# So neutral it must be natural!

This law neither I can prove right nor you can prove wrong.
This law leaves as trully neutral.

The law is:"Displacement always takes the direction of the force causing it"

It's the compromise between Newton 1 and Newton 2.

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I think this is exactly what te second law of newtons says.

Yes but it's only partially true. The difference is that constant force suggests constant displacement rate.

Doc Al
Mentor
Just plain wrong

deda said:
The law is:"Displacement always takes the direction of the force causing it"
Nonsense. First, force is not required to "cause" a displacement. (A net force causes acceleration.) Second, the displacement of an object need not be in the direction of the force on the object.
It's the compromise between Newton 1 and Newton 2.
Why compromise?

russ_watters
Mentor
deda said:
Yes but it's only partially true. The difference is that constant force suggests constant displacement rate.
Constant non-zero net force means constant acceleration in newtonian physics.

russ_watters said:
Constant non-zero net force means constant acceleration in newtonian physics.
I'll believe it when somebody will answer this:
www.geocities.com/dr_physica/fv.jpg

Doc Al said:
Nonsense. First, force is not required to "cause" a displacement. (A net force causes acceleration.) Second, the displacement of an object need not be in the direction of the force on the object.
Fine, when I push one object forward then I'll expect it to move backward!
That's some sense!

deda said:
Fine, when I push one object forward then I'll expect it to move backward!
That's some sense!
If it weren't true, we'd all be in the Sun by now.

Doc Al
Mentor
deda said:
Fine, when I push one object forward then I'll expect it to move backward!
That's some sense!
Expect anything you like. If you want to know what happens, learn some physics. A net force applied to an object produces an acceleration. To find the object's displacement, you need to know how it was moving as the force was applied.

Lonewolf's example of the Earth orbiting the Sun is apt: the force is toward the Sun, but the Earth moves sideways.

I don't know why the link doesn't work.
It's yahoo's problem.

I have attached the jpeg

1st the planets don't spin around the sun but the planets and the sun are spining around the center of the lever they all make together.
2nd the force of each IS in direction of its displacement. Not like Newton tells you.
3rd F * dR = - R * dF > 0 decides how the radius and the force of each object change each other!

Newton got it all wrong and you are doing the same mistake!
But have it your way any way. Why should I care?

where did the attachment go?

deda said:
where did the attachment go?
It must have been Newton!
I pushed the jpg in one direction it went in another.

Kidding, the jpg is to big.

try this:
www.geocities.com/dr_physica/fv.htm

Chi Meson
Homework Helper
THe jpg appears to show the force on an object and the velocity of an object as a (simultaneous, I assume) function of time.

But the force curve is a "bell" curve, such as you would get when you measure the actual force on an object during a collision, while the velocity function is an elementary "straight-line" slope suggesting constant acceleration. THis is of the type that you never get in an actual experiment.

Deda is pointing out the disparity between real experiment and idealized situations. I have never, in any of my computer-graphic analyzed collision labs, seen a v-t graph look like this one.

Doc Al
Mentor
deda said:
1st the planets don't spin around the sun but the planets and the sun are spining around the center of the lever they all make together.
They orbit about the center of mass of the system, which is pretty close to the center of the sun. There is no lever.
2nd the force of each IS in direction of its displacement. Not like Newton tells you.
Nope. The displacement follows the orbit (obviously), the force does not.
3rd F * dR = - R * dF > 0 decides how the radius and the force of each object change each other!
I have no idea what that could possibly mean.

Chi Meson said:
THe jpg appears to show the force on an object and the velocity of an object as a (simultaneous, I assume) function of time.
True.
I followed Newton's reasoning to come up with it.
Chi Meson said:
But the force curve is a "bell" curve, such as you would get when you measure the actual force on an object during a collision, while the velocity function is an elementary "straight-line" slope suggesting constant acceleration. THis is of the type that you never get in an actual experiment.
Ok, you give me better graphic that obeys Newton's laws.
Chi Meson said:
Deda is pointing out the disparity between real experiment and idealized situations. I have never, in any of my computer-graphic analyzed collision labs, seen a v-t graph look like this one.
How do you think it is?

Remember that if V=function_1(t), F=function_2(t) then F and V are directly related.

Is this F-V relation anything like Newton says?

Doc Al said:
I have no idea what that could possibly mean.
Imagine F - the force and D - the distance from the center are vectors then:
New(F) = cos(a) * F + sin(a) * (|F| / |D|) D
New(D) = cos(a) * D - sin(a) * (|D| / |F|) F
Here a is the same phase - difference os oscillation F and D have.
That is what F * dD = - D * dF > 0 stands for.

Chi Meson
Homework Helper
deda said:
True.
I followed Newton's reasoning to come up with it.

Ok, you give me better graphic that obeys Newton's laws.

How do you think it is?

Remember that if V=function_1(t), F=function_2(t) then F and V are directly related.

Is this F-V relation anything like Newton says?
No you followed your own incorrect intstinct to come up with it. Newton knew that the velocity fuction would be an integral of the acceleration function. And we now understand that acceleration is proportional to net force (as is one modern interpretation of Newton's 2nd law).

so in reality function v(t) = {integral} 1/m * F(t)/dt

All you need to do is pick up a college physics text book to show the actual curves produced by experiment

And by the way, when a force is acting perpendicularly to the direction of motion of an object, the result is a change in the direction of that object, but the direction of motion that results is not going to be the subsequent direction of force on the object. Only when an object is not moving will the subsequent direction of motion definately be the same as the direction fo the force.

I don't know a thing of physics.

Last edited:
Chi Meson
Homework Helper
deda said:
I don't know a thing of physics.
QUOTE]

You said it. Goodbye.

I was only fooling you.
I know if a = const then:
$$V = V_0 + at = V_0 + \frac {F}{m} t$$
So how does your graphic look like since its better?