Why isn't centrifugal force drawn on a free body diagram?

WBahn

Joined Mar 31, 2012
29,978
Railroad cars have mating buffers that absorb the push.
The coupling links are slack in such circumstance. They are never in compression.
How is that possible? When you have a long train with engines at both the front and the back, are you saying that the coupling between the rear locomotive and the car just ahead of it is not in compression?

I've only seen mating buffers (or what I think you mean by that term) on cars that have tensioned chain linkages. I don't recall ever seeing any such buffer on a typical railroad car that uses the normal coupling (or at least what I think of as the standard railroad coupling). For instance, consider the following image (snagged from Google):

upload_2015-4-7_15-15-18.png

Where is the mating buffer and how is this coupling not under compression if one car is pushing the other?
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
Wow, it looks like I got quite a discussion going.:rolleyes:
You are confusing centrifugal (outward acting) force with centripetal (inward acting) force. The force on the object holding it in orbit is centripetal force. There is NO centrifugal force acting on the object (in an inertial reference frame). If you want to talk about the force acting on the rope (or on the other object in the case of force at a distance) then it is reasonable to call that centrifugal force. But the TS is very clearly talking about the free body diagram of the object being swung on a rope, NOT on the rope and NOT on the object at the other end of the rope.
@WBahn: So then there really is such a thing as centrifugal force, but it acts on the rope and the object at the other end of the rope, not the object being swung? I knew there had to be some force acting outward in order for Newton's 3rd Law to hold true.
The force of the floor acting on the box very much is a real force. It is not a pseudoforce.
OK, but I don't quite understand that. How can an inanimate object exert a force?
 

studiot

Joined Nov 9, 2007
4,998
Perhaps that why I have been consistently talking about tensioned (chain) linkages or couplings.

Just the same way you were talking about a weight whirling on the end of a rope, not a rigid rod (which is also possible).

However that is all a diversion. It doesn't matter if the coupling is chain or rod, if it is in tension, the question remains unanswered.

Where does the other 10MN come from?
 

WBahn

Joined Mar 31, 2012
29,978
Going back to my example.

There is 5MN of rolling resistance so nothing happens until the tractor/loco exerts 4MN+.
????

Are you saying that the static friction is LOWER than the rolling resistance?

At this point extra applied (pull) force will start to accelerate the load (up to 10MN in my case).
So the maximum accelerating force is 10MN
In your original example you had a total rolling resistance and you had the tractor towing an airplane. Presumably the rolling resistance therefore applied only to the tractor. So assuming that 10MN is sufficient to overcome the static friction of the load (which, with it being an airplane presumable hovering somehow off the ground) is probably the case.

I'm just having a real hard time picturing what you have in mind.

Let's simplify things and forget about this airplane and have a locomotive and a rail car. Now, when you say the total rolling resistance, how much of that if for the locomotive and how much of that is for the car. It matters since the rolling resistance of the car has to be overcome via the coupling while the rolling resistance of the locomotive doesn't.

Then let's consider two extreme cases. In the first, the locomotive is extremely heavy and the rail car is extremely light. In this case, when the locomotive is exerting is maximum force it is not able to accelerate itself at as high a rate as 10MN would accelerate the car, so the locomotive ends up at full power and less than 10MN of tension appears in the coupling. At the other extreme you have an extremely light locomotive and an extremely heavy car (and let's assume that this doesn't result in drive wheels slipping). Now you are limited by the 10MN limitation on the coupling. But how much total pull is being exerted by the locomotive depends on how that 5MN of rolling resistance is split between the locomotive and the car. If the locomotive had 3MN of rolling resistance and the locomotive had 2MN of rolling resistance, then the car would see a net force acting on it of 8MN (the 10MN from the coupling forward less the 2MN from the rolling resistance pulling back). The acceleration would then be whatever 8MN could accelerate the (unknown) mass of the car at. However, the locomotive has to put out the 10MN that it is being pulled back by the coupling plus the 3MN of its own rolling resistance plus sufficient additional pull in order to accelerate its own mass at that same rate. If that ends up exceeding the total tractive pull limits of the locomotive, then you have just transitioned into the heavy loco, light car scenario.
 

studiot

Joined Nov 9, 2007
4,998
If I did it was a slip.

Here is my original statement in post#12
It is known that the total rolling resistance is 4MN.
This is meant to be the total of air, ground and any other resistance or dissipative factors such as axle friction.
The difference between static and dynamic is lost or ignored.
 

WBahn

Joined Mar 31, 2012
29,978
Perhaps that why I have been consistently talking about tensioned (chain) linkages or couplings.
You didn't specify a chained coupling until just now. However, my aside was the first mention of pushing instead of pulling. So you apparently had one particular coupling style in mind and I had another. I think we've resolved that, though I still wonder about how the clattering in a traditional coupling is dealt with. But we can put that aside.
 

WBahn

Joined Mar 31, 2012
29,978
If I did it was a slip.

Here is my original statement in post#12


This is meant to be the total of air, ground and any other resistance or dissipative factors such as axle friction.
The difference between static and dynamic is lost or ignored.
Okay. So my numbers need to be adjusted to reflect the original 4MN and not 5MN.

I have no problem with ignoring static/dynamic or lumping everything together into a composite "rolling resistance". But it is important how that total rolling resistance is split between the locomotive and the car.
 

studiot

Joined Nov 9, 2007
4,998
You are complicating things which is obscuring the meat of the issue.

After allowance for any of its operations or workings the loco can exert a maximum pull on its end of the coupling of 20MN.

This has nothing to do with the load, and would be the same for any reasonable load. This value is set by the coefficient of friction between the wheels and the ground and the weight of the loco.

In its turn the load pulls back or resists with a total combined resistance of 4MN, which we suppose to be constant at all speeds, including zero, within the range of consideration.

Nevertheless the coupling cannot be in internal equilibrium tension with 4MN applied at one end and 10 MN at the other.

So where does the difference come from?
 

WBahn

Joined Mar 31, 2012
29,978
Wow, it looks like I got quite a discussion going.:rolleyes:
And this surprises you how?? :D

@WBahn: So then there really is such a thing as centrifugal force, but it acts on the rope and the object at the other end of the rope, not the object being swung? I knew there had to be some force acting outward in order for Newton's 3rd Law to hold true.
[/QUOTE]

Your original question asked about the free body diagram of the object being swung and why the centrifugal force isn't drawn on that free body diagram for that object. The reason is that there is NO centrifugal force acting on that body.

Most people use the term "centrifugal force" to mean some mythical force that is pulling/pushing the object outward. There is no such force.

OK, but I don't quite understand that. How can an inanimate object exert a force?
Pinch your thumb and forefinger together. Is your thumb pressing on your finger or is your finger pressing on your thumb. Is the force felt by your thumb real and the force felt by your finger not, or is it the other way around?
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
Pinch your thumb and forefinger together. Is your thumb pressing on your finger or is your finger pressing on your thumb. Is the force felt by your thumb real and the force felt by your finger not, or is it the other way around?
Both fingers are pressing on each other with an approximately equal force, right? I don't see the comparison here. A 100 kg person standing on the floor doesn't exert 490.5 N on the floor and experience a normal force of 490.5 N exerted on them, do they? The total force comprised by their weight (981 N) isn't split equally between them and the floor, but the force comprised by pressing one's fingers together is (since two fingers each pressing against each other with a 10 N force create 20 N combined).
 
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WBahn

Joined Mar 31, 2012
29,978
You are complicating things which is obscuring the meat of the issue.

After allowance for any of its operations or workings the loco can exert a maximum pull on its end of the coupling of 20MN.

This has nothing to do with the load, and would be the same for any reasonable load. This value is set by the coefficient of friction between the wheels and the ground and the weight of the loco.

In its turn the load pulls back or resists with a total combined resistance of 4MN, which we suppose to be constant at all speeds, including zero, within the range of consideration.

Nevertheless the coupling cannot be in internal equilibrium tension with 4MN applied at one end and 10 MN at the other.

So where does the difference come from?
Okay, so if I understand what you are now saying is that the rolling resistance of 4MN is JUST for the car. That we assume that the locomotive has sufficient power above and beyond the 20MN to overcome its own rolling resistance and accelerate itself at whatever maximum rate the car can be accelerated at. Fine. Let's go with that.

The locomotive increases its pull until it is putting out 4MN to the coupling and, up to that point, everything is just sitting there and looking stupid. At that point the locomotive /car start to move and as the locomotive continues to increase its output they start to accelerate. At the point when the locomotive is pulling on the coupling with 10MN there is a net unbalanced force on the car of 6MN that results in acceleration. The coupling sees 10MN of tension on the front from the locomotive and 10MN of tension on the back from the car, 4MN of which is due to friction and 6MN of which is due to inertia and Newton's 2nd law.

What's the problem?
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
Both fingers are pressing on each other with an approximately equal force, right? I don't see the comparison here. A 100 kg person standing on the floor doesn't exert 490.5 N on the floor and experience a normal force of 490.5 N exerted on them, do they? The total force comprised by their weight (981 N) isn't split equally between them and the floor, but the force comprised by pressing one's fingers together is (since two fingers each pressing against each other with a 10 N force create 20 N combined).
OK, I think I get it now. Two fingers both pressing with a 10 N force against each other is comparable to a 100-kg person and a floor both pressing with a 981 N force against each other. Is that correct?
 
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WBahn

Joined Mar 31, 2012
29,978
Both fingers are pressing on each other with an approximately equal force, right?
Yep.

I don't see the comparison here. A 100 kg person standing on the floor doesn't exert 490.5 N on the floor and experience a normal force of 490.5 N exerted on them, do they?
Nope. The person exerts 981N on the floor and the floor exerts 981N on the person.

The total force comprised by their weight (981 N) isn't split equally between them and the floor, but the force comprised by pressing one's fingers together is (since two fingers each pressing against each other with a 10 N force create 20 N combined).
The combined forces of your fingers is zero since they are equal and opposite -- otherwise something would be accelerating in some direction. If you put a spring or a scale between your finger and thumb it would register 10N, but it can only do so (and remain stationary) if it is seeing 10N on each side (in opposite directions).
 

WBahn

Joined Mar 31, 2012
29,978
It is an interesting exercise for the purist to type

are inertial forces real?

into google or wikipedia.
One of the great unanswered questions (as far as I know) is whether gravitational mass is proportional to inertial mass.

As for whether inertial forces are real, if the force you feel when someone bounces a rock off your head is real, then inertial forces are real.
 

studiot

Joined Nov 9, 2007
4,998
I'm not sure if 'unification' of heat and work in thermodynamics or gravitational mass and inertial mass in mechanics came first but GUT theories are not new.

:)

Here are a couple of treatments of inertia 'forces'.

The first is from a very famous text, The Mechanics of Flight

inertia10.jpg

inertia3.jpg
 
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WBahn

Joined Mar 31, 2012
29,978
As far as I know, no GUT has been vetted and the specific question of whether gravitational and inertial mass are proportional has yet been answered. At some point that probably will be (and perhaps already has), but I think a lot more theoretical and experimental physics (along the lines of finding something akin to the Higgs boson) is required. The equivalence of the two is at the heart of Einstein's Equivalence Principle, but this is more of a postulate of his theory of gravitation as opposed to a proof that they are, in fact, equivalent. About five years ago there was some work in quantum theory that not only said that they could be different, but that they must be different. But what I don't know is whether that modification to quantum theory has stood up.
 

tonyStewart

Joined May 8, 2012
131
I think the confusion on the Free-body diagram is that it has a linear coordinate system where inertial references are excluded.


The so-called fictitious due to inertia such as concentric rotational inertia are not applied because this is an apparent force converting from one set of coordinates to another ( linear to polar ); Also the Coreollis Effects where a linear and Polar coordinates are imposed thus shifting the apparent position as the earth rotates when the rocket or other object moves thru an inertial reference and the ground shifts due to the change in radius from the centre of earth.

So we have both non-inertial references such as used in free-body diagrams and inertial references such as used in Lagrangian mechanics and the Physics of Actions and spacetime calculations.

For example we do not consider the inertia of atomic particles or the Solar System because it is outside our frame of reference, the linear coordinate system, which is used in the free-body diagram.

Likewise we must consider forces from stored energy in linear spring problems which add to the applied initial force.


Side Note

The energy stored in a spring is assumed lossless and thus like energy stored in an Inductor is not real energy but can cause real current losses by its flow. It also causes an apparent phase shift in current, which we call power factor. The total energy in electrical terms includes the Real energy and the Stored or Reactive or Ficticious Energy as we call it from Fictitious Forces in Inertial Systems.


Another Side Note.

Often this confusion in the Electrical world causes young engineers to think they can create an over-unity machine. But the reality is often due to neglected measurement errors and that the energy comes from somewhere like stored energy from permanent magnets.

Feel free to correct me. I'm still working on my next cup of coffee.
 
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