Easier description of fictitious forces?

[SpiderET]

It seems to me the current description of fictitious forces is too complicated. It is described as effect of accelerate non inertial frame and there are four types of fictitious forces: rectilinear acceleration force, centrifugal force, Coriolis force and Euler force.

Why can't we say that there is only one fictitious force and this force is acting on accelerating object in opposite direction to direction of acceleration?

Sounds much easier to me, covers all four types of fictitious forces and you don't need this complicated and redundand babble about reference frames. It would be much easier to understand.

 

[A.T.]

Why can't we say that there is only one fictitious force and this force is acting on accelerating object in opposite direction to direction of acceleration?

In an inertial frame there are no fictitious forces on an accelerating object.

 

[SpiderET]

In an inertial frame there are no fictitious forces on an accelerating object.

Could you please give me an example of that, because I think that any accelerating object is experiencing fictitious force.

 

[nasu]

A car accelerating on the road does not experience any fictitious force in the reference frame of the road. (the road is "inertial" with very good approximation).
You are confusing the acceleration of the object under study with the acceleration of the reference frame. The inertial ("fictitious") forces are related to the later one.

 

[SpiderET]

A car accelerating on the road does not experience any fictitious force in the reference frame of the road. (the road is "inertial" with very good approximation).
You are confusing the acceleration of the object under study with the acceleration of the reference frame. The inertial ("fictitious") forces are related to the later one.

Car accelerating on the road is experiencing fictitious force/inertial force acting against the accelerating car, otherwise you would need zero fuel in your car :)

The use of "non inertial reference frame" is just confusing things, which are pretty simple in reality.

 

[nasu]

You should read about the definition of fictitious force before making nonsense statements.
They have nothing to do whatsoever with fuel consumption.

You may be confusing "fictional: with "frictional"? (the inertia forces are usually called fictitious)
The frictional and resistance forces are what make a car to need an engine and use fuel even when it moves with constant velocity.

http://en.wikipedia.org/wiki/Fictitious_force

 

[A.T.]

The use of "non inertial reference frame" is just confusing things

Then don't use them and you won't have to worry about fictitious forces.

 

[SpiderET]

You should read about the definition of fictitious force before making nonsense statements.
They have nothing to do whatsoever with fuel consumption.

You may be confusing "fictional: with "frictional"? (the inertia forces are usually called fictitious)
The frictional and resistance forces are what make a car to need an engine and use fuel even when it moves with constant velocity.

http://en.wikipedia.org/wiki/Fictitious_force

When a car is accelerating, there is some little share of friction in wheel, but the main force is used to overcome inertia, that means fictitious force acting against the direction of acceleration. This is pretty basic physics and I have been reading the linked wikipedia article long ago and I can cite from it, because you have obviously not read it:
Figure 1 (top) shows an accelerating car. When a car https://www.physicsforums.com/wiki/Acceleration , a passenger feels like they're being pushed back into the seat. In an inertial frame of reference attached to the road, there is no physical force moving the rider backward. However, in the rider's non-inertial reference frame attached to the accelerating car, there is a backward fictitious force
End of citation.

So I am still right with my original and simple description which doesn't use non inertial frame at all:
There is only one fictitious force and this force is acting on accelerating object in opposite direction to direction of acceleration

 

[A.T.]

the main force is used to overcome inertia, that means fictitious force acting against the direction of acceleration.

No it doesn't meant that.

 

[SpiderET]

No it doesn't meant that.

Sure it does :)

 

[nasu]

When a car is accelerating, there is some little share of friction in wheel, but the main force is used to overcome inertia, that means fictitious force acting against the direction of acceleration. This is pretty basic physics and I have been reading the linked wikipedia article long ago and I can cite from it, because you have obviously not read it:
Figure 1 (top) shows an accelerating car. When a car https://www.physicsforums.com/wiki/Acceleration , a passenger feels like they're being pushed back into the seat. In an inertial frame of reference attached to the road, there is no physical force moving the rider backward. However, in the rider's non-inertial reference frame attached to the accelerating car, there is a backward fictitious force
End of citation.

So I am still right with my original and simple description which doesn't use non inertial frame at all:
There is only one fictitious force and this force is acting on accelerating object in opposite direction to direction of acceleration

The passenger experiences inertia forces in the reference frame of the accelerated car.The car is a non-inertial frame.
The car doesn't, in the reference frame of the road.

It's not good practice to pretend you said something else than you actually said initially.
"Car accelerating on the road is experiencing fictitious force/inertial force acting against the accelerating car, otherwise you would need zero fuel in your car :)"
But if you want to "be" right when ignoring basic definitions and facts, why not? :)

 

[Orodruin]

When a car is accelerating, there is some little share of friction in wheel, but the main force is used to overcome inertia, that means fictitious force acting against the direction of acceleration. This is pretty basic physics and I have been reading the linked wikipedia article long ago and I can cite from it, because you have obviously not read it:

All that you quoted was referring to the accelerated frame in which the car is at rest. In this frame, there is afictitious force acting on all objects, which is necessary for Newton's second law to hold. In an inertial frame there are no fictitious forces, it is as simple as that. The force accelerating the car is equal to its mass multiplied by its acceleration as per Newton's second law. There is also a force from the car on whatever object is pushing it and this force acts in the opposite direction and is a very real force, but it is not acting on the car (in the typical accelerating car, this is the friction force on the ground).

I am sorry, but even if you have read the wiki, it does not seem as if you have actually understood what it says.

 

[SpiderET]

All that you quoted was referring to the accelerated frame in which the car is at rest. In this frame, there is afictitious force acting on all objects, which is necessary for Newton's second law to hold. In an inertial frame there are no fictitious forces, it is as simple as that. The force accelerating the car is equal to its mass multiplied by its acceleration as per Newton's second law. There is also a force from the car on whatever object is pushing it and this force acts in the opposite direction and is a very real force, but it is not acting on the car (in the typical accelerating car, this is the friction force on the ground).

I am sorry, but even if you have read the wiki, it does not seem as if you have actually understood what it says.

Actually I think I understand the topic, but I am looking on it from broader perspective than others. I will explain it on example.

So we have accelerating car with driver. The car needs F=ma to accelerate and as m is calculated the weight of car including the weight of driver. Standard view is that the fictitious force is only acting on driver, not on car. And the fictitious force acting on driver is -ma, with m of driver, let's say 100 kg.

But let's have second example, where we have special the same driver driving a special light carbon bicycle weghting only 1 kg. So standard view is that to accelerate bicycle we need F=ma, where m is m of driver + m of bicycle. And the fictitious force acting on driver is -ma with m of driver 100 kg. But this time almost all F accelerating the bicycle is used to counter the -ma of fictitious force.

Lets have third example where the bicycle is so light, that we can ignore it. And we have F=ma with m of driver and -ma counter force which is fictitious / inertial force.

So what I am saying, that my definition is broader and doesn't use this ridiculous non inertial reference frame.
Just to repeat, how it is in reality:
There is only one fictitious/inertial force and this force is acting on accelerating object in opposite direction to direction of acceleration.

You will not be able to find example which is not covered by this easy definition, when we abstract from relativistic effects.

 

[Orodruin]

The car needs F=ma to accelerate and as m is calculated the weight of car including the weight of driver. Standard view is that the fictitious force is only acting on driver, not on car.

This depends on what system you are looking at things in. In the accelerating frame, there is a fictitious force on all objects. In an inertial frame there are no fictitious forces. You have clearly misunderstood what you are reading and I suggest going back to basics.

 

[PeterDonis]

Standard view is that the fictitious force is only acting on driver, not on car.

No, that is certainly not the standard view. The car is at rest in the non-inertial frame just as the driver is. Therefore the fictitious force that opposes the applied force of the engine through the wheels must act on the car as well as the driver to hold it at rest.

Your entire post, and indeed everything you have said in this thread as far as I can see, is based on this same misconception. Thread closed.