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

[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.

 

[Mark2020]

Let's agree on this: the nut acquires a constant acceleration in the linear actuator by increasing the input power (and force) constantly. While the nut travels from one end to the other, what will be the momentum of the rest of the system?

Will it follow or will it go against the nut? Please take into consideration there are no rectilinear forces along the length of the leadscrew (or the linear actuator).

 

[jbriggs444]

Let's agree on this: the nut acquires a constant acceleration in the linear actuator by increasing the input power (and force) constantly. While the nut travels from one end to the other, what will be the momentum of the rest of the system?

Will it follow or will it go against the nut? Please take into consideration there are no rectilinear forces along the length of the leadscrew (or the linear actuator).

You continue to post contradictions.

If there is no net linear force on the nut then the nut cannot have a constant non-zero linear acceleration. There is a net linear force from actuator on nut in this case. Otherwise it would remain motionless where it started at rest. I claim that this net linear force arises primarily from the normal force of lead screw threads on nut threads.

Let us assume that you are indeed ramping up power to the actuator, increasing the spin rate of the lead screw at a constant rate. Let us agree that this results in a constant non-zero acceleration of the nut down the axis of the actuator.

Then it follows that as the nut accelerates one way down the length of the actuator, the lead screw and the rest of the actuator mechanism will be accelerating in the opposite direction. This is called "recoil".

 

[Mark2020]

Then it follows that as the nut accelerates one way down the length of the actuator, the lead screw and the rest of the actuator mechanism will be accelerating in the opposite direction. This is called "recoil".

Again, that would be true if we had to do with collinear forces. In order this to take place you have to replace the leadscrew with an unthreaded rod. Then the "recoil" force will work that will be interpreted to system's opposite momentum.

In our case, the leadscrew turns just to one direction. The advancing of nut and the system as a whole is limited/depended on the direction of rotation of the leadscrew. So, a clockwise rotation of the leadscrew will force the nut to advance counterclockwise that appears moving to the right. This means the system cannot develop the known collinear "recoil" force but just over the leadscrew. Conclusively, since the leadscrew turns only to one direction (never both) at a time, then the "recoil" force that applies to the rest of the system will have the same direction as the force (non collinear force) makes the nut evolving to the right. In other words, the nut and the system will have the same magnitude and same direction of momentum.

 

[jbriggs444]

Again, that would be true if we had to do with collinear forces. In order this to take place you have to replace the leadscrew with an unthreaded rod. Then the "recoil" force will work that will be interpreted to system's opposite momentum.

Again, there is no magic that occurs when you change the name you choose to label a force with.

 

[Mark2020]

I just would like to thank you for your participation in this discussion. I was looking to test some of my ideas with the experts of this forum.

Thanks everybody!

 

[berkeman]

Thread closed temporarily for Moderation...

 

[Dale]

This thread will remain closed. We remind all readers that personal speculation is not permitted on PF