces. It's perfectly reasonable/logical. What is it that YOU don't understand about it?

You just seem to be in an endless loop of saying you don't understand or the explanation isn't good enough without saying what your problem is.
 
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  • #22
russ_watters said:
I don't see what's hard to understand about why I'm not accelerating down through the floor where I stand. The floor is solid, capable of resisting/providing a wide range of forces. It's perfectly reasonable/logical. What is it that YOU don't understand about it?

You just seem to be in an endless loop of saying you don't understand or the explanation isn't good enough without saying what your problem is.
My question was: why is the force the body exerts on flat surface equal to the force of gravity that body experiences in Earth’s gravitational field?
 
  • #23
adjurovich said:
My question was: why is the force the body exerts on flat surface equal to the force of gravity that body experiences in Earth’s gravitational field?
I'm aware that's the question (one of a couple of variants). We've given you the answer. Now YOU need to answer OUR question about what you don't understand/like about the answer. When you just repeat the question without directly referencing the answer it makes us wonder if you even read the answer.
 
  • #24
jbriggs444 said:
I want to be very sure that I understand what you are saying. You have tossed out three terms here:

"weight"
"gravity"
"normal reaction"

You've already indicated that "weight", for you, is a synonym for the normal up-force of surface on body. The "normal reaction" would then be the third law partner of that: the down-force of body on surface.

But then you say that this down-force of body on surface is equal and opposite to gravity and balances it?!

So "gravity", for you, must be the gravitational up-force of body on earth?! That hardly seems right.

Please provide the free body diagram.


I do not see the distinction between "make sense of" and "understand".

Science is about experiments and predictive models. If you have a model that works, that's all there is.
I’m not from USA and in our physics literature, weight and gravitational force are considered as two different forces. Gravitational force acting any body on Earth (or anywhere else) is basically described as the force with which Earth pulls objects towards itself, and it’s equal to the mass of an object multiplied by gravitational acceleration:

##F_g = mg##

Weight is defined as the force bodies exert on surface (basically a contact force). We’d consider normal force on an inclined plane to be weight, and it’s denoted by letter ##Q## . It’s said that when body is standing on a flat surface, the weight can (force it presses the surface with) can be calculated this way:

##Q = mg##

Normal reaction of surface would be action-reaction pair with weight. Sorry for confusion and misunderstanding.
 
  • #25
adjurovich said:
The only reasonable idea I had is that weight is just a different name of the same force (gravity). The bodies come in contact but they are still some distance ##r## apart if observed from their centers and bodies exert force of gravity on each other regardless of being in contact or not.
The attractive forces of the gravitational field are always present, regardless of solid contact or not.
Weight is only the vertical force (in the absence of any other vertical acceleration) that can be measured, for which that solid contact is necessary.

If linear acceleration, impact, or centripetal effect are also present, the measured contact force "(the force the object exerts on a surface/rope)" is usually greater than the weight and its direction may be different from vertical.

A bathroom balance would indicate temporary different values of body weight, in case of a person standing on it while inside an elevator that is being accelerated upwards or downwards.

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  • #26
russ_watters said:
I don't see what's hard to understand about why I'm not accelerating down through the floor where I stand. The floor is solid, capable of resisting/providing a wide range of forces. It's perfectly reasonable/logical. What is it that YOU don't understand about it?

You just seem to be in an endless loop of saying you don't understand or the explanation isn't good enough without saying what your problem is.
Thanks, I get it now! I thought about it for some time and it makes sense.
 
  • #27
adjurovich said:
Normal reaction of surface would be action-reaction pair with weight. Sorry for confusion and misunderstanding.
Not a problem at all. Language issues are no one's fault.

As I use the words, the force assignments are:

"Normal force": The perpendicular (upward in this case) component of the contact force of a flat surface on an object in contact with that surface. You appear to call this "normal reaction force".

"Normal reaction force": The third law partner of the normal force component. You appear to call this "normal force". You consider this downward contact force to be synonymous with "weight".

"weight": The downward inertial force of apparent gravity on the body. This includes not just Newtonian gravity, but also any other inertial forces arising from a decision to use an accelerating frame of reference. You appear to use the term "gravitational force" for this and are not concerned with any minor discrepancies arising from the Earth's rotation, atmospheric buoyancy and such.

"gravitational force": The downward force of Newtonian gravity on the body. This includes only Newtonian gravity (##F=G\frac{mM_e}{r^2}##) and does not include any correction for centrifugal force.

I do not know of any commonly used term for the upward gravitational force from the body acting on the Earth below.

Note that I do not disagree with your word usage at all. I am just trying to make my usage clear.
 
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  • #28
adjurovich said:
Why is the weight of an object (the force the object exerts on a surface/rope) equal to the force of gravity that the body experiences in Earth’s gravitational field?
The surface/rope deform until the resistance to deformation (contact force) becomes large enough to balance the gravitational force on the object. Such constraint forces are adjusting themselves to whatever is needed to prevent certain types of motion (object moving down).
 
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  • #29
adjurovich said:
Why is the weight of an object (the force the object exerts on a surface/rope) equal to the force of gravity that the body experiences in Earth’s gravitational field?
The weight is not, in all cases, equal to the force the object exerts on a surface/rope.
 
  • #30
Mister T said:
The weight is not, in all cases, equal to the force the object exerts on a surface/rope.
It is only equal in inertial frame of reference when there object is perpendicular to the flat surface or directly attached to rope.

I am not from the USA / UK and here we have kinda different terminology. Weight is usually considered to be the pulling force exerted on rope / force that body exerts on surface (the one of which reaction pair is the normal force ##\vec{N}##). However we have the other thing that I wouldn’t be sure how to directly translate, probably something like Earth’s force of gravity? The force that Earth exerts on all objects in its atmosphere. That’s why I started this thread quite some time ago
 
  • #31
Personally, I don't use the concept of weight in physics. You need the concepts of force and mass, but weight is is a superfluous and often ill-defined concept. It's possible of course to define the weight of a body of mass ##m## as simply ##W = mg##. Often, however, astronauts in the space station are described as weightless - which clearly contradicts this definition.

Note that Earth's gravitational pull extends beyond its atmosphere. In fact, theoretically, to infinity.
 
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  • #32
PeroK said:
Personally, I don't use the concept of weight in physics. You need the concepts of force and mass, but weight is is a superfluous and often ill-defined concept. It's possible of course to define the weight of a body of mass ##m## as simply ##W = mg##. Often, however, astronauts in the space station are described as weightless - which clearly contradicts this definition.

Note that Earth's gravitational pull extends beyond its atmosphere. In fact, theoretically, to infinity.
The idea used here is that “Earth’s force of gravity” works as a concept only very close to the surface of Earth. It happens to work this way:​

##F_g = mg##

where ##g## is the strength of Earth’s gravitational field:

##g = G \dfrac{M}{R^2}##
As long as there is no additional height ##h## added to the Earth’s radius, g is approximately 9.8 or 9.81 depending on where you measure it. But unique value can be chosen. I find it a bit odd to use two different forces to explain literally the same thing

When it comes to astronauts, my interpretation is that they in fact do not exert the force on other objects since they are floating so they are weightless,
however they do not truly experience 0 gravity state unless they are far enough from other planets / stars. However theoretically speaking it would be impossible, but in practice you can the neglect small quantities​
 
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  • #33
My interpretation is that "g" is whatever free fall acceleration you accept by adopting a non-inertial local reference frame.

Here on earth is is convenient to adopt a local frame where the soil is unmoving. In a space craft it is convenient to adopt one where the lab bench is unmoving.

An object's weight is its mass times the free fall acceleration that you have accepted.

Note that the free fall acceleration relative to the Earth's surface is not in general given by ##\frac{GM_e}{r^2}##.
 
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  • #34
adjurovich said:
When it comes to astronomers, my interpretation is that they in fact do not exert the force on other objects since they are floating so they are weightless,
This would be wrong. A spacecraft in orbit around a planet is clearly experiencing the force of gravity - that's why it's in orbit. But it, and its crew, are "weightless" as commonly described.

The point is that all of them are (approximately) affected the same way by gravity so they have zero acceleration with respect to each other so do not see effects we commonly think of as due to gravity. That's why they are called "weightless". But the "approximate" part is neglecting tidal forces, and sufficiently precise experiment can detect them. So even "weightless" you can, if sufficiently careful, detect the presence of gravity.
 
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  • #35
Ibix said:
This would be wrong. A spacecraft in orbit around a planet is clearly experiencing the force of gravity - that's why it's in orbit. But it, and it's crew, are "weightless" as commonly described.

The point is that all of them are (approximately) affected the same way by gravity so they have zero acceleration with respect to each other so do not see effects we commonly think of as due to gravity. That's why they are called "weightless". But the "approximate" part is neglecting tidal forces, and sufficiently precise experiment can detect them. So even "weightless" you can, if sufficiently careful, detect the presence of gravity.
I will point something out. Weight is defined, here where I live, as the force that object exerts on other objects when it’s standing (for example on soil or a box) / attached to rope. I don’t think astronomers can stand? In that case they do not exert “weight” on other objects.

However I don’t think I said they don’t experience gravity? They surely do. But what I meant to say is that if a space ship was placed in some spot ##B## that is very far from any planet / star it would almost experience zero gravity.
 

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