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this is mikau

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newtons second law for a system of particles

(for some reason, the login thing stopped working again! whats with that?)

I'm having some trouble understanding the rules for newtons second law for a system of particles.  Fnet = mass*(acceleration of center of mass). Sounds simple enough, but according to my book:

Fnet is the net force of all external forces that act on the system. Forces on one part of the system from another part of the system (internal forces)are not included.

That sounds simple too. But then I considered this homework problem:

here i am asked to find the acceleration of the center of mass. By the above law this should be:

Fnet = (net force on redblock) + (net force on blueblock) = mass*a = (0.6 + 0.4)a = a

now consider the net force on each object, call the net force on the blue block Fnet1 and the net force on the red block, Fnet2

Fnet1 =  (Frope)i
Fnet2 = (-0.4*9.8)j + (Frope)j

Now, the way I see it, Frope is a force on one part of the system (the blue block) from another part of the system (the red block) except this force is transferred through the rope. It should, by my understanding of my books statement, be ignored. But this leaves us with Fnet1 = 0 and Fnet2 = (-0.4*9.8)j 
which implies:
Fnet = (-0.4*9.8)j = a, thus there is only a downward acceleration, the horizontal velocity is constant.
But this does not agree with my books answer, and also doesn't make sense.

As it turns out, you DO have to include Frope in the net forces acting on the objects, this produces the correct answer. But to me, this is clearly an internal force. That being the case, i don't think I understand what the book means in that statement. I mean, I suppose it could mean forces that act directly against each other, like a box pushing another box, but they would cancel eachother out anyway and there is no need to exclude them from Fnet!

this is mikau

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Re: newtons second law for a system of particles

i forgot to state that the surface in the diagram is supposed to be frictionless. But you knew that.

mikau

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Registered: 2005-08-22

Posts: 1,504

Re: newtons second law for a system of particles

Okay, it seems the login is working again.

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A logarithm is just a misspelled algorithm.

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: newtons second law for a system of particles

If you are considering forces on the individual blocks, then you do need to take into account the tension in the rope, since the rope is “external” to each block.

If you are considering forces on the whole blocks+rope system, then the tension in the rope is not an external force.

However, in problems of this sort, it is often impossible to determine the acceleration just by looking at the system as a whole; we have to consider forces on the individual blocks instead.

Last edited by JaneFairfax (2008-03-15 11:23:59)

mikau

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Registered: 2005-08-22

Posts: 1,504

Re: newtons second law for a system of particles

Right! And all of what you say makes sense! But doesn't this defy the books statement? I mean we are talking about newtons second law for a system of particles, and we are told that the total mass, times the acceleration of the center of mass of the entire system is equal to the sum of the net external forces (with internal forces ignored).  If the force of the rope is internal for the entire rope-block system, then wouldn't that imply, as i explained earlier, that the acceleration of the center of mass is zero in the horizontal direction?

In otherwords, you said the force would be internal for the entire rope block system. Is that not the system we are considering? Isn't newtons second law for a system of particles used when you want to consider the entire system rather than the individual parts?

Last edited by mikau (2008-03-15 13:15:06)

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A logarithm is just a misspelled algorithm.

John E. Franklin

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Registered: 2005-08-29

Posts: 3,588

Re: newtons second law for a system of particles

Guessing:  Maybe you draw a topology around the parts you want to call internal.
Then the internal stuff might move within itself, like a living organism, very active
inside.  But also rotational and translational accelerations by external forces are possible,
plus compression, heating, etc.

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igloo myrtilles fourmis

mikau

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Registered: 2005-08-22

Posts: 1,504

Re: newtons second law for a system of particles

Maybe....
What i'm thinking right now is..

Jane, would it work just to sum all the parts of the system, internal and external? Because internal forces should cancel out anyway, right?

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A logarithm is just a misspelled algorithm.

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: newtons second law for a system of particles

Yes, you can do that – but why do you need to add together things which are only going to cancel out? Even external forces which cancel out don’t always need to be brought to attention. For example, in your particular problem, there are two other external forces on the 0.6kg block (besides the tension in the rope), namely its weight and the normal reaction of the table on it. These two forces balance out and contribute no net effect to the motion of the whole system; hence it was not even necessary to mention them.

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: newtons second law for a system of particles

Fnet = (-0.4*9.8)j = a, thus there is only a downward acceleration, the horizontal velocity is constant.
But this does not agree with my books answer, and also doesn't make sense.

Huh!? You mean the answer is answer is not 3.92 m s[sup]−2[/sup]? That was the answer I got; it’s not right?

mikau

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Registered: 2005-08-22

Posts: 1,504

Re: newtons second law for a system of particles

whoops! Sorry Jane, I forgot about this thread.

The acceleration is in 2 dimensions and therefore has two componants. The book gives the answer as (2.35)i - (1.57)j

You gave the acceleration of the boxes, they want the acceleration of the center of mass.

Anyway, the reason i asked about summing individual forces, is this. Yes, canceling forces are a waste of time, but that still takes care of them. So I can basically just note that cancelling forces can be ignored. However, i believe canceling forces do not necessarily mean the force is internal. Know what I mean?

In otherwords, I'm hoping if i'm unclear about what are internal or external, i can just sum all the individual forces to find Fnet and internal forces will be taken care of. However, I still think i need to be able to identify an internal force.

Right now I'm thinking maybe its two parts of the system acting directly against eachother, rather than a force on one part of the system thats acting on one part, and eventually acting on another part after its been altered.

Last edited by mikau (2008-03-19 08:54:53)

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A logarithm is just a misspelled algorithm.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: newtons second law for a system of particles

You gave the acceleration of the boxes, they want the acceleration of the center of mass.

Oh, of course! I feel so silly now.