Easy physics question

[Vega$ NSX]

We were assuming the car hits the wall at 100 mph (no acceleration - constant velocity). Therefore, there is no net unbalanced force on the car (otherwise it would be accelerating) before it hits the wall.

Yes I agree with this statement. There are no forces on the car at all at that point.

To stop the car, you must have an unbalanced force (from the car's perspective) to cause it to stop (ie, a deceleration).

I disagree. When the car hits the wall, it WILL generate a force. That force is called momentum (p=mv) and imulse. In order to stop that force, the wall must push back with an equal and opposite force equal to that momentum and impulse. If the wall was not anchored down tight enough (and could therefore be moved) then it could not generate enough force equal to an opposing force. The car's forces would be greater than the wall's forces and the car would move the wall.

Let me say that again - you MUST have an unbalanced force on the car to cause the car to stop.

Again see above. You must have an equal and opposite force to stop the car. (Newton's Law).

If there is an unbalanced force, there is also an acceleration (deceleration).

The problem with this statement is that you fail to understand that you can have a force and not have an acceleration. Not all forces are strictly mass times acceleration. The earth pushing back up on our feet has NO acceleration.

Once the car is stopped, it is not accelerating because the forces are balanced and it is at equilibrium (obviously). So acceleration might GO to zero but it goes through a really big number first.

Again, you have a delta v and a finite delta t. There's no way you cannot have an acceleration (deceleration).

Yes, the car decelerates. Deceleration is "losing acceleration". But deceleration is not negative acceleration (which you said earlier). There is a hugh difference and I believe that is what you are misunderstanding. When you say the car develops negative acceleration, you are saying that in actual the process of stopping it generates a negative force opposite of its direction of travel. That is incorrect. It is the wall that generates that force on the car which equals the mass times that negative acceleration (actually velocity, momentum) that causes the car to decelerate.

Plug the numbers in:

t = 0.1 s (random small number I made up to represent how long it takes for the car to come to a rest)
m = 1500 kg (3300 lbs)
vf = 0
vi = 44.44 m/s (100 mph)

You'll get a force of -666,666.7 N on the car. That means the acceleration is -444 m/s^2, or roughly 45 g's.

What you did was properly solve for the force that the wall applied to the car because that is what the wall happens to push back at. That is because the NORMAL force of an immovable wall is pushing back the same momentum and impulse force. But as I mentioned above what if the wall was moveable and could only push back at 50% of the force. Then the car would only lose half of its force but only because the wall can only push back at that much.

Fcar = Fwall * 50%

But note that it is the WALL that determines the force not the act of the car slowing down. The wall is providing the only stoping force and is dictated purely by its ability to push back on the car. The car itself does not generate any negative acceleration simply by slowing down.


Look, this is easy. Solve for a man with a constant force pushing a big wooden crate with 50% coefficient of friction. Man pushes with force (Fman=m x a). Box pushes back at 50% x Fman or 50% x m x a. The only thing that could possibly construed as "negative acceleration" is the "a" in the equation: 50% x m x a. However, that is NOT a negative acceleration. It is only part of the answer to solve for friction forces because it is dependant on the acceleration of the pushing force, but it isn't an actual negative force.

Try it with a man (you determine mass) with a 1 m/s running head start hitting a box (again you determine mass) to move it (you determine constant force) Use a 75% coefficient of static friction and a 50% of dynamic friction. You'll see the same thing.

 

[remotemonkey]

the lack of FBD's in this thread saddens me.:frown:

 

[JonBoy]

Yes, the car decelerates. Deceleration is "losing acceleration". But deceleration is not negative acceleration (which you said earlier). There is a hugh difference and I believe that is what you are misunderstanding. When you say the car develops negative acceleration, you are saying that in actual the process of stopping it generates a negative force opposite of its direction of travel. That is incorrect. It is the wall that generates that force on the car which equals the mass times that negative acceleration (actually velocity, momentum) that causes the car to decelerate.

I'm not saying the car generates a force or negative acceleration, I'm saying the car is acted on by a force (said it multiple times, in fact). With that force comes an acceleration opposite the direction of travel of the car (ie, negative). Looking back, "negative" is arbitrary unless a frame of reference is given, so that's where we're getting crossed up, possibly?

The car experiences (is acted on by) a force in the direction opposite its original direction of travel. In a mathematical sense, if the car was travelling in the positive direction, the force stopping it would be in the negative direction (using a typical sign convention).

The acceleration that is causing deceleration is a negative acceleration (relative to the frame of reference of the initial direction of travel of the car being positive). The force ACTING on the car does not have an acceleration and I've never said it did. I'm merely saying that the force acting on the car CAUSES THE CAR TO EXPERIENCE AN ACCELERATION.

The very definition of an acceleration - delta v / delta t - says that the car absolutely experiences an acceleration (whereas you say it does not).

The sign convention of that acceleration was termed "negative" by me, using a frame of reference of positive direction being that of the travel direction of the car prior to hitting the wall.

If ----> is positive then mathematically, <----- is negative. So, the force of the wall reacting to the car is negative (mathematically) from the frame of reference of the car.

Clear?

 

[Vega$ NSX]

I'm not saying the car generates a force or negative acceleration, I'm saying the car is acted on by a force (said it multiple times, in fact). With that force comes an acceleration opposite the direction of travel of the car (ie, negative). Looking back, "negative" is arbitrary unless a frame of reference is given, so that's where we're getting crossed up, possibly?

Sorry that is not what you initially said. Go back and look at your previous posts. Example of your previous statements:

Your first argument (05-12-2008 19:34):
"Except that hitting a brick wall, your acceleration doesn't go to zero, it goes way below zero."

Followed up by (5-13-2008 11:55):
"Acceleration GOES to zero, yes, I agree. The car, once it hits the wall and has been reduced to zero movement, has zero acceleration. When it hits the wall, though, it has an immediate (assuming an unmovable wall) reduction in velocity, which means that it has an immediate and massive negative acceleration, after which the acceleration obviously GOES to zero (since the car is not moving, ergo there is no delta v, therefore no acceleration)."

You did not say it experiences a negative acceleration, you say "the acceleration goes way below zero" (a la negative acceleration). You also say "it has an immediate and massive negative acceleration". It is clear that you initially were trying to say that the car actually goes negative in acceleration, which is clearly not true. Only in your most recent posts did you state that the car experiences a negative acceleration from the wall which is what I've been trying to explain to you all along and I stated is a HUGH difference from what you've been trying to say.

So now if you mean the acceleration is due to the force of the wall but in the opposite direction, then I can better understand what you are trying to say. You are trying to use the word "negative" as meaning "opposite direction". However, it would have been much clearer if you had stated that the car experiences an acceleration from the wall in the opposite direction of the acceleration of car, instead of saying the car has negative acceleration as you initially did.

Having said all that, you are still wrong! Sorry! :redface: :redface: You are one step closer but your terminology is still incorrect. The car doesn't experience a negative acceleration or an acceleration from the wall in the opposite direction as you keep stating. The car experiences a Force in the opposite direction. That is all, it is a reactionary force, nothing more and nothing less. It is not an acceleration, it is not a mass. It is only a force. The only reason mass and acceleration even comes into play is because in order to figure out what the value of that force is, you know it is equal to the mass and acceleration of the car, and you use those values to figure out the value of that force. As we just stated, the WALL is the object generating the force on the car. If that is the case, then why wouldn’t we take the mass of the wall, times the acceleration of the wall to calculate the force it generates on the car? Why do we use the mass and acceleration of the car? Because all we know is that the reactionary force of the wall just happens to equal the value of the force of the car. That is why we use the mass of the car and the acceleration of the car to calculate the force of the wall. Not because the wall has any acceleration at all or mass equal to the car. All we know is that it is equal in value to the force of the car in the opposite direction.

Understand it this way:
1) The force of the wall equals the force of the car (equal and opposite reaction: Newton's Law)
Analogy: I have the same exact amount of money as you.

2) The force of the car happens to equal mass times acceleration of the car.
Analogy: We happen to know you have 4 quarters which makes $1

3) Since we know that the force of the wall equals the force of the car, we can say that the value of the force of wall equals the value of the car in the opposite direction. That does NOT mean the wall has the same mass or acceleration as the car; only the same FORCE.
Analogy: Since we know I have the same money as you (and you have $1) then we know I have $1 too. It does NOT mean I have 4 quarters (I could have 10 dimes). All it means is I have the same total money as you.

Are you starting to understand? The WALL is exerting a force on the car. It is a force and only a force. The wall itself has no acceleration component. We only use the mass and acceleration of the car to help us figure out what the value of that force is. That is all. If you go back and re-read the posts you’ll start see that this is what I’ve been saying from the very start.

 

[JonBoy]

Semantics. My initial contention was that the car experiences a negative acceleration.

Delta v is negative. Delta t is positive. Ergo, a is negative.

That simple. You dragged in what was pushing what, what was reacting, etc, etc, but it's all just ground in between the beginning and ending of what I was saying.

If we can agree that the car experiences/goes through an acceleration that causes it to slow down, we're in agreement as far as I'm concerned. Whether you agree to term it "negative" or not is really a point of either semantics or different frame of reference.

I know the wall isn't accelerating, etc, etc, but the reaction of the wall has an immediate EFFECT (despite not having an acceleration in and of itself). What's that effect? The car stops because it experiences a massive acceleration (which I term negative, as a positive acceleration would cause it to speed up).

As I said.................semantics.

 

[Vega$ NSX]

No it's not semantics. A force is NOT an acceleration and an acceleration is NOT a force. There is a very distinct difference between the two that you are not understanding. A friction force is not an acceleration. A normal force is not an acceleration. A reaction force is not an acceleration. A car hitting a wall see a reactionary force, which is NOT an acceleration. You can't use the words interchangeably as you keep doing so.

To say a car hitting a wall sees a negative force is correct. To say a car hitting a wall sees a negative acceleration is incorrect.

Finish this phrase:

A book sitting on a table does not fall to the floor because the ____________ of the table pushes back up on the book.

Do you say "acceleration" of the table or the "force" of the table? You say force, because acceleration is wrong. According to you, the book does see a negative acceleration (negative gravity). But like I said before, the reason that is wrong is because it is not a negative acceleration it is seeing, it is a negative force. To get the value of that force we use the acceleration of gravity in our calculation, but it is incorrect to say that the book sees a negative acceleration based on that.

 

[JonBoy]

No it's not semantics. A force is NOT an acceleration and an acceleration is NOT a force. There is a very distinct difference between the two that you are not understanding. A friction force is not an acceleration. A normal force is not an acceleration. A reaction force is not an acceleration. A car hitting a wall see a reactionary force, which is NOT an acceleration. You can't use the words interchangeably as you keep doing so.

To say a car hitting a wall sees a negative force is correct. To say a car hitting a wall sees a negative acceleration is incorrect.

Finish this phrase:

A book sitting on a table does not fall to the floor because the ____________ of the table pushes back up on the book.

Do you say "acceleration" of the table or the "force" of the table? You say force, because acceleration is wrong. According to you, the book does see a negative acceleration (negative gravity). But like I said before, the reason that is wrong is because it is not a negative acceleration it is seeing, it is a negative force. To get the value of that force we use the acceleration of gravity in our calculation, but it is incorrect to say that the book sees a negative acceleration based on that.

If you drop the book, it sees an acceleration as it hits the table.

You're using a static example to try and define a dynamic situation. The car in our example is moving, then stops suddenly. In your example, the book is sitting still on the table.

As I said, my original point was that the car sees an acceleration (negative, ie, bringing it to a stop) because it has a negative delta v over a finite t.

The road you've taken to get here is a hodge podge of semantics, misunderstandings, and words being put in my mouth/posts. You're arguing points I never made.

I'm not saying the car will experience an acceleration because the wall puts an acceleration into the car, I'm saying the car will an experience an acceleration because it hits an unmovable object while moving at a speed greater than zero, causing said speed to drop to zero.

Again, negative delta v over finite t = negative acceleration. You still haven't addressed this point, instead running around about forces that I'm not debating.

 

[Vega$ NSX]

If you drop the book, it sees an acceleration as it hits the table.

You're using a static example to try and define a dynamic situation. The car in our example is moving, then stops suddenly. In your example, the book is sitting still on the table.

No it doesn't matter. Even if I don't drop the book, it still sees an acceleration. What is gravity? It's an acceleration isn't it? It has units of m/s/s? The only difference if I drop the book instead of sitting he book on the table is that now it has a momentum force on top of a gravity force.

Static book force: F=mg
Dynamic book force: F=mg + pv

Regardless the table pushes back up with a reactionary force equal to the total force. Again, not an acceleration.

The road you've taken to get here is a hodge podge of semantics, misunderstandings, and words being put in my mouth/posts. You're arguing points I never made.

I keep trying to tell you it's not semantics. We can't even begin to talk numbers if you aren't willing to get on the same page in terms of fundamental physics. You keep using the same equation over and over again, negative delta v over finite t = negative acceleration, but I keep trying to tell you that does not apply here. That is a correct equation in the wrong application. That is an equation of motion (particle motio), not a conservation of energy. This problem is a balance of energy an not a motion equation. For some reason you aren't seeing that and you keep thinking in terms of motions and using motion equations. When you interact and collide objects you need to use a balance of energy equations and look at it using a balance of forces.

I'm not saying the car will experience an acceleration because the wall puts an acceleration into the car, I'm saying the car will an experience an acceleration because it hits an unmovable object while moving at a speed greater than zero, causing said speed to drop to zero.

I understand what you are saying and again, the correct term is that the car will decelerate (the acceleration will go to zero) but it will not experience an acceleration. That is not semantics. It is a very importantant to distinguish that the car will not have negative acceleration. If we are looking at the motion of a particle, then yes, the car will have negative acceleration. However, that is not the case here and there is a huge difference in understanding why.

At this point, if you don't see it, then I don't think I will ever be able to get you to see it. I'm not trying to insult you, but I think you have just forgotten much from your engineering classes. You are still in the early level classes of particle motion which is only the basic level of describing dynamics. You need to be thinking in terms of conservation of energy and balance of forces. Again, I'm not trying to insult, because I'm positive you already know all this, but you just need to refresh your memory on some of this stuff again. You can think I'm just arguing semantics but if you are making statements that blatently incorrect, I don't see how you can think I'm just splitting hairs. As I said before, I don't believe I'll be able to change your opinion on this issue. I think the only way you'll be able to see this for yourself is to sit down with one of your old engineer professors and walk through it yourself.

 

[JonBoy]

Can you tell me what would define a negative acceleration?

 

[Briank]

Can you tell me what would define a negative acceleration?

Slowing down :biggrin:

 

[lowellhigh79]

Agree with all that no more force at point of clutch being disengaged equals no more acceleration. There is a slight sensation of lunging forward due to instantaneous weight transfer which could give one a sense of increased velocity, before drag begins the deceleration process. Apologies if someone added this in before.

Regards,

Danny

 

[JonBoy]

I understand what you are saying and again, the correct term is that the car will decelerate (the acceleration will go to zero) but it will not experience an acceleration. That is not semantics. It is a very importantant to distinguish that the car will not have negative acceleration. If we are looking at the motion of a particle, then yes, the car will have negative acceleration. However, that is not the case here and there is a huge difference in understanding why.

At this point, if you don't see it, then I don't think I will ever be able to get you to see it. I'm not trying to insult you, but I think you have just forgotten much from your engineering classes. You are still in the early level classes of particle motion which is only the basic level of describing dynamics. You need to be thinking in terms of conservation of energy and balance of forces. Again, I'm not trying to insult, because I'm positive you already know all this, but you just need to refresh your memory on some of this stuff again. You can think I'm just arguing semantics but if you are making statements that blatently incorrect, I don't see how you can think I'm just splitting hairs. As I said before, I don't believe I'll be able to change your opinion on this issue. I think the only way you'll be able to see this for yourself is to sit down with one of your old engineer professors and walk through it yourself.

I appreciate you not calling me a total idiot, given that we're at such extremes of opinion here.

That said, I think you're overthinking this problem. You're looking at it as a huge system, talking of balancing forces, no net acceleration, etc, etc, but you're ignoring the actual motion of the car. Despite the fact that this is a collision and conservation of energy applies, you cannot forget that the car does indeed experience a change in motion.

You have the car with momentum (no net force on the car, so constant velocity and mass) and a wall with no motion, only a reaction force applied over time (impulse) to stop the car. I already solved that equation for the force reacting to the car. While the force does not HAVE an acceleration, per se, the wall does CAUSE an acceleration on the car (a negative acceleration) because the motion/state of the car absolutely changes (velocity decreases very quickly). The wall is more or less a constraint on the car, which means the car is induced to change its state of motion.

You are basically saying that despite the car having a change in velocity, it does not experience an acceleration. That is a very basic principle and in all of your energy equations and force balances, I think you're missing that very simple point.

My dictionary says deceleration is a change in speed, not a change in acceleration. You're saying deceleration is a change in acceleration since acceleration goes to zero.

Acceleration is zero once the car is at rest but in the meantime, velocity is changing (deceleration), which implies an acceleration.

 

[Vega$ NSX]

OK!!! I think I've found our disconnect!:smile:

When you are using the term negative acceleration, you are using it in terms of a negative velocity change over a positive distance change. Obviously I was never saying the car didn't have a change in velocity. I think we can agree on that. However, my interpretation of negative acceleration was a velocity change over a negative distance change. For example: If the car was replaced by a rubber ball and you threw the ball against the wall, what would happen? It would rebound and actually bounce off the wall with negative acceleration so that it would accelerate away from the wall with increasing speed. To me, that is what negative acceleration is describing. The negative componenet of the acceleration coming from the delta x (distance) rather than velocity. I tried to use the term decelerate to denote the difference between the two because they describe two very different things. So in the case of a rubber ball, I would say, when the ball hit the wall, it would decelerate to a stop, then it would negatively accelerate away from the wall in the opposite direction. It appears you used the phrase negative accelerate for the entire process (which is correct too).

I think I can see your thinking on that, although IMHO I think it is very important to distingush the difference between the two processes. However, I can see your way of thinking on it and it's technically not incorrect either.

So does that sound about right? All that and it comes down to a nomenclature issue?!? :tongue: No wonder I didn't like English and Liberal Arts majors!

 

[JonBoy]

Yes! Exactly! Arghhhhhh!

 

[JonBoy]

I DID say (many, many, many, many times) that if you have a negative delta v over a positive delta t to get a negative a, so I figured it was obvious that I wasn't addressing a negative (relative) distance change at all.

I think you saw it so many times that you just dismissed it as out of hand.

Now I can go back to trying to get this FEA model to work.