[SpyRemo]

The example works for massless gremlins capable of exerting force in a vacuum (rocket packs?).

A simpler way to understand the crate example would be to ponder the answer to this question: when you're pushing a heavy crate so that it moves with a fixed speed, you still have to keep pushing it in order for it to move at that fixed speed, don't you?

[Andoror]

The example works for massless gremlins capable of exerting force in a vacuum (rocket packs?).

I was being fantastic. The problem I seem to be having is understanding dynamic equilibrium in ordinary circumstances.


A simpler way to understand the crate example would be to ponder the answer to this question: when you're pushing a crate so that it moves with a fixed speed, you still have to keep pushing it in order for it to move at that fixed speed, don't you?

But in order to set the crate in motion you have to overcome any opposing forces, how can those opposing forces be equal if the object is in motion?

If the opposing forces, your push and friction, are not equal then the crate is not in equilibrium.

If the forces are equal then the crate won't move.

[Vigeommighica]

Just remember Fugg's law: "If you push anything hard enough it will fall over."

[investmentonlinev2006x]

Separate your initial push from your continued push. The initial push to get it up to 1m/s velocity is the same as, in your example, whatever got the potato to move 1m/s in space. Something must've accelerated it in the first place. Your initial push does that. Now, once it's going 1m/s, you must apply a force identical to friction to keep it going at that rate. Two separate force vectors, even if they seem to be together. You don't have to push as hard once it's going as you do to get it going, after all.

I've been thinking again.

The initial push gets the crate into motion, at which point you can push it just enough to put it in dynamic equilibrium.

What if you push the crate at a constant speed of 100km/h? It takes an increase in force to cause the acceleration, but is an increase in force required to maintain the new speed?

[Elissetecausa]

Separate your initial push from your continued push. The initial push to get it up to 1m/s velocity is the same as, in your example, whatever got the potato to move 1m/s in space. Something must've accelerated it in the first place. Your initial push does that. Now, once it's going 1m/s, you must apply a force identical to friction to keep it going at that rate. Two separate force vectors, even if they seem to be together. You don't have to push as hard once it's going as you do to get it going, after all.

I've been thinking again.

The initial push gets the crate into motion, at which point you can push it just enough to put it in dynamic equilibrium.

What if you push the crate at a constant speed of 100km/h? It takes an increase in force to cause the acceleration, but is an increase in force required to maintain the new speed? Imagine a long train traveling at 100 km/h hitting a rail cart and you'll get it quickly.

[Avgustslim]

You can't push (exert force on) something with a speed (distance/time), you can only push it with a force (mass*distance/time^2).

This is of course true but I think he knows that, I understood his question differently.

What if you push the crate at a constant speed of 100km/h? ?

I asumed it was informal speak for:

-you hit the crate with a velocity of 100km/h
-dv/dt = 0 , your speed is constant



(that is why I gave the example of the train whose mass far exceeds the mass of the rail cart to give him an intuitive idea of what happens).

[Pasy]

Draw the force diagram.

---->|-||-||-||-||-|

[wepoiyub]

In other words, you don't have to increase force to maintain the velocity at higher and higher levels, until the point where drag from the atmosphere comes into play.

[cbUDaNFRu]

In that case, then, if you add some force beyond that of friction, the cart will continue accelerating and reach equilibrium, unless you factor in air resistance (drag).

??? Please rewrite this so it makes sense.

[ingeneensueva]

In other words, you don't have to increase force to maintain the velocity at higher and higher levels, until the point where drag from the atmosphere comes into play.

Ignoring all forces other than the push and friction from the object's contact with the ground.

If it takes 1N at low speed to achieve dynamic equilibrium, then at very high speed it would also take 1N to achieve dynamic equilibrium?

[BenWired306]

Ignoring all forces other than the push and friction from the object's contact with the ground.

If it takes 1N at low speed to achieve dynamic equilibrium, then at very high speed it would also take 1N to achieve dynamic equilibrium?

If I read what he wrote correctly, then yes, that's what KH is saying (friction is effectively a constant force, so if 1N is the force of friction, you need to apply a constant 1N at any given speed to maintain a constant speed).

[secondmortgagek]

OK, now I'm confused. I know that power and force are different, but what exactly do you mean there? Do you mean it's harder [takes more energy over time] to apply 1N of force at 100m/s than at 1m/s?

[avaiguite]

OK, good. Just wasn't sure if you were implying something more.

[dabibibff]

Really? Hmmm, obviously confused then. I thought friction increased with velocity (not a lot, but a little).

In that case, then, if you add some force beyond that of friction, the cart will continue accelerating and never reach equilibrium, unless you factor in air resistance (drag).

Kinetic friction is proportional to the contact surface, the weight of the object 'above' and the chemical bonding between the surfaces.
It has nothing to do with velocity.

For drag... well it's a little more complex, it depends on linear or turbulent flux, blah-blah-blah...

[rozalinasi]

From the perspective of an observer outside the train (standing on the ground), the situation is more complicated:

1) He sees you pushing the box with force 1N at 100m/s.
2) He sees the train pushing your feet with force 1N at 99m/s (each foot is at rest relative to the train as it pushes)

So this observer says to himself "the guy on the train is putting 100W into the box and getting back 99W from the train, so he's putting in 1W himself".

If the guy is a physicist, he just notice that the guy on the train is in an inertial frame of reference, so he can do his calculations from the observer on the train point of view.
And the same maybe applied to the space potato and the 2 gremlins. You can see it move at constant 1m/s in some inertial frame of reference, but you can also see it not moving at all in another, or see it move at 100m/s in a third one.
Makes no difference.

[CealialactBek]

Wheels tend not experience that much kinetic friction with the ground, unless they are skidding.

[HaremShaih]

Driving full stop requires friction. Heck, walking requires friction - mostly not kinetic.

[nancywind]

If the guy is a physicist, he just notice that the guy on the train is in an inertial frame of reference, so he can do his calculations from the observer on the train point of view.
And the same maybe applied to the space potato and the 2 gremlins. You can see it move at constant 1m/s in some inertial frame of reference, but you can also see it not moving at all in another, or see it move at 100m/s in a third one.
Makes no difference.

Gee, thanks. I didn't realize you could do that. I certainly wasn't trying to explain, from a mechanical perspective WHY that works properly.

[tipokot]

Ah, right. I am not 100% clear on the difference between kinetic and static friction, I think...

Skid marks..

[Asianunta]

Ok, that makes sense, I wasn't thinking about it quite that way. What's the 'one level up' terms explanation of kinetic friction, then? I know that static friction derives simply from the normal force and the coefficient of friction of the contact surface; why is kinetic friction different from that [ie, from what I understand, it's similarly calculated from the normal force and a (different) coefficient of friction; why is that coefficient different?]