How Does the Action-Reaction Principle Apply to a Screw Fastener and Nut?

[Mark2020]

A screw fastener advances into a nut. Assuming there are no frictional forces (or negligible):

a) Is the action force tangential to the male helical thread (screw) and the reaction tangential (in opposite direction to action) to the female helical thread (nut)
b) or does the action-reaction pair appears along the length of the screw (while advancing into the nut)?

According to my understanding, (a) justifies screw's motion (helical) and should be the answer, otherwise the screw wouldn't be able (due to screw threads) to advance into the nut.

Could someone help me understand the above? (a drawing would be very welcome!)

 

[jbriggs444]

A screw advances into a nut. Assuming there are no frictional forces (or negligible):

a) Is the action force tangential to the male helical thread (screw) and the reaction tangential (in opposite direction to action) to the female helical thread (nut)
b) or does the action-reaction pair appears along the length of the screw (while advancing into the nut)?

According to my understanding, (a) justifies screw's motion (helical) and should be the answer, otherwise the screw wouldn't be able (due to screw threads) to advance into the nut.

Could someone help me understand the above? (a drawing would be very welcome!)

You are trying to characterize a force that exists all up and down and around the threads of a screw and nut and treat it as if it were a single force applied in a single direction?

That's what vector sums are for.

I have a pet peeve about classifying forces as "action" and "reaction". Those are empty adjectives. They mean nothing.

 

[sophiecentaur]

Assuming there are no frictional forces (or negligible):

If there were no friction forces, all you'd be doing is giving the nut angular and linear acceleration. There will be a force / couple pushing the nut along / around and there will be a reaction force / couple against the wrench. This is exactly the same principle (it always is) at work when you push a block across a flat, horizontal and frictionless table.

As @jbriggs444 implies, first deal with your concerns about what's action and what's reaction (arbitrary choice really) then get to solving any problem with a Forces diagram.

 

[Mark2020]

@sophiecentaur,
Certainly, there has to be a wrench (or something else) to initiate nut's motion, however my question points to where the action-reaction pair appears between the male and female threads.

So, If I understood correctly from your answer, does what I marked as (a) is the correct answer?

 

[jbriggs444]

If there is no friction, how would the force end up being tangential?

 

[Mark2020]

Just to make the question clearer for all of us: I just would like to know in terms of the action-reaction principle, if the nut has to go around the screw thread over an action force (tangential to helix trajectory) then in order this motion to occur is required a reaction (tangential to helix trajectory) force (a pushing back) on screw's thread. Is this correct?

 

[Lnewqban]

@sophiecentaur,
Certainly, there has to be a wrench (or something else) to initiate nut's motion, however my question points to where the action-reaction pair appears between the male and female threads.

So, If I understood correctly from your answer, does what I marked as (a) is the correct answer?

A screw-nut system is equivalent to an inclined plane wrapped around a cylinder.
It is still a simple machine to do work with.
Just replace the weight of the sliding block with axial load.
Friction is what prevents actual systems from unscrew themselves under axial load.

This is the case of your problem:
https://en.wikipedia.org/wiki/Inclined_plane#Frictionless_inclined_plane

 

[Mark2020]

If there is no friction, how would the force end up being tangential?

I am speaking about an ideal situation where there are no energy losses due to friction. That was my initial thought.

 

[jbriggs444]

I am speaking about an ideal situation where there are no energy losses due to friction. That was my initial thought.

The question remains. We have a bolt that is freely rotating as it advances into a nut. There is no friction. Why would you expect any tangential force? Why would you expect any force at all?

When you say "tangential", which direction is that? Along the helical track of a thread? Or around a circle at a fixed radius from the center line?

[In the U.S. dialect at least, the distinction between a "screw" and a "bolt" is that screws are self-threading. A "screw" does not use a nut. It cuts or squishes into the material (typically wood, plastic or sheet metal) making the threads with which it meshes. Bolts, by contrast use nuts or pre-threaded ("tapped") holes. This fact is reflected in the design attributes of the two fasteners. "Screws" tend to have pointed tips. "Bolts" tend to have flat tips.]

 

[Mark2020]

The question remains. We have a bolt that is freely rotating as it advances into a nut. There is no friction. Why would you expect any tangential force? Why would you expect any force at all?

[The distinction between a "screw" and a "bolt" is that screws are self-threading. A "screw" does not use a nut. It cuts into the material (typically wood, plastic or sheet metal) making the threads with which it meshes. Bolts, by contrast use nuts or pre-threaded ("tapped") holes]

You are right, I messed it up a little. What I am speaking about is an ideal situation where there are no energy losses due to friction and we have a screw fastener and a nut. The initial thought was about a frictionless mechanical advantage. It sounds it contradicts in principle since the screw and the nut work based on frictional forces. Just for the sake of the argument we keep these frictional forces at play but without consuming energy. I just would like to know if what I share on Comment #6 is correct.

When you say "tangential", which direction is that? Along the track of a thread? Or around a circle at a fixed radius from the center line?

Along the track of the thread.

 

[jbriggs444]

Along the track of the thread.

Then the absence of friction means the absence of tangential force.

Now you have to decide whether there is tension on the screw/bolt.

 

[Mark2020]

Then the absence of friction means the absence of tangential force.

Now you have to decide whether there is tension on the screw/bolt.

And with tension what is the result?

 

[jbriggs444]

And with tension what is the result?

Still no tangential force. But what about the normal force?

 

[Mark2020]

Then what I wrote in Comment #6 should hold over the normal force or not? The nut has in a way to push the screw fastener (or vice versa) in order to advance. But the normal force would be perpendicular to the track of the thread, right?

 

[jbriggs444]

Then what I wrote in Comment #6 should hold over the normal force or not? The nut has in a way to push the screw fastener (or vice versa) in order to advance. But the normal force would be perpendicular to the track of the thread, right?

Right. The normal force at any point along the thread is perpendicular to the track of the thread. It has a non-zero component parallel to the axis of the bolt.

Picture a bolt inserted through this spring and threaded into the nut.

 

[Mark2020]

If the normal force is perpendicular to the track of the thread then, how the nut may push the screw fastener (or vice versa) in order to advance? This is the reason what I shared on my first post that the action and reaction forces should tangential to the threads of the nut and the screw fastener. Otherwise, motion cannot occur.

 

[jbriggs444]

If the normal force is perpendicular to the track of the thread then, how the nut may push the screw fastener (or vice versa) in order to advance? This is the reason what I shared on my first post that the action and reaction forces should tangential to the threads of the nut and the screw fastener. Otherwise, motion cannot occur.

Motion does not need force. Acceleration needs force. You provided the force to cause the initial acceleration when you started the bolt spinning. In the absence of friction and tension, that is all that is needed. The nut can spin onto the bolt indefinitely with no further force required.

Now we need to work through the consequences of tension and normal force.

 

[Mark2020]

Motion does not need force. Acceleration needs force.

In order the screw fastener to advance into the nut (with frictions but we assume they consume negligible energy), a continuous force is required. Stopping the action (or the wrench), the screw fastener will stop advancing. In this particular case force is associated with motion.

 

[jbriggs444]

In order the screw fastener to advance into the nut (with frictions but we assume they consume negligible energy), a continuous force is required. Stopping the action (or the wrench), the screw fastener will stop advancing.

Either it is negligible energy and no force is required or it is non-negligible energy and a force is required.

Pick one. You cannot have it both ways.

 

[Mark2020]

OK, I pick the second one.

 

[jbriggs444]

OK. So you have friction and you need a tangential force. So you need a wrench. What's the problem?

 

[Mark2020]

OK. So you have friction and you need a tangential force. So you need a wrench.

Does it mean what I write in Comment #6 is correct?

 

[Dale]

A screw fastener advances into a nut. Assuming there are no frictional forces (or negligible):

a) Is the action force tangential to the male helical thread (screw) and the reaction tangential (in opposite direction to action) to the female helical thread (nut)
b) or does the action-reaction pair appears along the length of the screw (while advancing into the nut)?

I also am having a hard time wrapping my mind around a bolt and a nut without friction. So I will simply answer with regards to an ordinary bolt and nut that do have friction under normal fastening conditions.

The threads of the bolt and nut form a pair of inclined planes that are in contact all along their length. So you don't have a force per-se but rather a force density. The technical term is a stress. There are components of the stress directed perpendicular to the inclined plane surface and components of the stress directed parallel to the inclined plane section.

If you take an arbitrary section of the nut thread surface then you can integrate the stress over that section to get the force acting on that section of the nut thread surface. By Newton's third law there will be a corresponding section of the bolt thread surface, and if you integrate the stress over that section of the bolt thread surface then you will find that the force is equal and opposite the corresponding nut thread force.

I don't know how to translate that to a frictionless case, but there it is for the usual case.

 

[jbriggs444]

Does it mean what I write in Comment #6 is correct?

Yes, the "action reaction principle" holds for every incremental section where screw thread contacts nut thread. The tangential/frictional force of the one on the other is equal and opposite to the tangential/frictional force of the other on the one.

In practical terms, this means that you need two wrenches. One to drive the screw and one to hold the nut.

 

[jbriggs444]

I also am having a hard time wrapping my mind around a bolt and a nut without friction.

Most easily visualized for me as a wing-nut on a smooth bolt. Give that wing nut a good flick and it'll spin a long ways down the bolt.

 

[Mark2020]

@jbriggs444. Thank you very much for your confirmation. I was really confused as you saw.

An interesting observation on the above (your confirmation) is: Seeing the screw fastener and the nut from a distance, let us say the screw and nut threads are not visible (from a naked eye): Along the length of the screw fastener there is not any rectilinear reaction but just a "virtual" rectilinear action force (nut advances over the screw fastener or vice versa) that as seen from a distance pushes the nut (or the screw fastener) to one direction. Is this true or false?

 

[jbriggs444]

@jbriggs444. Thank you very much for your confirmation. I was really confused as you saw.

An interesting observation on the above (your confirmation) is: Seeing the screw fastener and the nut from a distance, let us say the screw and nut threads are not visible (from a naked eye): Along the length of the screw fastener there is not any rectilinear reaction but just a "virtual" rectilinear action force (nut advances over the screw fastener or vice versa) that as seen from a distance pushes the nut (or the screw fastener) to one direction. Is this true or false?

I do not know what a "virtual rectilinear action force" is. The momentum of the advancing nut is unchanging over time. It needs no force to allow it to advance.

However, if there is resistance from friction and if the nut continues to advance with unchanging momentum then it follows that there must be a matching "second law partner force" that counters that resistance: A wrench applied to the nut. [Be very wary of using the phrase "second law partner" in public]

Newton's second law says that $\sum F = ma$. If we know that $ma$ is zero, it follows that $\sum F = 0$. If there are just two forces $F_1$ and $F_2$ then it is clear that $F_1=-F_2$. These two forces might be ironically called "second law partners"​

We might examine the tangential frictional force in more detail. It has a component in the direction of a circle around the bolt's center axis. It has a component in the direction parallel to the bolt's center axis. Along the length of the thread it will have a net that is in the axial direction together with a torque about that axis. Yet we counter both net effects with just a wrench. No axial force required. How can that be?

You need to consider the normal force as well.

 

[Mark2020]

However, if there is resistance from friction and if the nut continues to advance then it follows that there must be a matching "second law partner force" that counters that resistance: A wrench applied to the nut.

I speak about the quoted part, above. By "virtual rectilinear action force" I mean the advancing of nut (or screw fastener) is not caused by e.g. contact force (like directly pushing the nut or the screw fastener along its length) but through a "virtual" rectilinear force (as seen from distance without noticing the screw threads) that essentially is an induced force (not a contact force). Something similar to electromagnetic induction. Would that be a valid description of what we observe?

 

[jbriggs444]

I speak about the quoted part, above. By "virtual rectilinear action force" I mean the advancing of nut (or screw fastener) is not caused by e.g. contact force (like directly pushing the nut or the screw fastener along its length) but through a "virtual" rectilinear force (as seen from distance without noticing the screw threads) that essentially is an induced force (not a contact force). Something similar to electromagnetic induction. Would that be a valid description of what we observe?

Again, there is no change in momentum. No net force needed to advance the nut.

The net force from screw threads on nut threads is most definitely a contact force. If you want to summarize it as a net linear force plus a net torque, that is fine. [One can always summarize a set of forces acting on a rigid body into a single linear force plus a single torque that will produce the same effect. The calculation process is straightforward. The result is uniquely determined].

But labeling this net force as "virtual" and then making an analogy to electromagnetic induction. That strikes me as unnecessary.

 

[Mark2020]

jbriggs444 said:
Again, there is no change in momentum. No net force needed to advance the nut.

But didn't we say a real world screw fastener and nut with frictions, requires a constant wrench force in order the nut to advance or not?

jbriggs444 said:
The net force from screw threads on nut threads is most definitely a contact force. If you want to summarize it as a net linear force plus a net torque, that is fine.

I don't speak about what happens between the threads. You perfectly clarified that and thank you! I am just addressing the following situation. I will express it in a series of short statements:
1.Fastener screw advances into a nut because of force coming through a wrench
2.Imagine (1) as a picture without the wrench in it (however is still acting)
3.Imagine (2) from a large distance where the screw and the nut threads are not visible (due to distance)
4.Because of (3) the observer (from a distance) sees a screw fastener advancing into a nut or vice versa
5.We know what happens between the threads, when we are very close to the fastener screw-nut system
6.Because of (3) the observer (from a distance) has the impression that a rectilinear contact force applies along the rotation axis of the screw fastener or along the rotation axis of the nut. This happens because he cannot see the screw threads from that distance. Or simply we cover the threads from the naked eye and we stay close to the screw fastener and nut.

Is it clear up to now or am I wrong somewhere in between?

to be continued...

 

[Dale]

Most easily visualized for me as a wing-nut on a smooth bolt. Give that wing nut a good flick and it'll spin a long ways down the bolt.

Yeah, I just can't visualize the forces in that case. In principle there shouldn't be any since there is no acceleration. Maybe some forces opposing gravity? But that gets complicated. I am more confident with friction here than without.

 

[Mark2020]

7.Since the action-reaction principle applies just between the fastener screw and nut threads (not visible by a naked eye) then because of (6) it looks like having a not real (therefore "virtual") rectilinear contact action force exerted upon the nut. Again, because the observer sees no screw threads.
8.Due to (6) and (7), there is not any rectilinear contact reaction force exerted upon the fastener screw.

 

[jbriggs444]

7.Since the action-reaction principle applies just between the fastener screw and nut threads (not visible by a naked eye) then because of (6) it looks like having a not real (therefore "virtual") rectilinear contact action force exerted upon the nut. Again, because the observer sees no screw threads.
8.Due to (6) and (7), there is not any rectilinear contact reaction force exerted upon the fastener screw.

I do not know what you are talking about.

The frictional force is there whether you can see the threads or not. The constraint imposed by the threads (that the rotation rate correlates with advance rate) is there whether you can see the threads or not. There is no physical property of the system that changes when you label the interaction between the screw and nut as any of:

"virtual"
"rectilinear"
"contact"
"reaction"

So why bother with any of those labels. They do not convey any actionable information.

 

[jbriggs444]

I don't speak about what happens between the threads. You perfectly clarified that and thank you! I am just addressing the following situation. I will express it in a series of short statements:
1.Fastener screw advances into a nut because of force coming through a wrench
2.Imagine (1) as a picture without the wrench in it (however is still acting)
3.Imagine (2) from a large distance where the screw and the nut threads are not visible (due to distance)
4.Because of (3) the observer (from a distance) sees a screw fastener advancing into a nut or vice versa
5.We know what happens between the threads, when we are very close to the fastener screw-nut system
6.Because of (3) the observer (from a distance) has the impression that a rectilinear contact force applies along the rotation axis of the screw fastener or along the rotation axis of the nut. This happens because he cannot see the screw threads from that distance. Or simply we cover the threads from the naked eye and we stay close to the screw fastener and nut.

Is it clear up to now or am I wrong somewhere in between?

I may be coming to an understanding of what you are trying to say. Let me try to put it differently.

We are trying to describe the nut and screw interaction as something of a black box. We cannot see the details of the interface. The only things we can observe is the resulting behavior of the objects.

We may begin by considering the bolt to be fixed (perhaps its head is clamped in a vise).

We have a nut threaded onto the bolt. We have a torque wrench applied to the nut.

From a distance, we can observe the position of the nut on the bolt, the rotation angle of the nut and the reading on the torque wrench. And perhaps we are allowed to give orders to the fellow with the wrench.

We may observe, for instance, that the nut is constrained so that it orientation is always parallel to the bolt. Its position is always centered on the bolt. Importantly, its rotation angle is a fixed multiple of its linear position along the axis of the bolt. We may not know whether we have a thick bolt or a thin. We may not know whether we have course threads or fine. But we do know the ratio of rotation to linear displacement.

Let me add something to the system. A spring and a washer. As the nut advances onto the bolt, it moves the washer which in turn compresses the spring. From our far away vantage point however, we know nothing about the nature of the spring.

We can watch as a fixed torque is applied and the nut makes steady progress down the shaft.

We can compute the rate at which energy is supplied to the system -- power = torque times rotation rate.

If there were no friction, we would conclude that mechanical energy must leave the system somehow. It might be going into the spring. It might be dissipated as friction in the threads.

I will make a claim here: With the measurements available to us from our far away vantage point we cannot distinguish between the two possibilities.

 

[Mark2020]

So why bother with any of those labels. They do not convey any actionable information.

Although, obvious I was waiting to be confirmed before I proceed with what I have in mind. Then, let's say we have a construction that is based on the fastener screw-nut mechanism, namely a linear actuator suspended by a strand.

We have attached a small mass (0.5 Kgr just to give more momentum) to the moving part (nut) that can be transferred from one end to the other. Based on your comment above, all those terms I used convey no actionable information.

So, we power the actuator (it uses a motor) and when the moving part starts moving then due to the momentum conservation, the entire construction (linear actuator) will start moving in the same direction as the moving part, right?

 

[jbriggs444]

Although, obvious I was waiting to be confirmed before I proceed with what I have in mind. Then, let's say we have a construction that is based on the fastener screw-nut mechanism, namely a linear actuator suspended by a strand.

We have attached a small mass (0.5 Kgr just to give more momentum) to the moving part (nut) that can be transferred from one end to the other. Based on your comment above, all those terms I used convey no actionable information.

So, when the moving part starts moving then due to the momentum conservation, the entire construction (linear actuator) will start moving in the same direction as the moving part, right?

If I understand the setup properly, we have a nut and bolt suspended by a thread. The nut is made more massive by adding some mass to it. In my mind's eye, we have welded some wings onto it and balanced them nicely.

We have some sort of self-contained motor arrangement on the pair so that we can press a button and have the nut spin itself from one end of the bolt to the other.

Now you claim that when the nut spins down the shaft to the right, that the bolt + motor also moves to the right?! That would not conserve momentum.

I begin to have grave misgivings about the prospect that we are being asked to confirm the design of a reactionless drive -- something that magically converts torque to linear impulse.

Edit: Misgivings confirmed -- and I'm out as of #55.

 

[Mark2020]

Now you claim that when the nut spins down the shaft to the right, that the bolt + motor also moves to the right?! That does not conserve momentum.

The momentum conservation should always agree with Newton's 3rd law or not. Since in the direction of motion (to the right) there isn't any rectilinear reaction force then the system will follow the momentum of the nut, like having being pushed by a virtual rectilinear action force (the reason I used the definition "induced" or "virtual" since both do not justify a real rectilinear force).

 

[jbriggs444]

The momentum conservation should always agree with Newton's 3rd law or not. Since in the direction of motion (to the right) there isn't any rectilinear reaction force then the system will follow the momentum of the nut, like having being pushed by a virtual rectilinear action force (the reason I used the definition "induced" or "virtual" since both do not justify a real rectilinear force).

Words, words and more words signifying nothing.

Rectilinear: Meaningless.
Reaction: Meaningless.
Virtual: Meaningless.

Momentum is conserved. If the nut goes one way and no external force is applied, the bolt + motor will go the other. You cannot cancel out forces by slapping labels on them creatively.

 

[Mark2020]

Momentum is conserved. If the nut goes one way and no external force is applied, the bolt + motor will go the other. You cannot cancel out forces by slapping labels on them creatively.

That would be true in the case of a linear actuator (not a real one), where the leadscrew would be replaced with a non-threaded rod. In this particular case you have an action-reaction pair where these forces are actually collinear and apply along the length of the linear actuator.

 

[jbriggs444]

That would be true in the case of a linear actuator (not a real one), where the leadscrew would be replaced with a non-threaded rod. In this particular case you have an action-reaction pair where these forces are actually collinear and apply along the length of the linear actuator.

It is always true. Momentum conservation holds. No fancy mechanisms, no colorful labels and no creative summarizations applied to forces change that. You have an action reaction pair that are equal and opposite. The net force of screw on nut is equal and opposite to the net force of nut on screw.

This can be easily seen. The forces at each point of contact are equal and opposite. So the forces summed (or integrated) over all of the points of contact have to add to equal and opposite totals. Newton's third law holds both for the individual contributions and for the aggregated whole.

 

[Mark2020]

You have an action reaction pair that are equal and opposite. The net force of screw on nut is equal and opposite to the net force of nut on screw.

Nobody argues against that. However, as we discussed above and as you confirmed it, the action-reaction principle appears just between the threads or better saying tangential to the track of the threads (from the nut and from the fastener screw side). This action-reaction collinear pair is perpendicular to the motion of the nut as advances to the right (advances horizontally). Apparently, there isn't any action-reaction collinear pair along the direction the nut moves (horizontally).

 

[jbriggs444]

essentially perpendicular

"essentially perpendicular" is not the same as "perpendicular". Pay attention to the details so that they do not render your conclusions invalid.

You still have paid no heed to the normal force, choosing instead to restrict your attention to the frictional force -- the very force which you began by claiming that it was negligible.

 

[Mark2020]

Thanks, I corrected it to just "perpendicular".

 

[jbriggs444]

Thanks, I corrected it to just "perpendicular".

The tangential force is not perpendicular to the axis of the screw.
The normal force is not perpendicular to the axis of the screw.

In steady state and assuming no axial force from the actuator mechanism (pure torque only), the net axial component of the contact force is zero.

But if you want to move a stationary nut from one end of a linear actuator where it starts at rest to the other end of a linear actuator where it ends at rest, you will have to have something different from a steady state at both the beginning and the end of the exercise.

Momentum will be conserved throughout. But any momentum change in the nut will be matched by an equal and opposite momentum change in the screw + actuator body.

 

[Mark2020]

The tangential force is not perpendicular to the axis of the screw.

Then it is almost perpendicular due to the helix topology (ascribes a circle while the center of it advances in the z-axis). But this cannot invalidate the conclusion presented in post #41.

 

[jbriggs444]

Then it is almost perpendicular due to the helix topology (ascribes a circle while the center of it advances in the z-axis). But this cannot invalidate the conclusion presented in post #41.

Incorrectly reasoned.

First, the frictional force has a non-zero net component parallel to the axis. If there is any friction at all (and we've agreed that there is), it will be directed parallel to the threads. The direction parallel to the threads is not perpendicular to the axis. It has a non-zero axial component. There is a net frictional force on the nut retarding its forward motion.

But we've not considered the normal force. If the actuator mechanism exerts no axial force on the nut (well lubricated slides) then the normal component of the contact force must be working to advance the nut onto the screw. Has to do so. It's the only other force that can. F=ma. a=0. We have second law partners here.

But this is not terribly relevant. I think we can agree that there is zero net force between screw and nut during the steady state condition while the nut is advancing from one end to the other. The normal force and the frictional force end up adding to a linear total of zero. [They also add up to a non-zero torque -- hence the need for a motor]

But what about the start event and the end event? Normal force rears its head there. It takes normal force from screw on nut to accelerate the nut into motion. It takes normal force from screw on nut to decelerate the nut to a stop. Both have a net that is parallel to the axis of rotation and is not matched by an equal and opposite frictional force. Both result in a momentum change by the screw and actuator body that is equal and opposite to the change by the nut.

 

[Mark2020]

I think we can agree that there is zero net force between screw and nut during the steady state condition while the nut is advancing from one end to the other.

When one reads what is the mechanical advantage is about, it is a ratio of forces equals to a ratio of lengths or speeds that means as long as the nut moves, a force is always at play. No force, no motion. From my understanding (by seeing the mechanical advantage) the nut acquires a constant acceleration, meaning a constant changing in momentum (even if it is practically negligible).

 

[jbriggs444]

long as the nut moves, a force is always at play. No force, no motion.

Wrong, wrong, wrong. That is Aristotle talking.
Let me be more charitable. We have non-zero friction. If there is motion, there must be at least one force. If there is constant speed motion, there must be at least two forces. But we have two forces, so that's fine.

The nut eventually acquires a constant velocity. Which means a constant (zero) acceleration. A constant acceleration of zero means that momentum is not changing.

Newton tells us:

No net force => no acceleration.
No net force => continuing motion at a constant speed.
No forces => no net force (this is part of linearity and is not one of the three laws).

If you have motion and you have constant velocity, what you can conclude is that the forces add to zero.

 

[Mark2020]

Wrong, wrong, wrong. That is Aristotle talking.

No net force, no acceleration.
No net force, continuing motion at a constant speed.
No forces, no net force.

You are confusing free rectilinear motion (according to Newton) with how a linear actuator works. If the linear actuator starts running and you disconnect .e.g its power, it will stop immediately (even if the leadscrew is contactless (electromagnetic) that means no contact between the threads).

 

[jbriggs444]

You are confusing free rectilinear motion (according to Newton) with how a linear actuator works. If the linear actuator starts running and you disconnect .e.g its power, it will stop immediately (even if the leadscrew is contactless (electromagnetic) that means no contact between the threads).

Nothing is immediate. [Well, maybe particle decays, but those get shrouded in quantum uncertainty]

There will be a short period [perhaps very short] during which the mechanism is decellerating to a stop. This may involve large accelerations and large forces. Momentum is conserved during such events regardless of the magnitudes of the forces and accelerations.