Newton's Laws, not sure which law this is, also FBD

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flyingpig
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Homework Statement



[PLAIN]http://img576.imageshack.us/img576/5773/67450733.jpg

Suppose three blocks as stacked as shown on a floor which I forgot to draw and assume it is frictionless. Draw a free-body diagram for each mass

Now I know my FBD is wrong because it is a misconception of Newton's third law (ah that's what it was, the third law!), but for some reason it make sense to me...

[PLAIN]http://img408.imageshack.us/img408/6897/87576200.jpg I didn't bother drawing FBD for m3 because I know m2 is already wrong
 
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Let's stick with m1 for the moment. How many forces act on it? How would you label them? Which direction do they act?
 
hi flyingpig! :wink:

sorry, but these are completely wrong

a https://www.physicsforums.com/library.php?do=view_item&itemid=100" for a particular body must have only the forces directly on that body

for example, the weight m2g will be on the FBD for m2, but not for m1 …

all m1 gets from m2 is a reaction force, which you should call something like N2

try again :smile:​
 
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tiny-tim said:
hi flyingpig! :wink:

sorry, but these are completely wrong

a https://www.physicsforums.com/library.php?do=view_item&itemid=100" for a particular body must have only the forces directly on that body

for example, the weight m2g will be on the FBD for m2, but not for m1 …

all m1 gets from m2 is a reaction force, which you should call something like N2

try again :smile:​

[PLAIN]http://img109.imageshack.us/img109/4426/74220089.jpg

It can't be this can it...?
 
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tiny-tim said:
very nearly …

but there are three forces on m1, aren't there? :wink:​

But I had three forces in the first diagram.
 
tiny-tim said:
now I'm confused :confused: …

wasn't your first m1 diagram the left diagram?

that had only two forces

Gravity and normal? Oh that's what you meant?

[PLAIN]http://img46.imageshack.us/img46/548/40905270.jpg
 
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But I don't get it, isn't m2 and m3's weight part of m1's weight?
 
flyingpig said:
But I don't get it, isn't m2 and m3's weight part of m1's weight?

no no no no no!

that is not the way a free body digram works …

you only include the forces directly on the body …

m2g and m3g only act on m2 and m3 respectively …

all m1 gets is N12 :smile:
 
flyingpig said:
But I don't get it, isn't m2 and m3's weight part of m1's weight?

If you were considering the system of m1, m2 and m3 together, the weight of the system would be the combined weight of all of the blocks. In this case, you're thinking about m1 as a body independent of m2 and m3 which is interacting with the other blocks.
 
But m2 is sitting on m1, isn't that a direct force?
 
flyingpig said:
But m2 is sitting on m1, isn't that a direct force?
Yes, the normal force.
 
flyingpig said:
But m2 is sitting on m1, isn't that a direct force?

yes, and that's what N12 is :smile:

the reaction force between the two surfaces is a direct force, the weight of one is not a direct force on the other :wink:
 
tiny-tim said:
yes, and that's what N12 is :smile:

the reaction force between the two surfaces is a direct force, the weight of one is not a direct force on the other :wink:

Yes, but why? It looks very direct to me, in fact, I actually feel that N12 is more indirect than the combined weight force
 
flyingpig said:
Yes, but why? It looks very direct to me, in fact, I actually feel that N12 is more indirect than the combined weight force
The weight of an object is a force on the object due to the Earth's gravitational pull. So the weight of m2 only acts on m2, not on m1. Of course, there will be a normal force exerted on m1 by m2, which will depend on the weight of m2 (among other things).
 
What about centripetal force? Isn't it a fictitious force? I see many FBD includes them too
 
flyingpig said:
What about centripetal force?
How is that relevant here?
Isn't it a fictitious force?
No. Perhaps you're thinking of centrifugal force.
 
Doc Al said:
How is that relevant here?

No. Perhaps you're thinking of centrifugal force.

But I have been told that I shouldn't consider centripetal force as a real force and instead as the sum of the vectorial forces, doesn't that make it a fictitious force as well?
 
flyingpig said:
But I have been told that I shouldn't consider centripetal force as a real force and instead as the sum of the vectorial forces, doesn't that make it a fictitious force as well?
No. The term "fictitious force" has a specialized meaning in physics. But it's true that 'centripetal force' is just a term for the sum of forces in the radial direction; nonetheless, the forces that contribute to that sum are usually quite real.
 
Doc Al said:
No. The term "fictitious force" has a specialized meaning in physics. But it's true that 'centripetal force' is just a term for the sum of forces in the radial direction; nonetheless, the forces that contribute to that sum are usually quite real.

But...but...so real that we don't consider it like all the other forces...? Something is not right!
 
flyingpig said:
But...but...so real that we don't consider it like all the other forces...? Something is not right!
:confused:

Rather than speak in generalities, pick a specific problem where centripetal acceleration is relevant that confuses you.
 
But isn't that true for all cases? Like we don't add vectors of centripetal forces like other forces
 
flyingpig said:
But isn't that true for all cases? Like we don't add vectors of centripetal forces like other forces
Pick a specific example.
 
Let's just say I am a Ferris Wheel and it is moving, my car (it's called a car right?) is at the to of the Ferris Wheel
 
flyingpig said:
Let's just say I am a Ferris Wheel and it is moving, my car (it's called a car right?) is at the to of the Ferris Wheel
OK. What would you like to discuss? The forces acting on you while you're sitting in your car at the top of the Wheel? Since you are executing circular motion, the net force on you will be towards the center of the Wheel, which is vertically downward. If you draw a FBD, you'll only show the individual forces acting on you, such as gravity and the normal force of your seat. You would not show anything labeled 'centripetal force' on the FBD diagram.
 
Yes, that's what I am getting at. The centripetal is "real", yet we don't put it on a FBD because sum of the normal force and gravity turns out to be the same.
 
On your free-body-diagram for anyone of the masses, you should only show the forces that are acting on it.

It seems like you have made a fair bit of progress for m1...you noted that its interaction with the Earth is a non-contact force given as m1*g, that its contact with m2 is a contact force (maybe called N21, the normal force exerted by 2 ON 1), and that its contact with the floor is a contact force (maybe called Nf, the normal force from the floor).

Without worrying about centripetal forces (not applicable for this problem), I think your major confusion is labeling N21 as m2*g. If you label it just as N21 and likewise draw a force on m2 (on a separate F-B-D) as N12 (the Newton's third law pair), and continue in that fashion for m2 and m3, you might have some insight as to why simplistically labeling it as a generic contact force is beneficial for solving the problem. Eventually, you'll be able to use the idea of balanced forces in a state of equilibrium to solve for N21, N12, N32, N23 and see how the weights of the other blocks play into the problem. But you need to go through this process first, and having the correct F-B-D is absolutely vital, not just because your homework asks you for it, but to understand other problems as you progress.
 
flyingpig said:
Yes, that's what I am getting at. The centripetal is "real", yet we don't put it on a FBD because sum of the normal force and gravity turns out to be the same.

To be exceedingly clear, it's not just that the sum of the forces somehow happens to equal the centripetal force, but rather — by definition — the centripetal force is the sum of the forces in the radial direction. Saying the centripetal force is 'fictitious' is like saying that the x-component of some force is 'fictitious'. Clearly this is wrong. (By the way, there are no fictitious forces. There are inertial forces, which arise in non-intertial reference frames, but they are real forces to observers in those reference frames.)