Formulas needed for a dynamic problem

[Femme_physics]

(or "circular motion" problem)

Okay, our lecturer pretty much has thrown us into cold water here. My big problem is getting the right list of formulas.

I'm not sure this is a homework problems since I'm stuck in preliminary stages.

Homework Statement

http://img850.imageshack.us/img850/1654/carsab.jpg

Bridge AB is supported at A and B on immobile pin support, location on the same horizontal plane. The bridge has an arch shape with a radius of 30m (see drawing)

A car whose weight is 28 kN goes up on a bridge on a constant speed of 36 km/h.

A) Calculate the max force in which the car presses against the bridge in both cases

- When the bridge is arched up (drawing A)
- When the bridge is arched down (drawing B)

In which of the two cases the wheels of the cars may cut off from the road, and in what speed will it happen?

Homework Equations

Like I said, I'm pretty much stuck at "relevant equations".

May you can help tell me if I'm right:

http://img804.imageshack.us/img804/9303/formulasmine.jpg

Is that it?

 

[I like Serena]

Hi FP :)

There are 2 forces on the car.

Gravity:

F = m g​

The centripetal force:

F = m v² / R​

In the one case they add up, in the other they subtract.
(You decide which is which ).

The centripetal force is the force required to keep an object in a circular trajectory.

 

[Femme_physics]

A) Thanks :)

B) That's interesting. I remember when taking a medic's course and learning about car accidents the instructor mentioned "centripetal force", he said that it's because of centripetal force that the car veers off track when it does a sharp turn with too much velocity.

So it's not because there is too much centripetal force that the car in his example veers off track. It's because there's not enough centripetal force, right?

 

[I like Serena]

A) Thanks :)

B) That's interesting. I remember when taking a medic's course and learning about car accidents the instructor mentioned "centripetal force", he said that it's because of centripetal force that the car veers off track when it does a sharp turn with too much velocity.

So it's not because there is too much centripetal force that the car in his example veers off track. It's because there's not enough centripetal force, right?

Yes. Very true!

So are you also aware of the so called centrifugal "pseudo-force"?

 

[Femme_physics]

So are you also aware of the so called centrifugal "pseudo-force"?

Not sure what that means

But while I got your attention, I'll steal another question. :)

What is the "position vector"? And how in the world can position have a vector? Position is just position. It's just a force.

http://img811.imageshack.us/img811/3079/positionvector.jpg

 

[I like Serena]

Not sure what that means

But while I got your attention, I'll steal another question. :)

What is the "position vector"? And how in the world can position have a vector? Position is just position. It's just a force.

Stealing questions now? I like that! :)

Uhh. Position is not a force.

A position is usually identified by an x and y coordinate.
There- you have a vector.
It's a pointy thingy that starts at the origin of your coordinate frame, and that ends at the position.

In your drawing the position vector starts at the center of the circle and ends at some point of the circle.EDIT: What's ?

 

[Femme_physics]

Oops-- yea, I mistyped, I meant to say exactly what you said: "position is just position, it's NOT a force"

Gah.

I am aware where the position vector starts and where it ends. But what's the point of it? Does it signify the exact location of the velocity vector?

 

[I like Serena]

Oops-- yea, I mistyped, I meant to say exactly what you said: "position is just position, it's NOT a force"

Gah.

I am aware where the position vector starts and where it ends. But what's the point of it? Does it signify the exact location of the velocity vector?

Oh goody!

Suppose we start with the blue position vector in your drawing.
It identifies the location of the object at a certain time.

1 second later it will have changed. The object will be a little along the circle.
At this point it's identified by a new position vector.

The difference of these 2 position vectors is the "change in position" in a certain time interval.

Hey, isn't that just what a "speed vector" is?
(Or more accurately, the change in position divided by the time interval.)

 

[Femme_physics]

Oh goody!

Suppose we start with the blue position vector in your drawing.
It identifies the location of the object at a certain time.

1 second later it will have changed. The object will be a little along the circle.
At this point it's identified by a new position vector.

The difference of these 2 position vectors is the "change in position" in a certain time interval.

Ohhhhhhhhh!

I get it.

[ So glad to have you here now :) ]

So the change in position is described by an angle, yes?

http://img221.imageshack.us/img221/87/v1v2change.jpg

I made a drawing here showing angle alpha is our change in position based on the change in velocity (delta V2 divided by delta V1? I think...)

Hey, isn't that just what a "speed vector" is?
(Or more accurately, the change in position divided by the time interval.)

Right, the change in position in a rate of time is velocity! :)

 

[I like Serena]

Ohhhhhhhhh!

I get it.

[ So glad to have you here now :) ]

So the change in position is described by an angle, yes?

I made a drawing here showing angle alpha is our change in position based on the change in velocity (delta V2 divided by delta V1? I think...)Right, the change in position in a rate of time is velocity! :)

alpha is not "the change in position".
alpha is an angle that does indeed "describe" the change in position.

The change in position is a vector. It starts at the first position and ends at the second position.
If the change in position is sufficiently small the speed vector coincides with this change-in-position vector.

If you divide alpha by the time, you won't get the speed vector, but you'll get the so called angular velocity "omega" (the rate of change of the angle per unit of time).

 

[Femme_physics]

Hmm, so angular velocity is measured by

Rad/Sec

Correct?

alpha is an angle that does indeed "describe" the change in position.

Fair enough

If the change in position is sufficiently small the speed vector coincides with this change-in-position vector.

What's considered "sufficiently small"?

 

[I like Serena]

Hmm, so angular velocity is measured by

Rad/Sec

Correct?

Yes! :)

What's considered "sufficiently small"?

Oh, say 1 or 2 degrees.

Actually, the smaller the angle the more accurately will the change-in-position vector coincide with the speed vector.
In the limit, they are identical!


Btw, what is ?

 

[Femme_physics]

Btw, what is ?

LOL it's like one big eye one small eye for a shocked/awed expression. It's like saying "holy **** wtf am I seeing?" just in smiley's version. :)

Yes! :)

Hmm... uniform circular motion we only care about "angular velocity", not about the velocity vector? Or, is the velocity vector important too? Or, is it only important so we could derive angular velocity from it?

BTW, if I recall correct angular velocity is notated with a greek W-like letter.

Oh, say 1 or 2 degrees.

Per second?

Truth is, I rather lose you at "... the speed vector coincides with this change-in-position vector."

How can they coincide if they always point at different directions? (position vector - from the center outwards). (velocity vector - tangent to the circle)

 

[I like Serena]

Hmm... uniform circular motion we only care about "angular velocity", not about the velocity vector? Or, is the velocity vector important too? Or, is it only important so we could derive angular velocity from it?

At this stage it's mostly a matter of definitions, so we can name stuff and can refer to it.
But yes, we can derive angular velocity from it:

ω = v / r

BTW, if I recall correct angular velocity is notated with a greek W-like letter.

Yes, the greek W-like letter is called omega (ω) and represents angular velocity.

Truth is, I rather lose you at "... the speed vector coincides with this change-in-position vector."

How can they coincide if they always point at different directions? (position vector - from the center outwards). (velocity vector - tangent to the circle)

At the chance of confusing you more, I'll try to explain anyway.

Here's a picture that I drew to see if I can make it clearer.

In the top drawing I've drawn 2 subsequent position vectors (blue).
Between them I've drawn the change-in-position vector (green).
The change-in-position vector is the difference of the 2 position vectors.

In the bottom drawing, I've drawn the same initial position vector (blue).
And the speed vector that the object has at that point (green).

As you can see, the change-in-position vector points the same direction as the speed vector.
The difference is that the speed vector is longer.
That is because we only go into the direction of the speed vector for a short time interval.

After that time interval the speed will have changed a little under the influence of the centripetal acceleration.

 

[Femme_physics]

Makes perfect (physics) sense :) Yes, I do get it now, and I've seen a drawing similar to yours just now in mechanics class... in fact. Thanks for all your effort, ILS, you're amazing :)

I was in fact rather surprised that all I needed to do was the same equation as in statics just equate it to ma as opposed to zero, and start solving! And, I could theoretically solve this problem. Now, we solved it in class and he gave us two questions to be counted as a small part of our grade, so I'm going to take a crack at it tomorrow :) \

You rock!

 

[I like Serena]

Makes perfect (physics) sense :) Yes, I do get it now, and I've seen a drawing similar to yours just now in mechanics class... in fact. Thanks for all your effort, ILS, you're amazing :)

I was in fact rather surprised that all I needed to do was the same equation as in statics just equate it to ma as opposed to zero, and start solving! And, I could theoretically solve this problem. Now, we solved it in class and he gave us two questions to be counted as a small part of our grade, so I'm going to take a crack at it tomorrow :) \

You rock!

Rock with me baby! Yeah!

And yes, those sexy statics problems just started moving!

 

[Femme_physics]

I think I've grown up a lot since, look what I was able to achieve :)

I solved the problem for case 1. However, for case two I don't see what I should do differently. Can I get a hint at least, please?

http://img9.imageshack.us/img9/1135/ma1kd.jpg

http://img39.imageshack.us/img39/3861/ma2ny.jpg

 

[I like Serena]

I think I've grown up a lot since, look what I was able to achieve :)

Very good!

Do note that you wrote 2800 where it should say 28000, although the result is correct.

And I have another note here, but let's first look at case 2.
(I'm going to keep a secret now! Just to avoid getting side tracked! )

I solved the problem for case 1. However, for case two I don't see what I should do differently. Can I get a hint at least, please?

Draw an FBD with the two external forces?
(Oh and preferably with lengths that are not equal to better understand what's going on!)

 

[Femme_physics]

Do note that you wrote 2800 where it should say 28000, although the result is correct.

Oops. Yea, just a typo. :) Supposed to be 28000.

Draw an FBD with the two external forces?

(Oh and preferably with lengths that are not equal to better understand what's going on!)

Just 2? What about centripetal force (tiny-tim's wrath is increasing for me saying "centripetal force" ;) )

Here's my idea:

http://img593.imageshack.us/img593/7294/fbd20202020.jpg

Hmm...this time normal force and Fc get the same sign. So, that means ma must be in a minus this time! Right?

And I have another note here, but let's first look at case 2.

The "notemaster"! :)

Kinky.

 

[I like Serena]

Just 2? What about centripetal force (tiny-tim's wrath is increasing for me saying "centripetal force" ;) )

Tiny-tim's point was that centripetal force is not an external force, and it shouldn't be in an FBD.
To avoid confusion his suggestion was not to use the word or the symbol (F_c) at all, but instead use only "centripetal acceleration".
So I'm going to try and do that.

Hmm...this time normal force and Fc get the same sign. So, that means ma must be in a minus this time! Right?

Assuming you choose your y coordinate up and your "ma" up, "ma" would be positive in this case.
But yes, the plus and minus signs are different in this case! EDIT: I just saw that in your calculation of case 1, you had the sign of the centripetal acceleration wrong. So you have the wrong result there.

The "notemaster"! :)

Kinky.

:D

 

[Femme_physics]

Tiny-tim's point was that centripetal force is not an external force, and it shouldn't be in an FBD.
To avoid confusion his suggestion was not to use the word or the symbol at all, but instead use only "centripetal acceleration".
So I'm going to try and do that.

So centripetal force = centripetal acceleration? It's the same thing?

I'm confused.

I thought centripetal acceleration is v^2/R and that centripetal force is mass times v^2/r.

Assuming you choose your y coordinate up, "ma" would be positive in this case.
But yes, the plus and minus signs are different in this case!

Different when they're on the same side of the equation. The same when they're at the other sides of the equation.

Would it be proper then to write it out like that?

http://img18.imageshack.us/img18/9664/beproper.jpg

 

[I like Serena]

So centripetal force = centripetal acceleration? It's the same thing?

I'm confused.

I thought centripetal acceleration is v^2/R and that centripetal force is mass times v^2/r.

As you can see from the formulas that you mention, they're not the same thing.
The one has mass in it, and the other doesn't.

Centripetal acceleration is always about the result, in this case a circular motion.

Whereas a centripetal force can be a lot of things.

For starters, as you use it, it's not an external force.

Second, it has no place in the summation of forces.

Third, the term does not uniquely identify it.
All the word "centripetal" actually says, is that it's directed inward.
So the normal force is a centripetal force as well, just not the centripetal force.

Would it be proper then to write it out like that?

I'm assuming we're talking about case 1 here?

Still, no.

Nitpicker that I am, you can't change the sign of "ma" between the first and second line, since you're using the same symbols.

However, you can let the value be negative as you have it on the third line.

 

[Femme_physics]

(Sorry for the delay -- I actually had some work to do!)

All the word "centripetal" actually says, is that it's directed inward.

That's a marvelous fact, actually :)

So the normal force is a centripetal force as well, just not the centripetal force.

I'm not able to relate the difference between "a centripetal" force and "the centripetal" force. Aren't we over-defining stuff here?I thought this stuff is pretty clear. Centripetal acceleration is just v^2/r whereas centripetal force is just mass times that. That, in fact, represents mass times acceleration. The other side of the equation simply has all the forces in the equation: in our cases it's mostly weight and normal/tension forces. Can't we keep it like that for simplicity's sake? :(

I'm assuming we're talking about case 1 here?

Still, no.

Nitpicker that I am, you can't change the sign of "ma" between the first and second line, since you're using the same symbols.

However, you can let the value be negative as you have it on the third line.

I'm not sure I understand how to write the solution to that, then :( Can you show me, maybe?

EDIT: I just saw that in your calculation of case 1, you had the sign of the centripetal acceleration wrong. So you have the wrong result there.

Noted, I'll see about that. Gotta do your job now and help students in mechanics.

Thanks ;)

 

[I like Serena]

I'm not able to relate the difference between "a centripetal" force and "the centripetal" force. Aren't we over-defining stuff here?

I thought this stuff is pretty clear. Centripetal acceleration is just v^2/r whereas centripetal force is just mass times that. That, in fact, represents mass times acceleration. The other side of the equation simply has all the forces in the equation: in our cases it's mostly weight and normal/tension forces. Can't we keep it like that for simplicity's sake? :(

Yes! Let's try to keep things simple!

I'm not sure I understand how to write the solution to that, then :( Can you show me, maybe?

All right, but it feels a bit silly! ;)

$$\sum F_y = m a$$
$$-W + N = m a$$
$$-28000 + N = -9514$$
$$N = 18486 \text{ [N]}$$

 

[Femme_physics]

Yes! Let's try to keep things simple!





All right, but it feels a bit silly! ;)

$$\sum F_y = m a$$
$$-W + N = m a$$
$$-28000 + N = -9514$$
$$N = 18486 \text{ [N]}$$

Thanks :) Yes, now it all makes perfect sense. I'll review it and reply later. Gotta run, thanks! :)

 

[Femme_physics]

Then I solved for

case 1:

N1 = 18486 [N]

Case 2:

N2 = 37514 [N]

All true according to manual :)

Now they ask me for something really really weird.

"At which of those two cases the wheels of the car may be derailed and what is the speed in which this would happen?"


Okay, when the car is NOT on the road that means there's no normal force. So N1 = 0. That means when the bridge is arched up the car will just fly off ahead (which can't happen in an arched-down bridge). So,

http://img43.imageshack.us/img43/9272/5477z.jpg

Hmm...they have 17.16 [m/s] from some reason

 

[I like Serena]

Now they ask me for something really really weird.

"At which of those two cases the wheels of the car may be derailed and what is the speed in which this would happen?"

That's not weird. That is fun! Wheeee!

Okay, when the car is NOT on the road that means there's no normal force. So N1 = 0. That means when the bridge is arched up the car will just fly off ahead (which can't happen in an arched-down bridge). So,


Hmm...they have 17.16 [m/s] from some reason

Perhaps because mass is not the same as weight?

 

[Femme_physics]

Why... *smacks forehead!* do I keep *smacks forehead!* doing *smacks forehead!* the same mistake! *Smacks, smacks, smacks!*

Thanks :) w00t, got it!

 

[I like Serena]

Why... *smacks forehead!* do I keep *smacks forehead!* doing *smacks forehead!* the same mistake! *Smacks, smacks, smacks!*

Thanks :) w00t, got it!

So now you can fly? That is, from a bridge with no arches?

 

[I like Serena]

Actually, I think there's a mistake in your solution manual.

The answer to case 1 is not quite right.

They ask for the max force that the car presses against the arch.
The actual max force in case 1 is greater if the car is not in the middle of the bridge... ;)

 

[Femme_physics]

Ah, but they appear to have defined the car at the center of the bridge according to the diagram (the question only says that the car is on the bridge, not where). Hmm.

Are you getting smart with me? ;)

 

[I like Serena]

Ah, but they appear to have defined the car at the center of the bridge according to the diagram (the question only says that the car is on the bridge, not where). Hmm.

Are you getting smart with me? ;)

Yes. Very suggestive to draw the car in the middle of the bridge.

Let me ask you this then.
Suppose you're building bridges and want to know whether your cartoon man will be able to cross the bridge without it collapsing.
Do you think it only matters what the max force is in the middle of the bridge?

 

[Femme_physics]

Alright, fine, so I killed 500 cartoon men to date! I said I'm sorry to their families! Stop alluding to my negligence!

http://www.ludoflash.com/wp-content/thumbs/cargo-bridge_v3_img4.jpg




//
But yea, I do realize stress is not only important when the cartoon man/car/elephant is at the middle :) thanks.

 

[I like Serena]

Alright, fine, so I killed 500 cartoon men to date! I said I'm sorry to their families! Stop alluding to my negligence!

//
But yea, I do realize stress is not only important when the cartoon man/car/elephant is at the middle :) thanks.

Bwahaha!

To be fair, I think the person who stated the problem didn't expect you to go that far either.