Force and rate of change of momentum

Click For Summary
Force is defined as the rate of change of momentum, specifically the net force acting on an object. While individual forces can be seen as rates of momentum transfer, only the net force correlates directly with the total momentum change. In most educational and engineering contexts, force is often simplified to a push or pull, neglecting the nuanced relationship with momentum. The confusion arises from the traditional understanding of force as F=ma, which applies when mass is constant. Ultimately, while force and momentum change are related, they are not identical; force causes momentum change, making them distinct yet interconnected concepts.
  • #31
alkaspeltzar said:
But can you explain why certain physics books write F=dp/dt...that would make me think it is a rate.
Force can indeed be thought of as the rate of momentum transfer, and has the appropriate units. The units for the rate of momentum transfer would be momentum/time. In SI that unit is 1 N.
 
Physics news on Phys.org
  • #32
Dale said:
Force can indeed be thought of as the rate of momentum transfer, and has the appropriate units. The units for the rate of momentum transfer would be momentum/time. In SI that unit is 1 N.

But why is it from my high school physics class thru college it was never treated or thought of as a rate. Just 100lbs, or 10 N, as in a simple push or pull. Rate was not even thought of.

And am I wrong with my way of thinking? F=ma, force is a push/pull. Is it just in the end, what I know as force can also be thought of as a rate but most of the time isn't due to the fact it isn't necessary?
 
  • #33
alkaspeltzar said:
But why is it from my high school physics class thru college it was never treated or thought of as a rate.
I think you would have to ask those teachers.

I assume that acceleration was described as the rate of change of velocity, so multiple of acceleration should already be thought of as a rate. They may have in fact taught that, but the mental connection just took some time to establish.
 
  • #34
Dale said:
I think you would have to ask those teachers.

I assume that acceleration was described as the rate of change of velocity, so multiple of acceleration should already be thought of as a rate. They may have in fact taught that, but the mental connection just took some time to establish.

how come modern units are not rate units then? lb, kg, ton, dyne...where as unit of rate for speed are typically m/s, inch/sec, feet/sec??
 
  • #35
kg and ton are not forces. The imperial system with pounds, pound-force, slug and whatever is horrible so I won't comment on that. Dyne is just 10-5 N, same units.

1N = 1kg m/s2 - there is your "per second". You can also write it as (1kg m/s)/s where the numerator is momentum and the denominator is time.
 
  • #36
mfb said:
kg and ton are not forces. The imperial system with pounds, pound-force, slug and whatever is horrible so I won't comment on that. Dyne is just 10-5 N, same units.

1N = 1kg m/s2 - there is your "per second". You can also write it as (1kg m/s)/s where the numerator is momentum and the denominator is time.

a ton is not a unit of force? https://en.wikipedia.org/wiki/Ton-force
 
  • #37
"Ton-force" is not "ton". And again, that is some weird imperial unit stuff I won't discuss further. 95% of the world population uses the SI system for many very good reasons.
 
  • #38
not in the engineering world they don't
 
  • #39
The article about ton-force doesn't exist in any other language than English, apparently because no one needs it outside the US, and I have never seen those "[mass unit]-force" units used anywhere outside the US.
Anyway, those units have the same unit as a Newton, so apart from weird numerical prefactors nothing changes.
 
  • #40
alkaspeltzar said:
how come modern units are not rate units then? lb, kg, ton, dyne...where as unit of rate for speed are typically m/s, inch/sec, feet/sec??

You have a severe misunderstanding of calculus, and "rate of change" in general.

A velocity is defined as the rate of change of displacement, i.e. v=ds/dt. However, velocity ITSELF need not change over time, i.e. it can be a constant! In this case, there is no time rate of change of velocity, yet, there is a rate of change of displacement.

Force is defined as the rate of change of momentum, i.e. F = dp/dt. However, as in the case of velocity, one CAN have a constant force over time, i.e. the force doesn't change. We commonly see this in intro physics classes.

Zz.
 
  • #41
No, it is not that I don't understand calculus and rate of change. I understand that the rate of change can be constant like your velocity example, that is not my question.

My question is this: Force was defined as F=MA, a vector quantity, based on the units of a Newton, described as that force required to accelerate 1kg mass 1 m/sec^2. Force as never described as a rate and in general is not defined as one in most books. Simply a push or a pull.

Now am I seeing force is a constant rate, as F= dp/dt. My questions is how, when it is not taught or understood that way? And esp when working with the unit of force, typically rates involve some time, and in the case of all but the Newton, none do. So that is my question.
 
  • #42
alkaspeltzar said:
No, it is not that I don't understand calculus and rate of change. I understand that the rate of change can be constant like your velocity example, that is not my question.

My question is this: Force was defined as F=MA, a vector quantity, based on the units of a Newton, described as that force required to accelerate 1kg mass 1 m/sec^2. Force as never described as a rate and in general is not defined as one in most books. Simply a push or a pull.

Now am I seeing force is a constant rate, as F= dp/dt. My questions is how, when it is not taught or understood that way? And esp when working with the unit of force, typically rates involve some time, and in the case of all but the Newton, none do. So that is my question.

But F=ma is a part of F=dp/dt!

In classical physics, since p=mv, then

F = d(mv)/dt = m dv/dt + v dm/dt.

For a system where the mass doesn't change, then dm/dt is zero, which leaves

F = m dv/dt.

But dv/dt = a, and you get back

F = ma.

And you are wrong in the sense that this isn't taught. This may not be taught YET to you, but we certain teach that in college intro physics. The idea of "Impulse", which is nothing more than a change in momentum, i.e. Δp = ∫F⋅dt comes from the definition of F=dp/dt. It may not be taught to you in the VERY beginning, because the concept of force and kinematics are the main focus, and the concept of momentum hasn't yet been introduced. But it certainly WILL be taught.

And this WILL be a common theme if you are going to be studying physics for any considerable length of time. In advanced mechanics, you will be told that we don't even have to deal with the concept of "force" to solve for the equation of motion!

Zz.
 
  • #43
alkaspeltzar said:
how come modern units are not rate units then?
They are. I don't know what makes you think that they are not. The unit of force in every system of units has the same units as momentum/time.

alkaspeltzar said:
My question is this: Force was defined as F=MA, a vector quantity, based on the units of a Newton, described as that force required to accelerate 1kg mass 1 m/sec^2. Force as never described as a rate and in general is not defined as one in most books.
Acceleration is clearly a rate so multiplying acceleration by mass clearly gives a rate also. Were you never taught that acceleration is the rate of change of velocity?
 
  • #44
Please try looking harder at the units you are describing. The unit of force is not a rate of force change (because it isn't over time), but it is a rate of momentum change. You're cris-crossing those concepts: keep them separate and look at them for what they are. Force is force. Momentum is momentum. Force is not momentum.

You can of course have situations where force is changing and describe that as a rate of force change.
 
  • #45
Recall that Newton is merely a special name we give to the derived unit kg-m/s2. The rate of change with respect to time is still in there.

You often see weight expressed in grams or kilograms. Scientifically, weight is measured in Newtons, but coloquially we often say an object has a weight of "x" kilograms, which is understood to mean the gravitational force an object with "x" kg of mass would experience near the Earth's surface.
 
Last edited:
  • #47
Okay, i understand how we mathematically get from F=MA to F=dp/dt. So in that sense we can mathematically equate Force and rate of change of momentum.

But what gets me is isn't Force and the rate of change of momentum still in the physical world two distinctly separate things? I mean we don't think or work with a force as rate, and isn't it the force on an object that causes a rate of change in momentum?

I was taught F=MA, force is a push or a push, and interaction that causes an acceleration, not that it is a rate, so how can we call it that?

Just confused. Thanks

PS: I guess the only thing that made sense to me is that force is it own thing, not a rate. It is the push or pull that causes this rate of change, hence abstractly in math we can equate them to help solve for one of the other.
 
  • #48
alkaspeltzar said:
Okay, i understand how we mathematically get from F=MA to F=dp/dt
no it is the other way round . we get f=ma from f=dp/dt (when m const)
alkaspeltzar said:
So in that sense we can mathematically equate Force and rate of change of momentum
we don't equate them as a mathemtical result. dp/dt is given a special name of force in the same way dx/dt is given a special name of velocity
alkaspeltzar said:
I was taught F=MA, force is a push or a push, and interaction that causes an acceleration, not that it is a rate, so how can we call it that?
it was observed the with no ext interaction (mass *velo of com) was const and more the interaction more was rate of change of (mass *velo of com)
so force was defined as rate of change of momentum
alkaspeltzar said:
not that it is a rate
force is a rate . since nonzero rate of net momentum change is impossible with zero ext interation
 
  • #49
How come my college physics book doesn't even describe it as a rate, just F=MA? I have never seen this till now and it just confuses me.

and in general, if force is a rate, then why don't the units of measure for a force have time associated with them?
 
  • #50
alkaspeltzar said:
How come my college physics book doesn't even describe it as a rate, just F=MA? I have never seen this till now and it just confuses me.

and in general, if force is a rate, then why don't the units of measure for a force have time associated with them?

The units of force are ##MLT^{-2}##. That's the rate of change of something that is already itself a rate of change.

##F = ma## explicitly defines force as a rate, as acceleration is rate of change of velocity, so force is rate of change of mass x velocity, which is rate of change of momentum.
 
  • Like
Likes BiGyElLoWhAt
  • #51
well I have talked to other people here at work, other engineers and it is all foreign to us. But maybe it is true and we just don't understand it that way. I guess I will just have to accept it and move on.

At the end of the day I know how to calculate force and know how to use the relationship F=MA and that is all that matters

Thanks
 
  • #52
alkaspeltzar said:
well I have talked to other people here at work, other engineers and it is all foreign to us. But maybe it is true and we just don't understand it that way. I guess I will just have to accept it and move on.

At the end of the day I know how to calculate force and know how to use the relationship F=MA and that is all that matters

Thanks
In mechanical statics or dynamics, sure F=ma, is good enough concept to begin with.
But, for example,
Force being the rate of change in momentum is a basic starting point for fluid dynamics, where we have changes in velocity from one point in the flow to the next.
 
  • #53
So I suppose you can start with position, take it's rate of change with respect to time, and get velocity.
##\frac{dS}{dt}= v##
then we can take that and multiply it by mass, and get momentum.
##p = m \frac{dS}{dt}##
So momentum itself is already a rate of change, if for some reason we were to define the quantity ##mS## or mass times position, momentum would be the rate of change of this quantity.
It's not taught this way in lower level physics classes, as it's easier to work with the latter F=ma. However, this is also a differential equation, which it's not taught as in lower level physics classes.
Look at a mass on a spring.
Force of spring = -kx
##\sum F = m\frac{d^2x}{dt^2} = -kx \therefore m\frac{d^2x}{dt^2} + kx = 0##
So I suppose the moral of the story is walk before you run, and this is employed in physics classes.

Edit*
If you were looking at a situation such as a block sliding across a surface that has a couple pivot points like such (see attached), and you wanted to know at what time the object was going to tip, it might be useful to define the quantity ##mS## with S being measured from the pivot. Then you could look at it's rate of change and use that to solve for t.
 

Attachments

  • mass on pivots.png
    mass on pivots.png
    632 bytes · Views: 481
Last edited:
  • #54
alkaspeltzar said:
I mean we don't think or work with a force as rate
How can you not? Acceleration is a rate and force is proportional to acceleration, so it must be a rate also.

This is the second time recently that this confusion has come up.

EDIT: actually this is the third thread you have started recently on the same topic. I have merged them, please don't start more threads on the same topic
 
Last edited:
  • #55
Dale said:
How can you not?
I think it's in the way that it's taught in introductory mechanics.
You are taught Newton's laws, and you really focus on N2L. Then you talk about energy, and relate it to F.x, but momentum tends to be taught via conservation and lumped in more so with energy than forces. Just my experience, as I took intro and intermediate mechanics not more than 3 or 4 years ago. The difference is night and day.
 
  • #56
alkaspeltzar said:
Sorry bear with me but I am just not getting it.

So are you saying that the physical force(what we long ago defined as a push or pull) is exactly the same as rate of change in momentum, aka the rate of momentum transfer?

Don't you have to have a force to have a rate of change in momentum? Part of me thinks if a body has acceleration, then there must be a force. Likewise, if a body has a rate of change in momentum, it must have a force causing it...so aren't they two separate things, just related since one can't exist without the other?

And if they are the same, why don't we use one name...why say force if it is really a rate of momentum transfer or visa versa?

I guess I m looking for a simple explanation, please no math at this point.
I think I see the issue.
Momentum is a property of an object, namely ##mv##
Force is an interaction quantity, but the magnitude of the net force is equal to the magnitude of the rate of change of momentum. So, in a sense, they are two different "entities".
Momentum is, as stated above, a defined quantity, and net force = ma is derived from that definition, and the fact that they are equal quantities.
They are distinct in the sense that, just to reiterate, momentum is "had" by and object, in other words, momentum is a quantity that describes an object, whereas Force is a quantity that describes an interaction. When a nonzero net force is applied, you get a change in momentum, and the time rate of change is equal to the net force.
 
  • #57
I'm really not seeing why this is a problem. Force itself is just Newtons (N) and isn't a rate. Momentum change (kg-m/s2) is something you can do with force and is a rate. The units are equal as simple derivations with the definition equations will tell you. I don't see why this should be so difficult to believe/accept.
 
  • #58
russ_watters said:
I'm really not seeing why this is a problem. Force itself is just Newtons (N) and isn't a rate. Momentum change (kg-m/s2) is something you can do with force and is a rate. The units are equal as simple derivations with the definition equations will tell you. I don't see why this should be so difficult to believe/accept.

That is what I was asking all along.

Force is a push or pull, measure in Newtons and it is not a rate. The change in momentum per time is kg-m/s^2--is rate. Force can create momentum change. Therefore, they are physically two different things, but because one relates directly to the other, we can mathematically write an equation stating the magnitude of one equals the other, just in pure calculation.
 
  • #59
BiGyElLoWhAt said:
I think I see the issue.
Momentum is a property of an object, namely ##mv##
Force is an interaction quantity, but the magnitude of the net force is equal to the magnitude of the rate of change of momentum. So, in a sense, they are two different "entities".
Momentum is, as stated above, a defined quantity, and net force = ma is derived from that definition, and the fact that they are equal quantities.
They are distinct in the sense that, just to reiterate, momentum is "had" by and object, in other words, momentum is a quantity that describes an object, whereas Force is a quantity that describes an interaction. When a nonzero net force is applied, you get a change in momentum, and the time rate of change is equal to the net force.

okay, well that makes sense. So are what you saying is that force and rate of change of momentum are physically different things. But because of the relation one has to another, we can mathematically state they are equal, treat as equal for calculations, yet know if when we work with a force is more less the interaction, that push as I have always been taught.

Sorry I do not have the depth as others in physics. This stuff is hard sometimes to grasp which is why I keep asking
 
  • #60
A force is essentially a push or a pull, but a push or a pull (non-zero) always creates a change in momentum, and the product F*t (for constant force).

So basically:
Force between two objects, acts for a time interval, and the change in momentum is equal to F*t for constant forces.
Interaction quantity for a time interval -> change in property quantity.
Force * time -> change in momentum
 

Similar threads

High School Questions about momentum and force as well as normal force/friction
  • · Replies 35 ·
2
Replies
35
Views
4K
Cannot understand dL/dt (angular momentum changes)
  • · Replies 34 ·
2
Replies
34
Views
6K
High School The physics of flywheel launchers (like tennis ball shooters)
  • · Replies 60 ·
3
Replies
60
Views
5K
Undergrad Precession of a gyroscope in freefall
  • · Replies 10 ·
Replies
10
Views
3K
High School Understanding Momentum: The Relationship Between Mass, Velocity, and Force
  • · Replies 29 ·
Replies
29
Views
3K
Why Does Force Impact Momentum More Directly Than Kinetic Energy?
  • · Replies 17 ·
Replies
17
Views
3K
Undergrad Question about Momentum vs. Kinetic Energy vs. Deforming In Collisions
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Lagrangian coordinate transformation (Noether)
  • · Replies 11 ·
Replies
11
Views
2K
Mass, acceleration and velocity
  • · Replies 3 ·
Replies
3
Views
2K
Understanding Momentum & Force: Self-Studying Physics Tips
  • · Replies 7 ·
Replies
7
Views
2K