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Can you give me an example of how you'd use F=ma?

Paul, why is it so hard to understand?

before entering the turn the mass has a speed of +40m/s to the RIGHT, after the turn the mass has a speed of +40m/s to the LEFT.
Instead of LEFT and RIGHT we are using signs:
+40m/s (to the RIGHT), -40m/s (to the LEFT).

The difference is a velocity of 80m/s.
Because velocity has a direction, its a vector.

As kallog stated several times: depending on how much time you allow for this velocity change of 80m/s the force applied to the mass can be anything.
You chose 1 second, so the mass (100kg) will get a force of 8000N applied.

I tell you what is so hard to understand , its the
extra 4000N he is claiming.

that is nonsence , you cannot get -8000N force from a -4000N force simply by turning the mass around , it will change its direction but it will not apply a force of -8000N to the pipe.

change in velocity is not a force

and change in velocity cannot be used to accelerate the pipe.

my calculations?

F=ma

the 100kg mass applies a force of 4000N for 1 second.
since force is determined by m*a
and acceleration is the change in velocity over time

so force = d/(dt)*mv

F=40m/(40m*1s)*mv
F=40m/40 * mv
F=1*mv
F=1*100kg*40m/s
F=1*4000
F=4000N

that 4000N force is Applied to the pipe
and it is applied to the mass.

so the mass still has +4000N
and the pipe has recieved -4000N

you were getting 8000N but the only way you can get
a 8000N force is to calculate the force by using a
200kg mass , if you keep the time and the velocity the same.

F=40m/(40m*1)*mv
F=40m/40 * mv
F=1*mv
F=1*200kg*40m/s
F=1*8000
F=8000N

the mass did not increase to 200kg so your assumption that the mass would apply a -8000N force to the pipe is incorrect.

the above is a force calculation not a momentum or direction change equation , it would apply to the amount of force that the pipe will recieve as the mass passes through the turn , where a change of direction calculation can only calculate the amount of the change in direction.

a force is required to change the direction of a mass and
that force is 4000N supplied by the pipe.

stop using momentum calculations to make it appear that
the pipe will have a larger force applied to it.

use the same force calculation as above that calculates FORCE.



FORCE

A Force of 4000N applied for 1 second on a 100kg mass with a velocity of 40m/s against its direction of movement, will result in the mass stopping.

But you want the mass to move again with 40m/s in the other direction.

The velocity difference between 40m/s to the RIGHT and 40m/s to the LEFT is 80m/s.

not if the mass is turning around !
on a track.

just like a pendulum.

the 100kg mass will not just stop as it turns around.
but it does apply a 4000N force to the pipe as it turns around.

you must have just dropped in.

you should read up on the issue before replying.

yes , the velocity difference or the change in velocity
is 80 m/s , I never questioned that.
I questioned kallogs use of the 80m/s * 100kg to get
a force of 8000N , that he applied to the pipe to cause
the pipe to stop and then move in the other direction.

I agree the pipe just stops with the force of 4000N

but a change in velocity is like this
---------------- -40 m/s ------------>
<--------------- +40 m/s -------------

it turned around add them up the total change

IN VELOCITY / DIRECTION is 80 m/s , thats not a FORCE
its a change in direction.

or you can do it like this
--------------- 40 m/s North------------>
<-------------- 40 m/s South-------------

total change in velocity is 80 m/s
its still not a force , its a change in direction.

its like driving your car at 50 miles per hour
for 30 minutes to point A
and then driving it back at 50 miles per hour
for 30 minutes to point B

the total change in velocity durring the 1 hour
is 100 miles per hour , but you never drove your
car over 50 miles an hour.

the change in velocity of the car required no extra gasoline than it would if you just
kept driving north for an entire hour.

do you have the comprehension ability to comprehend that?

because kallog couldnt ever grasp that.

or he was just using that to try to make it appear that the pipe would just go forward
then backward , resulting in the pipe never moving further than 16 meters in any direction.

Where did the first 4000N come from? It came from F=mv which we know is wrong.

If you really insist on 4000N, all you have to do is change the turnaround time from 1s to 2s and it's back to 4000N again. Would that make you happy?

Or we could use 0.1s and it becomes 80,000N. Or 10s and it's 800N.

We're treating it as a 1-dimensional turnaround. Like a spring or something. Alternatively it's the same as having a proper turning track but only considering the longitudinal component of the velocity. In both cases the velocity we use does indeed go to zero - the mass stops.

If you absolutely must analysise it in 2D, then do that. Of course then there'll be sideways forces too, which you must include. Just makes it harder. And it won't change the fact that the velocity of the 100kg mass changes by 80m/s during 1 second. So it'll end up exactly the same as my calculations.



He's hit the nail on the head.


I didn't do that. Check my calculation again. I also divided it by the turnaround time. Then I got the 8000N.


Who said it was a force? Only you. It's not a force, it's a change in velocity. It does cause a force to be applied, but we have to actually calculate that force, it's not 80N.



Sure. And I never said the mass travels over 40m/s at any point. Where did you get that idea from?


The mass's turnaround requires no burning of gas or use or electricity or anything. It's a passive device too. Again I never said we need to put energy into it to make it turn.

In the car example you could turn off the engine and let it cost around a 180deg curved track. In the absence of friction it'll get a 50mph change in velocity too. And of course it'll apply a force to the track, despite not burning any fuel and not losing any speed.

Correct. And 80m/s/s is not a change in velocity, it's an acceleration. The change in velocity is only 80m/s.




In this statement of Newton's 2nd law, d is not distance, is part of the derivative symbol.

F=d/dt (mv)
m is constant so
F=m d/dt v
F=m dv/dt
dv/dt is, by definition, equal to acceleration
F=ma
Then continue as I've already shown recently.



Here's a quote from the very same page you referenced. Didn't bother to read any of the text with that equation?

"
the use of the constant factor rule in differentiation allows the mass to move outside the derivative operator, and the equation becomes
F=m dv/dt
By substituting the definition of acceleration, the algebraic version of Newton's second law is derived:
F=ma
"

the 1000kg pipe has a speed of 4m/s

in order to stop the pipe in 1 second you must apply a force
of 4000N for 1 second.

the turn is 40 meters long.
the half way point in the turn is 20 meters.

the 100kg mass travels to the 180 degree position in
0.5 seconds because it has a speed of 40 m/s.
the 100kg mass only applies its force to the pipe for 0.5 seconds
before it reaches the half way point.

40m/s is 20meters/half second
its velocity is 20 meters / half second

the below force calculation uses acceleration not velocity
acceleration is determined by distance over time

acceleration is (((( DISTANCE / ((((DISTANCE * TIME ))))
acceleration is 20 m / (20 m * 0.5 seconds)
acceleration is 20 m / 10
acceleration is 2 m/s/s

the pipe has a mass of 1000kg

2000N = 1000kg * 2m/s/s

the pipe slows down durring the first half of the 1st turn from
4 m/s to 2 m/s
it does not stop durring the first half of the 1st turn.

the end of the second half of the 1st turn
is when the pipe stops.


I think your problem is that you dont include time.
or when you do you dont take into consideration that
half of a second is a big difference when using a
time frame of 1 second.

I agree.


Yes. And it also applies a force to the pipe during the 2nd half of its travels out the other side.



You love inventing dimensionally inconsistent equations to serve your pre-assumed result, don't you? That isn't acceleration.




How about trying to understand my calculation? - the one in bold I posted a couple of days ago is probably the most complete and up-to-date. Then you can actually identify what my problem is, rather than guessing.

I consistently do exactly that for you. See in this message I understood that you used a wrong formula for acceleration and pointed it out. You can try to defend it, but good luck finding any websites or books or experiments or people or anything that support the equation a = distance / (distance * time)

I didnt make that up.



that is how you determine acceleration given that your
knowns are
the distance and the time that the distance was traveled

and the knowns for the force calculations above are

v=20 m/s
t=0.5 seconds
d=20 meters
m=100kg mass

use the above FORCE equation and you will
get 4000N

---------- WHERE IS THE FORCE EQUATION THAT YOU USE?-------------------
or are you using a MOMENTUM CHANGE equation
or a VELOCITY CHANGE equation to find for FORCE?

or is there one , I would like to see it.

--------------------------- distance ---------------------------



Displacement is not Distance as is evident above.
traveling a distance is going from point (a) to point (b) along a given path.

Displacement is the shortest distance between two points.

traveling that same distance in 1 second is going from
point (a) to point (b) in 1 second

if that distance is 40 meters and you travel that distance
in 1 second you have traveled 40 m/s

---------------------------- acceleration -----------------------

if you accelerate from 0 m/s/s to 40 m/s/s in 1 second
your acceleration is 40 m/s/s
and you traveled 20 meters in that 1 second.

when I use Distance / Distance*time it is acceleration

a 1kg mass with a acceleration of 1m/s/s requires a force of 1N
if that mass travels a distance of 0.5 meters in 1 second it has traveled 0.5m/s
that means that mass had a acceleration of 1m/s/s

lets check that with a physics math formula used for finding acceleration
a=f/m
1m/s/s=1N/1kg

lets use the formula D/dt to check the acceleration result

0.5 meters distance / 0.5 meters distance * 1 second = 1m/s/s acceleration

Those formulas are correct, but as I said (and you ignored), you're misinterpreting them - grossly.

d isn't distance. It isn't any number, it's part of the differential operator d/dt. d/dt means "rate of change with respect to t of...". I'm not going to try to teach you calculus, but it's not needed. Those formulas above are identical to

F=ma
and
a=F/m

The proof is shown on the Wikipedia force page you linked to before.

You'll never understand things if you make an effort to ignore anything you don't understand. When I don't understand a word, I don't pretend it wasn't there, I go and look it up, or at least make an educated guess and keep in mind that I might've been wrong in case it leads to problems later.






You've seen them many times. They're clearly spelt out in bold in post #35266.


I'm getting tired of this Paul. Maybe you're just playing the fool to be irritating, or maybe you're afraid to do anything correctly because you realize it could prove your idea wrong. If you post one more wrong equation I'm giving up.

they seem to work well for me , whats the problem that you have with them.

how do you determine acceleration if you dont use
distance and time , where do you start?
suppose you dont know what the force is , or you dont know what the mass is?

what do you do then?

----------------

could you please put your equation up again , I cant find it.

and Im not playing a fool , Im genuinely concerned with how
you get 8000N from 4000N

forces dont just appear , they must come from somewhere
and I want to find out where your getting the extra
4000N from.

They work well because you invent suitable units and meanings that lead to the 4000N force you somehow decided you should be getting. Doing it correctly, you have no choice about units, you have no choice what the answer's going to be, it just comes out one way, like it or not.




If you know enough other variables, you don't need to know distance. In this case we know the time it takes, we know the velocity change it experiences, we know it's mass, so we can calculate the force as below.






The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N





Maybe your stumbling block is assuming it somehow 'started' as 4000N then increased to 8000N. There was never any 4000N to begin with.

I found your error...
you use the mass*acceleration to get the 8000N
then you use the same to get another 8000N to act against the pipe.

the mass and the pipe both experience the same 8000N
you cannot seperate this action and then get two seperate forces from it.



the 8000N is divided between the mass and the pipe.
4000N goes to the pipe as I have always said.

you dont get your extra free force to move the
pipe in the opposite direction.

in order for you to apply a 8000N force to the mass and a 8000N force to the pipe , you would need 16,000N

so the pipe stops , and your bulls&^$t stops also.
im pretty much tired of your scammy ways kallog

end of our discussion.

Yes, this is a direct application of Newton's 2nd law: F=ma


I don't use F=ma for that, I use Newton's 3rd law:
3rd law on Wolfram Alpha


In some fundamental way they're the same, that's why the have the same magnitude. But the direction is opposite. You push on your desk, the desk will experience a force and may move because of it. You hand will also experience a force, the same force, but in the opposite direction, you'll feel it pushing back at you. You don't feel nothing, and you don't feel the desk pulling your hand away from you.

centripetal acceleration

centripetal acceleration calculator

radius of turn = 12.732 meters
velocity is the tangential velocity = 40 m/s
ac=v^2/r

ac=125.6676m/s/s

centripetal force

mass = 100kg
Fc=mac=m*(v^2/r)
Fc=100kg*125.6676 m/s/s = 12566.76N

Fc=((40m/s*40m/s)/12.732m radius) * 100kg = 12566.76N

so the outward force is the same as the inward force

so the pipe really does have a 12,566N force pushing
it in the opposite direction.

I find that hard to believe but I recall that that is
the exact principle that another invention I had opperated on.

so from a 4000N force we get an extra 12,566N force

amazing , I apologize to you , I was wrong.

from what I have read up on , the magnatude of the force
can be greater with a shorter turn and less with a larger turn.

interesting, I can see possibilities here.

anyway in 0.6368 seconds the pipe slows to 0m/s velocity.
while it is slowing it continues to move for 4 meters in the
(+) direction.
so now it has traveled +16 meters & +4 meters = +20 meters

after the 1 second has passed the pipe has reversed direction and is now moving in the opposite direction at a velocity of 8.566 m/s

it has traveled a distance of 2.283 meters in the opposite direction while the mass completes the 1st turn , it free floats the 160 meters for 18.678 seconds.

the pipe has a velocity of 8.566 m/s so it travels a distance of 8.566 m/s * 18.678 seconds = 159.995 meters in the opposite direction.

you would need to apply a force of 8566N to stop the pipe in 1 second.

the mass then enters the 2nd turn , having the same results as in the above so the pipe continues for a distance of 4 more meters in the (-) direction as it slows to 0m/s.

the mass applies a force of 12566.76N to the pipe and
the pipe slows to 0m/s

the pipe then moves in the (+) direction for
a distance of 2.283 meters before the mass enters the accelerator again.

the mass enters the accelerator at a velocity of 8.566 m/s
it accelerates the 100kg mass to 40.907 m/s in 6.4682 seconds

it now enters the 1st turn with a force of 4090.7N

so you were right again , it does go further backwards.

+16m
+4m
-----------> +20m
-2.283m
-159.996m
-4m
<----------- -166.273m

once again , I apologise.

you were right.

sorry.

but it will work and that is the important part.

Haha, amazing indeed! I have to frame that and hang it on my wall



Not exactly. That's the outward force. It's not always directed along the pipe, so the actual average longditudinal force during the turn will be different. Not sure exactly how to calculate it at the moment.




Jumping the gun again? :P

I noticed that I used the pipe velocity for the
mass velocity in my previous post as the mass moved
into the accelerator again , it was late
and it was friday.

so dont go framing up anything yet.

like you said that is the outward force,LOL.

I might have to figure out the total linear force
to see if I was right or if you were right.

LOL

I believe if I use a 180 degree turn I can divide the
1 second by 180 , since the mass is traveling in a circular path I can use the linear angle and the fc angle to find the amount of linear force for each degree of rotation , then add up the forces as I said before.

there will be 180 fc forces in the outward direction
and 180 linear forces in the linear direction.

starting at 270 degrees the linear force would be
very small , then at 180 degrees the linear force would
be very large then at 90 degrees the linear force would be very small again.

since the longitudinal forces outnumber the linear forces
2-1

and 12,000 / 3 = 4000

I seem to think I was right all along.

there are 8000N longitudinal
there are 4000N linear

so for now I recant my apology , LOL.

but save the frame you may need it later.