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

[flyingpig]

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

 

[Doc Al]

Let's stick with m1 for the moment. How many forces act on it? How would you label them? Which direction do they act?

 

[tiny-tim]

hi flyingpig!

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 ​

 

[flyingpig]

hi flyingpig!

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 ​

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

It can't be this can it...?

 

[tiny-tim]

very nearly …

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

 

[flyingpig]

very nearly …

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

But I had three forces in the first diagram.

 

[tiny-tim]

But I had three forces in the first diagram.

now I'm confused …

wasn't your first m₁ diagram the left diagram?

that had only two forces

 

[flyingpig]

now I'm 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

 

[tiny-tim]

that's it!

ok, now m₂ and m₃ ​

 

[flyingpig]

But I don't get it, isn't m2 and m3's weight part of m1's weight?

 

[tiny-tim]

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

 

[capandbells]

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.

 

[flyingpig]

But m2 is sitting on m1, isn't that a direct force?

 

[capandbells]

But m2 is sitting on m1, isn't that a direct force?

Yes, the normal force.

 

[tiny-tim]

But m2 is sitting on m1, isn't that a direct force?

yes, and that's what N₁₂ is

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

 

[flyingpig]

yes, and that's what N₁₂ is

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

Yes, but why? It looks very direct to me, in fact, I actually feel that N₁₂ is more indirect than the combined weight force

 

[Doc Al]

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).

 

[flyingpig]

What about centripetal force? Isn't it a fictitious force? I see many FBD includes them too

 

[Doc Al]

What about centripetal force?

How is that relevant here?

Isn't it a fictitious force?

No. Perhaps you're thinking of centrifugal force.

 

[flyingpig]

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?

 

[Doc Al]

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.

 

[flyingpig]

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!

 

[Doc Al]

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

Rather than speak in generalities, pick a specific problem where centripetal acceleration is relevant that confuses you.

 

[flyingpig]

But isn't that true for all cases? Like we don't add vectors of centripetal forces like other forces

 

[Doc Al]

But isn't that true for all cases? Like we don't add vectors of centripetal forces like other forces

Pick a specific example.

 

[flyingpig]

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

 

[Doc Al]

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.

 

[flyingpig]

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.

 

[Apphysicist]

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.

 

[capandbells]

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.)

 

[apst]

hello flyingpig.the figure is ok.but you did not mention whether the 3 contact surfaces are frictionless or not.if the surfaces are also frictionless:
consider m3 first.the forces on it are its weight and the reaction offered by m2.thats all.there is no other force on it.these 2 act in the same line in the opposite direction at the same point.reply if you understand this and i will explain about the others.i am a new user and i don't know how to get the figure drawn.

 

[vela]

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.

Say you were just looking at a block on an incline. You identify the forces on the block: the weight, the normal force, and friction. When you add those up, you get a resultant force, but you don't then draw the resultant force on the FBD and say that it's yet another force on the block, right? Yet that's essentially what you're saying you should do with the centripetal force, which is just a particular type of resultant force.

This confusion is the reason some textbooks avoid the notion of a centripetal force and stress the concept of centripetal acceleration instead. When an object undergoes circular motion, it has a centripetal acceleration that's due to the forces acting on it. In your Ferris wheel example, the weight and normal force add up to give you a centripetal acceleration equal to v²/r, which you know you have because you're moving in a circle.

 

[flyingpig]

Say you were just looking at a block on an incline. You identify the forces on the block: the weight, the normal force, and friction. When you add those up, you get a resultant force, but you don't then draw the resultant force on the FBD and say that it's yet another force on the block, right? Yet that's essentially what you're saying you should do with the centripetal force, which is just a particular type of resultant force.

This confusion is the reason some textbooks avoid the notion of a centripetal force and stress the concept of centripetal acceleration instead. When an object undergoes circular motion, it has a centripetal acceleration that's due to the forces acting on it. In your Ferris wheel example, the weight and normal force add up to give you a centripetal acceleration equal to v²/r, which you know you have because you're moving in a circle.

Is there a reason why we don't treat it as so? I mean if we were to find the sum of acceleration in circular we add up the tangenital acceleration (which is very real) and the centripetal acceleration as vectors to form a resultant vector. In that case we are treating the centripetal acceleration as a real thing.

I am interchanging acceleration and force here because they are proportionally related.

This also brings me to another question, is centrifuge force an example of Newton's Third Law?

 

[Doc Al]

Is there a reason why we don't treat it as so?

Do you mean is there a reason why we don't treat 'centripetal force' as if it were another force in addition to gravity and the normal force? (In your Ferris wheel example.) Sure: If you did, you'd be counting it twice.

I mean if we were to find the sum of acceleration in circular we add up the tangenital acceleration (which is very real) and the centripetal acceleration as vectors to form a resultant vector. In that case we are treating the centripetal acceleration as a real thing.

Centripetal acceleration is real.

I am interchanging acceleration and force here because they are proportionally related.

Don't do that.

It's really simple. Newton's 2nd law says ΣF = ma. ΣF means add up the actual forces acting on the body; a is the actual acceleration. For uniform circular motion, a = a_c.

This also brings me to another question, is centrifuge force an example of Newton's Third Law?

Not sure what you mean, but no.