# So neutral it must be natural!

1. Apr 6, 2004

### deda

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.

2. Apr 6, 2004

### Peterdevis

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

3. Apr 6, 2004

### deda

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

4. Apr 6, 2004

5. Apr 6, 2004

### Staff: Mentor

Just plain wrong

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.
Why compromise?

6. Apr 6, 2004

### Staff: Mentor

Constant non-zero net force means constant acceleration in newtonian physics.

7. Apr 7, 2004

### deda

I'll believe it when somebody will answer this:
www.geocities.com/dr_physica/fv.jpg

8. Apr 7, 2004

### deda

Fine, when I push one object forward then I'll expect it to move backward!
That's some sense!

9. Apr 7, 2004

### Lonewolf

If it weren't true, we'd all be in the Sun by now.

10. Apr 7, 2004

### Staff: Mentor

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.

11. Apr 7, 2004

### deda

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?

12. Apr 7, 2004

### deda

where did the attachment go?

13. Apr 7, 2004

### deda

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

14. Apr 7, 2004

### Chi Meson

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.

15. Apr 7, 2004

### Staff: Mentor

They orbit about the center of mass of the system, which is pretty close to the center of the sun. There is no lever.
Nope. The displacement follows the orbit (obviously), the force does not.
I have no idea what that could possibly mean.

16. Apr 8, 2004

### deda

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?

17. Apr 8, 2004

### deda

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.

18. Apr 8, 2004

### Chi Meson

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.

19. Apr 8, 2004

### deda

I don't know a thing of physics.

Last edited: Apr 8, 2004
20. Apr 12, 2004

### Chi Meson

21. Apr 12, 2004

### deda

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?