Friction in Everyday Life: Walking & Knots

[fog37]

Hello everyone,

I recently realized that knots in ropes and shoe laces are possible because of the friction between the rope fibers. I guess if friction was not present the knot would just not work, i.e. a knot would still form but not be tight...

What are some other interesting examples of the action of friction in everyday life/physics? Walking in surely possible only thanks to static friction. What else?

Thanks!

 

[jedishrfu]

Wheel traction: If you were on ice you can see why friction is important.

Straps used in holding down items on a flatbed truck use friction to keep from loosening.

 

[fog37]

Thanks.

Quick comment about the wheel traction: if the wheel ends up on the ice already rolling and keeps moving at constant speed, no friction will be needed for the rolling.

As far as knots, goes is it really true that friction is necessary? Or would the interlocking between the rope sections or shoe laces still provide a firm junction?

 

[jbriggs444]

What else?

Nuts and bolts, screws, nails. It's also really hard to start a fire with a bow drill in the absence of friction. And the wind blows pretty fast.

 

[Drakkith]

What are some other interesting examples of the action of friction in everyday life/physics? Walking in surely possible only thanks to static friction. What else?

Sitting in a chair. Driving my truck. Turning many doorknobs. Without friction I'd slide right out of my chair (unless it's a strange chair or a chair that's tipped backwards), fall to the ground, and be stuck there since I'd never be able to get up. Nearly everything you do in your everday life depends on friction, so much so that I'm having a hard time even coming up with ideas of things which don't rely on friction. Even the things which don't depend on friction still wouldn't be able to be done since you'd need friction just to get up and move around to do them.

 

[A.T.]

As far as knots, goes is it really true that friction is necessary? Or would the interlocking between the rope sections or shoe laces still provide a firm junction?

What interlocking? In theory, you could design ropes with special surfaces that provide some resistance to sliding, without relying on friction, similar to hook-and-loop-fasteners.

 

[anorlunda]

With monofilament fishing line, some traditional knots work fine while others don't.

I'm sure friction plays a role but it is not the whole story.

 

[fog37]

As far as the screw goes, is friction really necessary? The screen is twisted inclined plane. Once it is inside the material, does friction really need to be there to help fasten?

 

[Drakkith]

As far as the screw goes, is friction really necessary? The screen is twisted inclined plane. Once it is inside the material, does friction really need to be there to help fasten?

Certainly. Without friction, the screw would very easily unfasten. What do you think keeps the screw from turning backwards?

 

[Nugatory]

As far as the screw goes, is friction really necessary? The screen is twisted inclined plane. Once it is inside the material, does friction really need to be there to help fasten?

The inclined plane geometry is an ingenious way of developing static friction between the face of the screw threads and the material around the screw. Turning the screw clockwise increases the normal force between the shoulders of the male and female threads, and hence the static friction that stops the screw from backing out.

If there were no friction, the same inclined plane would force the screw to turn backwards until it was loose as soon as you stopped forcing it in.

 

[fog37]

I see, thanks Nugatory and Drakkith. In essence, without static friction, the screen would quickly turn backward like when, in a linear frictionless ramp, we push an object up the ramp (or viceversa) and the object promptly slides back down. The static friction on the helical ramp is due to the pressure of the material in which the screw is screwed into because the ramp is being pushed, as the screw is turned, more and more into the material which causes a resistance. Would that be a compression type of resistance or shear?

 

[Asymptotic]

Plastics extruders rely primarily upon frictional heating for polymer melting.

 

[fog37]

Thanks Asymptotic. But, in the case of extruders, aren't high temperatures (softening point) used?

 

[Asymptotic]

Thanks Asymptotic. But, in the case of extruders, aren't high temperatures (softening point) used?

Yes. External heating is used when the screw is at rest, but most of the heating during processing is due to friction between polymer and barrel/screw. In fact, once screw speed increases beyond a certain point usually the problem is one of cooling the polymer sufficiently to prevent degradation.

General purpose screws are compromise designs to accommodate a range of different polymers and operating rates, and require heating and cooling "trimming" along their length. When a screw can be designed for a specific polymer and operation within a narrow speed range it becomes possible to maintain barrel zone temperature setpoints with very little or no external heating and cooling inputs, a condition once commonly (and erroneously) called 'adiabatic extrusion'.

 

[fog37]

A wedge is a single or a double inclined plane that gets pushed into a solid material. A knife and a chisel would be a type of wedge. The force on the wider area of the wedge produces a pressure which gets transmitted to the thinner area producing a larger force over the small area. So far there is no friction involved in this process. It seems that friction is not necessary for a wedge to do its job.

I am still not sure how to view the the situation from a force standpoint. Assuming an input vertical force on the top of the wedge trying to penetrate the horizontal surface of the material. I guess the vertical forces pushes the material along the two inclined planes of the wedge. Without the wedge, the material would just be compressed in, correct? The wedge inclined surfaces produce forces directed perpendicular to the the surface itself. How does that make it easier for the wedge to enter the material?

 

[jbriggs444]

A wedge is a single or a double inclined plane that gets pushed into a solid material. A knife and a chisel would be a type of wedge. The force on the wider area of the wedge produces a pressure which gets transmitted to the thinner area producing a larger force over the small area. So far there is no friction involved in this process. It seems that friction is not necessary for a wedge to do its job.

I am still not sure how to view the the situation from a force standpoint. Assuming an input vertical force on the top of the wedge trying to penetrate the horizontal surface of the material. I guess the vertical forces pushes the material along the two inclined planes of the wedge. Without the wedge, the material would just be compressed in, correct? The wedge inclined surfaces produce forces directed perpendicular to the the surface itself. How does that make it easier for the wedge to enter the material?

"easier" is perhaps not the best or most quantitative term to describe the "mechanical advantage" gained when using a wedge.

Suppose that you are driving a wedge downward into a piece of material. You apply a small force downward on the top of the wedge. The material is subject to a large force outward from the faces of the wedge. In the absence of friction, mechanical energy is conserved. The work done applying a small downward force over a large downward displacement on the top of the wedge is equal to the work done as the faces of the wedge apply a large normal force over a small normal displacement.

 

[fog37]

Yes, in the case of a wedge, friction is not helpful but only subtracts energy aways.

In screws instead, whose purpose is fastening, static friction is essential for keeping the screw inside the material.

 

[A.T.]

How does that make it easier for the wedge to enter the material?

- A sharp tip applies more pressure than a blunt one, given the same force.
- The pressure forces on the sides of the wedge partially cancel each other.

 

[fog37]

Thanks. So, conceptually, the large input force directed vertically down inside the material is converted into two sides forces normal to the wedge surface. These two side forces push the material sideways. Does pushing the material sideways that make the penetration of the wedge easier?

 

[jbriggs444]

Thanks. So, conceptually, the large input force directed vertically down inside the material is converted into two sides forces normal to the wedge surface. These two side forces push the material sideways. Does pushing the material sideways that make the penetration of the wedge easier?

Again, the word "easier". Again, not the best choice.

You can view the situation using the rules of vector addition of forces. The sum of two large and nearly horizontal forces can be a small vertical force.

You can view the situation using the notions of work and energy. The product of a large force and a small displacement is equal to that of a small force and a large displacement.

Perhaps there is another viewpoint -- what is resisting the penetration of the wedge?

1. The tearing of the material at (or in advance of) the point of the wedge.
2. The compression of the material as it is displaced by the wedge.

A wedge with a thinner and sharper point does not need to mash as much material downward.

A wedge with more nearly parallel faces does not need to do as much compression as it moves through the material.

 

[fog37]

Thanks jbriggs444.

So, from the force standpoint, the effort force directed vertically into the material to make the wedge enter the material to a depth $d$, is lowered because the the two larger and nearly horizontal forces produce a small vertical force acting vertically towards the outside of the material. However, I guess it is better to have two large horizontal forces pushing the material laterally than a single force pushing the material downward. I am looking for even more conceptual insight. It is like entering a material by pushing material away sideways instead of just pushing down (compressing) on it.

 

[A.T.]

Does pushing the material sideways that make the penetration of the wedge easier?

Draw a vector diagram.

 

[jbriggs444]

I am looking for even more conceptual insight. It is like entering a material by pushing material away sideways instead of just pushing down (compressing) on it.

You can cut a stick of soft butter in two with either the point of a butter knife or the flat of the blade. One way takes less effort and makes less mess.

 

[fog37]

Sure. the mechanical advantage of a wedge increases with decreasing the vertex angle. I guess a the blade of a sharp knife is a like a wedge with zero vertex angle. The blade does not push sideways on the material. It is just exerting a large pressure - hence large force - downward. For the blade to move down, I guess the material must be pushed somewhere else to get out of the way and allow for the blade to make its way. In the case of a soft stick of butter, I guess the butter stick gets easily deformed. In the case of wood, it is better to use a wedge though...

 

[jbriggs444]

I guess a the blade of a sharp knife is a like a wedge with zero vertex angle.

The bulk of a knife blade may or may not have a taper. Let us agree that we are talking about butter knives in which the two flat sides are parallel. Zero vertex angle. In the absence of friction, sliding the body of the blade through a slab of butter (or wood) would take zero force.

But there is always a taper on the cutting edge of a sharp knife. The vertex angle is always non-zero. There is a trade-off between a sharp angle where the cutting edge may become overly fragile and a blunt angle where slicing fails and crushing takes place instead.

 

[A.T.]

It is just exerting a large pressure - hence large force

A large pressure despite a small force.

 

[fog37]

Hello, I was re-reading these thread and the comments about wedges. Wedges are pointy, hence they produce large pressure (force per unit area) at the point of contact. People often state that a knife is an example of a wedge. But I don't see sloping sides on a knife but just a sharp blade which leads me to think that a knife only cuts due to the large pressure and not due to the wedge shape (sloped sides) design.

A wedge is essentially two inclined planes and is used to split apart objects. The applied force applies force to the wedge along both its sloping sides and these two forces causes the object to split apart. By vector equivalence, the large vertical force is converted into two side forces have small components downward and large components perpendicular to the wedge sloped planes.

When cutting an object, is it easier to to cut it be separating apart laterally than pushing vertically through it? Mere pushing, even with a large pressure, produces deformation, fusion, etc. The wedge proves that, by converting a single vertical force into two side forces, it is easier to enter a material by pushing it sideways than straight into it...

 

[jbriggs444]

Hello, I was re-reading these thread and the comments about wedges. Wedges are pointy, hence they produce large pressure (force per unit area) at the point of contact. People often state that a knife is an example of a wedge. But I don't see sloping sides on a knife but just a sharp blade which leads me to think that a knife only cuts due to the large pressure and not due to the wedge shape (sloped sides) design.

Have you ever tried to cut a crisp carrot with a knife? It is not just the point that does the work. Once the cut is started, the sides force the material apart and advance the break in front of the point. Same for splitting wood.

In an axe used cross-cut to shear chips off of a trunk or limb, you are correct that it is the point that achieves the desired effect, tearing fibers with extreme local pressure.

 

[A.T.]

I don't see sloping sides on a knife but just a sharp blade.

I don't see a forest, just lots of trees.

https://en.wikipedia.org/wiki/Knife#Blade

 

[fog37]

Thank you.

I think I am getting it now. I think I found out that even a zipper is based on the principles of the inclined plane...

When an inclined plane is at rest and an object of weight $mg$ is pushed up the slope, the effort to take it to a certain height $h$ is less than $mg$, i.e. $mg sin\theta$.
When the wedge is used as a door stopper, it is necessary for the base of the wedge to have some static friction. The door exerts a horizontal force on the wedge but this force, I guess, is less than if the stopper did not have a slope. A smaller force is applied to the wedge and the required static friction is less to keep the wedge from moving. I am still not sure why the horizontal force that the door applies to the wedge while the door tries to close is a smaller force than the force that would be applied if the stopper did not have a sloped surface.

 

[Nugatory]

I am still not sure why the horizontal force that the door applies to the wedge while the door tries to close is a smaller force than the force that would be applied if the stopper did not have a sloped surface.

The horizontal force the door applies to the stopper must be exactly balanced by the friction force between stopper and floor - if the forces aren't balanced the stopper would move, and if it's doing its job it's not moving.

And how great is that force? It's a frictional force, so it's equal to the coefficient of friction times the normal (vertical!) force between stopper and floor. Now we need to know what that normal force is, and that's going to be the weight of the stopper plus the vertical force of the door on the stopper. That vertical force is what is affected by the slope of the wedge.

You may find it helpful to consider a few extreme cases. First, suppose that the stopper is very heavy - we're using a concrete block or an iron anvil as a doorstop. The doorstop will work just fine even though it's not wedge shaped at all; the normal force from its weight is sufficient to generate enough friction with the floor to hold the door. Second, consider a more normal wedge-shaped doorstop, but imagine what would happen if the door really wanted to swing; not just a powerful wind pushing on it, but a very strong person trying to shove it open. Because of the wedge shape, the harder we push the door against the stop, the greater the vertical force between door and stopper and between stopper and floor (these forces must be balanced because the stopper isn't moving) so the greater the frictional force with the floor that keeps the stopper from moving sideways and allowing the door to open. Push hard enough, and you will see the door being forced upwards as the vertical forces increase - and it's the wedge shape that allows the horizontal push to generate a vertical force.

 

[A.T.]

I am still not sure why the horizontal force that the door applies to the wedge while the door tries to close is a smaller force than the force that would be applied if the stopper did not have a sloped surface.

The horizontal force by the door isn't smaller, but the vertical force by the door allows more static friction at the base, so the wedge can resist more horizontal force by the door than a block of the same weight.

 

[fog37]

Ok, I see:

the horizontal force exerted by the door produces a normal and vertical reaction force on the slope of the wedge. This vertical normal force, which would not exist if if the wedge slope was completely vertical, contributes to the available frictional force. When we push horizontally on a slope surface, part of the push is transferred by the sloped surface downward. This would not happen if the surface was perfectly vertical like a wall...

 

[A.T.]

the horizontal force exerted by the door produces a normal and vertical reaction force on the slope

Please, forget this "force produces another force" stuff. There is a contact force by the door on the wedge which can be decomposed into components, in different ways:
normal and tangential
vertical and horzontal

This vertical normal force,

The normal force on the wedge top is not vertical
normal : perpendicular to the contact surface
vertical: perpendicular to the floor

When we push horizontally on a slope surface, part of the push is transferred by the sloped surface downward.

Again, the horizontal force not reduced by "transferring part of it downward". But it can be balanced better thanks to the vertical force by the door, which allows more horizontal friction with the floor.

 

[fog37]

Thanks A.T.

To paraphrase you, the normal (reaction force) force on the sloped surface of the wedge can be decomposed into a vertical upward component and a horizontal component directed towards the door. The vertical force component adds to the normal force that the weight of the wedge stopper produces. This increases the available friction from the floor (since the overall normal force has been increased) allowing the stopper to be more effective at keeping the door from moving.