Science Question - Argument with wife.

@maxdancona,

maxdancona wrote:

First of all, please give a definition of the word "acceleration"?

A vector measure of a change in velocity with respect to time.

In short form: a = dV/dt

maxdancona wrote:

1) Is is possible for an object that is remaining at rest to have a non-zero acceleration?

Yes.

As I already pointed out, take Newton's second law:

F=m*a
F(weight) = m*g;

Apply your logic and plug zero in for the acceleration for a body at rest:

F=m*0;

So an object at rest has no force from weight? No. Acceleration due to gravity is never zero. It is a uniform force. Gravity is a conservative force.

$http://upload.wikimedia.org/math/7/1/2/712e364804373e76a3cd2dcb11a48dc3.png$
then plug that into...
$http://upload.wikimedia.org/math/e/9/9/e99c07a53446be85ca7fb0c903a84541.png$

maxdancona wrote:

2) Is it possible for an object that has multiple forces acting on it to have a no (or zero) acceleration?

Yes.

Start with a free body diagram. Apply as many vectors as you please. Was this second question a typo?

You're thinking of acceleration due to gravity in a kinematic way. I understand what you're trying to say, but you can't zero out acceleration due to gravity when at rest. Other contributing forces that might change the acceleration vector will go to zero at rest, but not gravity. Remember what defines weight in a physics sense here. Weight is only one component of the net force on a body. Those other forces (like friction, buoyancy, lift, etc) will move the vector of acceleration, but they will never change the weight.

A
R
T

1 Reply

@dadpad,

Me and a mate in a pub noticed a fly that was taking a long time to be destroyed by the bug zapper it had flown into...we had a bet...it took over 40 mins to stop zapping....my mate was closest with 5 minutes.....so never bet on flies !

0 Replies

@maxdancona,

maxdancona wrote:

Quote:
$http://upload.wikimedia.org/math/c/b/8/cb8a5631c35649416a009b36f8298639.png$

This is the correct equation for net force. This is why they made the variable name Fnet. The net force is the total of all forces on an object which in this example (and you really should have understood the situation the equation was explaining before you posted it) is 0 which is simply means that it is not accelerating.

I know what the net force is. That's why I keep calling it the "net force" and referencing it frequently as such. Did I meantion it's called the net force? For brevity, Fnet can be used if you don't want to keep typing "net force."

What I've been trying to communicate to you is that Weight is only one component of the net force on a body (like a jar with or without air). That specific component which is named specifically in the title of this thread: "Weight" is defined by the attraction of two bodies of mass over a distance, modelled by Newton's second law in the form of:

W = mg; Where g is never equal to 0. Ever.

Gravitational acceleration is acceleration.

maxdancona wrote:

The "buoyant" force (which is what we are discussing) is the second term in that equation. You will note that this is determined by density of air outside the jar (the f stands for fluid) times the volume of air displaced times the acceleration of gravity (not again that nothing is accelerating at this rate, this is just the rate an object would acceleration in the even that we dropped it with no other forces).

I already said this. You stated it doesn't matter what is in the jar.

maxdancona wrote:

What you will note about the buoyant force (i.e. the second term in the equation you posted) is that The only thing that matters it the density of the air outside the container, and the Volume of the container (since the acceleration of gravity at a point on Earth doesn't change)

Also something I already said.

maxdancona wrote:

So, as I said. The contents of the container, in a rigid container such as a glass jar where the volume doesn't change, doesn't have any effect on the buoyant force.

False. Even the same chemical compound being in both inside and outside can produce a difference in density and displacement ergo buoyancy. The volume doesn't have to change, there are other ways to increase buoyancy. This is why I made the example using the air, helium, and vacuum.

maxdancona wrote:

Of course, as you put more air or helium in a balloon, it gets bigger (i.e. a greater volume and more air displaced). So this is an important way that a balloon is different than a glass jar in this problem.

Yes a balloon is different. Of course. What is your point? Put .25 grams of air in one balloon and then put .25 grams of helium in another, what will be the difference in volume? The rubber in the balloon will stretch to reach equilibrium with both the fluid on the outside and inside.

In the case of the jar. it's volume is fixed, so buoyancy is entirely a matter of the density ratio.

maxdancona wrote:

Of course the first term of the equation is the container's weight (disregarding buoyant force). This of course is effected by the contents of the container as it adds mass. But that just tells us that a jar filled with air will have more mass (and more weight when it is on Earth) than a jar with a vacuum. And that is the whole point of this discussion.

You're repeating what I've already posted. It's not only that the jar with air has a great weight, but that the jar with a vacuum has a greater buoyancy. these forces act vector opposite.

maxdancona wrote:

For an expanded explanation of buoyancy, I will refer you to to the quite good wikipedia article which succinctly states

Quote:
Buoyancy = weight of displaced fluid.

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

I'm literally three posts ahead of you. If you source the equation images I used three posts ago, they are from said entry.

A
R
T

1 Reply

@failures art,

Failure's Art.

You are contradicting yourself (in the issue that I think is the key part of our disagreement).

First you say (correctly) that acceleration is "a vector measure of a change in velocity with respect to time".

Then you claim that an object that is remaining at rest can have a non-zero acceleration.

Do you see the problem here? How can an object change its velocity while remaining at rest? (Hint: It can't be done).

1 Reply

@failures art,

Art,

I think our disagreement is caused by a confusion between buoyant force, and real weight. They are completely different things (I use the term "real weight" to mean mg. There is also "apparent weight" which takes buoyant force into account).

But let's just consider buoyant force alone to see if we can reach agreement.

My reasoning is as follows. Please tell me what you disagree with (and remember that for this part of the discussion, we are only considering buoyant force).

1) The buoyant force of an object (i.e. a glass jar) in a fluid (i.e. the air) is equal to the weight of the air displaced by the jar. Any scientific explanation of buoyancy internet (including wikipedia) will state this principle.

2) The second term of the formula you linked to says this rho-f is the density of the air so rho-f*Vdisp*g is the weight of the displaced air (which has nothing to do with the contents of the jar.

3) The amount of air displaced by the jar only depends on the volume (size) of the jar). This means that a jar full of lead will displace the same amount of air as a jar full of feathers a jar full of air or a jar "full" of vacuum.

4) This means that the buoyant force on has nothing to do with the contents of the jar.

Remember that here I am only talking about the buoyant force, not the net force.

Are you with me so far? If not, tell me what step you have a problem with.

1 Reply

@maxdancona,

The acceleration to which you refer is in kinematic terms. You need to get past that.

N =! 0; normal force

W =! 0l weight

0 = N + W; set equal

N = W; plug in equations

m*a = m*g; divide by mass

a=g;

QED.

Also, since we are talking about acceleration, did you know you can maintain the same speed, and accelerate at the same time? Do you know how? There's a hint in the wording of my question.

A
R
T

1 Reply

@failures art,

Art,

I have physics degree. Yes, I understand this stuff very well. Yes, you are really wrong. Acceleration is a "kinematic" term. It has a well understood and agreed upon definition-- which as you yourself said is dV/dt-- which you can't go changing as it suits you, especially when you yourself don't want to use your own statement of the definition.

To your brain teaser, there is more then one answer. The simple one is circular motion.

Your statement that " a = g" is ridiculous. Where I am sitting the value for "g" is 9.8 m/s^2. Yet this is not my acceleration. Go figure.

2 Replies

@maxdancona,

maxdancona wrote:

1) The buoyant force of an object (i.e. a glass jar) in a fluid (i.e. the air) is equal to the weight of the air displaced by the jar. Any scientific explanation of buoyancy internet (including wikipedia) will state this principle.

Equal at rest, that is. If the body is in motion, then the buoyancy force is not equal to the force of weight. Also, weight is not just the air in the jar, but the jar itself. The sum or the mass includes the jar and what is inside the jar.

maxdancona wrote:

2) The second term of the formula you linked to says this rho-f is the density of the air so rho-f*Vdisp*g is the weight of the displaced air (which has nothing to do with the contents of the jar.

Agree.

maxdancona wrote:

3) The amount of air displaced by the jar only depends on the volume (size) of the jar). This means that a jar full of lead will displace the same amount of air as a jar full of feathers a jar full of air or a jar "full" of vacuum.

Agree.

maxdancona wrote:

4) This means that the buoyant force on has nothing to do with the contents of the jar.

Agree. I see your point here in that you're talking solely about one component of the net force.

A
R
T

1 Reply

@maxdancona,

So...
Your accelleration is zero.
Your mass is known to you.
What force is NOT acting on you?

F=m*a=m*0; ergo F=0?

You press into the earth as hard as it is pressing into you.

A
R
T

2 Replies

@failures art,

Quote:

Equal at rest, that is. If the body is in motion, then the buoyancy force is not equal to the force of weight. Also, weight is not just the air in the jar, but the jar itself. The sum or the mass includes the jar and what is inside the jar.

Good, I am glad we are coming to an understanding at the end, but I must clarify this part.

The principle is equal, not equal at rest. Again we are only talking about the buoyant force, not the net force. The speed of the object has nothing to do with the buoyant force (moving objects still have the same buoyant force as still objects).

Also when calculating the buoyant force we use the weight of the displaced air. This is the air that would have been occupying the space the jar is not occupying had the jar not been there. It is only the air outside the jar (not inside) that can be displaced.

1 Reply

@failures art,

I get that the net force = 0, but again, we are talking about weight. That's just one component of that.

A
R
T

0 Replies

@maxdancona,

maxdancona wrote:

The principle is equal, not equal at rest. Again we are only talking about the buoyant force, not the net force. The speed of the object has nothing to do with the buoyant force (moving objects still have the same buoyant force as still objects).

Your wording made that unclear. I thought you were saying that buoyancy is always equal to the weight of the body. My confusion. I'm not suggesting a body in motion has a different buoyancy (although as a body ascended the density profile would change in the outer fluid).

maxdancona wrote:

Also when calculating the buoyant force we use the weight of the displaced air. This is the air that would have been occupying the space the jar is not occupying had the jar not been there. It is only the air outside the jar (not inside) that can be displaced.

I concede this point. What is inside the jar only contribute to weight and the net force.

A
R
T

0 Replies

@failures art,

You are getting hung up on the equation F=m*a. This is a perfectly correct (and important) equation, but you are misusing it. Think of it this way. Does force cause acceleration? Or, does acceleration cause force?

Let's start with the basic core principle. You can not have acceleration without having a change in velocity (as you clearly understand this is different then a change in speed). This is the very definition of acceleration. So it is absolutely true that for an object that is remaining stationary, there is no acceleration.

The simplest way to use this equation is to consider the net force and the net acceleration. The common way to use this is to consider the sum of all the forces on an object and how it correlates to the objects change in velocity. This is easy to visualize and to calculate.

So where are you going wrong?

Well you are correct that for my pen which is lying motionless on my desk, there are two forces acting on it. There is its weight, and the Normal force. And (with a slight correction) N = -W. And of course, substituting m*g for W is a normal and perfectly reasonable thing to do.

Where you are wrong is insisting on substituting m*a for N. Well, it is and odd thing to do. I guess it this isn't exactly wrong as long as you keep in mind what you are doing. In this case you have invented a new "a". It isn't the acceleration of my pen (as I can clearly see because my pen is remaining motionless). "A" is now a kind of "partial acceleration" (I don't have a ready terms since we generally don't do this substitution). It is the "theoretical" acceleration that the pen would experience if we removed the force of gravity while maintaining the normal force.

But as I can see (since I am sitting here looking at it) the pen remains motionless. It is definitely not accelerating. I could calculate this "theoretical" or "partial" acceleration if I wanted to. A student of physics would generally not do this because, as you showed, it is always exactly equal to negative g (and as such isn't a very useful quantity).

1 Reply

@maxdancona,

Quote:

Does force cause acceleration? Or, does acceleration cause force?

Forces cause accelerations.

Gravity causes acceleration between matter.
Friction causes acceleration opposing the motion.
Buoyancy causes acceleration opposing gravity.
etc.

As for the theoretical 'a' it's not so unusual in my opinion.

Wouldn't the theoretical 'a' simply represent the potential energy from gravity in the form of:

Gravitational constant/(r^2)

I find this kind of thing to be a useful expression.

(Also, thanks for catching to dropped negative sign. I worked the night shift... sleepy Art...)

A
R
T

1 Reply

@failures art,

failures art wrote:

Gravitational constant/(r^2)

Actually, since acceleration is a vector measure, I think the form would be would be: -(G/(abs(R)^2))*runitvector
R
T

0 Replies

@maxdancona,

maxdancona wrote:

Your statement that " a = g" is ridiculous. Where I am sitting the value for "g" is 9.8 m/s^2. Yet this is not my acceleration. Go figure.

Only because there happens to be a chair where you sit. It exerts a force onto your butt that accelerates you away from the Earth. Moreover, the acceleration from the force of the chair just happens to be -a, which exactly offsets your acceleration from gravity's pull. Without that chair, your acceleration would exactly equal 9.8 m/s^2. Go figure! Smile

EDIT: I notice Art already made this point. I forgot to refresh before answering.

1 Reply

@Thomas,

You aren't helping Thomas, and I think you know better. We were discussing whether an object which remains at rest can have an acceleration.

Sure you can talk about net acceleration and do all the calculations in accelerations rather than forces. But this is far more confusing which is why no one who is not trying to be difficult ever does this.

The point is that there is no net acceleration.

Would you care to read the discussion and be helpful with elucidating the basic concepts?

The original post is here
http://able2know.org/topic/170755-3#post-4579686

My original response is here
http://able2know.org/topic/170755-3#post-4579990

And the heat of the discussion (in my opinion) is here
http://able2know.org/topic/170755-3#post-4580046

And the challenge (as is the challenge with anyone trying to educate about science) is reach a solid understanding of how the concepts work (rather cute word play and throwing equations around).

1 Reply

@maxdancona,

Maxdancona wrote:

You aren't helping Thomas, and I think you know better. We were discussing whether an object which remains at rest can have an acceleration.

Honestly, I think that this is no longer a serious discussion, and has become a free-for-all in silliness. But if you insist on an honest answer, then it's "no". No, a body at rest does not accelerate, by definition.

0 Replies

@dadpad,

dadpad wrote:

Look! over there, two flies crawling up a wall!

Wanna argue about them?

They're obviously not Australian, dp!

1 Reply

@margo,

Swats up?

0 Replies

Science Question - Argument with wife.

Related Topics

Quick Links

My Account

able2know