Can Scientists Prove Newton's Laws?

[Alem2000]

Hello fellow scientists! I wanted to know if anyone could prove Newtons laws to for me. I remember a friend did once but i wasnt paying attention at the time. What i remember was him telling me that F=ma could be broken down into a more complex equation with partial darivatives to solve more complicated problems. Can anyone help?

 

[hazakate]

Newtons first law
a body will remain in its state of rest or in uniform motion unless acted upon by and unbalanced external force
f(net)= ma
Newtons second law
the vector sum of all the unbalanced forces is proportional to the product of the mass and the acceleration for constant mass.
Newtons third law
for every action there is an equal (in magnatued) and opposite (in direction) reaction.

 

[Gza]

Some friend you have! F=ma IS a fundamental law. It basically gives a mathematical definition of what a force IS. If an object accelerates, it necessarily MUST have a force that acted on it. It can't be derived. Your friend probably showed you its differential equation form which is useful for solving problems where different time derivatives of position are related to each other, such as in my example a simple spring system, whose acceleration is a function of position. recall hooke's law which is the force of the spring as a function of its displacement from its equilibrium position:

<br /> F = -kx<br />

Inserting the definition of force into the left side of the equation yields:

<br /> m\frac {d^2x} {dt^2} = -kx<br />


Dividing through by m:

<br /> \frac {d^2x} {dt^2} = -\frac {k} {m} x<br />

voila! a second order differential equation from F=ma. This type of problem solving comes in handy when trying to find the function of motion for a particular system(position as a function of time). In this case analysing the diff equation from inspection yields the resultant solution to be cos() since it is the only equation you can diff twice and receive itself with a negative sign appended. Of course something else is needed for the functions argument to fully satisfy the equation, but any good book will show you all this. I don't know if this even came close to helping you out, but hey you hopefully learned a little physics, so that's never a loss

 

[Kalimaa23]

Time for a pet peeve of mine. F = m.a is NOT a fundamental law. The correct form of Newton II is

<br /> \frac{d\vec{p}}{dt}=\vec{F}<br />

When the mass is a constant, this reduces to F = ma

<br /> \vec{F}=\frac{d(m.\vec{v})}{dt}=m. \frac{d\vec{v}}{dt} + \vec{v}. \frac{dm}{dt}= m.\vec{a} + 0<br />

 

[Alem2000]

Thanks DIMITRI, I think that was what I was looking for.

 

[rattis]

I thaught that F=ma is derived from aceleration being proportional to Force, and accereration being inversly proportional to mass. This gives acceleration being proportional to, Force multiplied by one divied by mass (A is proportional to F x 1/m).

Giving, Force = mass multiplied by acceleration.

 

[baffledMatt]

In a sense, you can derive Newton's laws from Hamiltonian mechanics. I say 'in a sense' because that only works if you consider the Lagrangian formulation to be more fundamental.

Matt

 

[Gza]

Time for a pet peeve of mine

no need to be peeved, its just that here in the US (or in my school at least) F=ma tends to get burned into your mind as fundamental. Problems involving changing masses get lumped into a different subset of problems involving variable mass, such as fuel rockets. I can't say that the momentum form of F=ma is the first thing that pops into my head when trying to find the force on an object with a stable mass.

 

[Parth Dave]

I thaught that F=ma is derived from aceleration being proportional to Force, and accereration being inversly proportional to mass. This gives acceleration being proportional to, Force multiplied by one divied by mass (A is proportional to F x 1/m).

Giving, Force = mass multiplied by acceleration.

If you say force is proportional to acceleration and also mass you would have a formula of f = kma. They are simply proportional, not equal. You would have to than prove that k = 1 to show that f = ma. You can however prove that using simple experiments.

And yes, in NA it is sinked into your head that f = ma. I never found out that it was really dp/dt until i saw it on the web. I really don't understand why either, its not a very hard concept to understand.

 

[JohnDubYa]

Although Newton first described his law in terms of momentum, I fail to see much of a distinction. The only instance where there would be a difference would be if the mass of the object in question changed in value. But barring nuclear physics (which isn't applicable to Newton's second law anyway), I am unaware of such a system.

So is there really a difference between the two expressions?

 

[baffledMatt]

Rockets are also such a system where the mass of the object is changing.

 

[JohnDubYa]

What is the object? The rocket? Its mass doesn't change. The fuel? Its mass doesn't change either.

Newton's second law says that the vector sum of all forces acting on a system equals the mass of the system times its acceleration. So what exactly is the system at which you are applying the forces?

 

[baffledMatt]

The principle of rocket propulsion is based on the fact that it gains momentum by throwing fuel very quickly out the back (which is why you burn it - hot things travel faster). Hence, as fuel is used the rocket will get lighter.

Matt

 

[JohnDubYa]

Again, what is the system? Rocket? Fuel? Rocket plus fuel? In all three cases the mass stays constant.

 

[Gza]

I think the principal relies on the fact that the elements of mass consisting as the fuel are "thrown" out of the system, therefore only considering the mass of the rocket, which appears to get lighter.

 

[arildno]

I think the principal relies on the fact that the elements of mass consisting as the fuel are "thrown" out of the system, therefore only considering the mass of the rocket, which appears to get lighter.

The system doesn't just appear to get lighter; it does get lighter.
At any given instant, the system to be considerered is rocket+remaining fuel.

An example of a system gaining mass is an open van driving in rain. The added water adds to the mass of the system.

 

[TALewis]

Again, what is the system? Rocket? Fuel? Rocket plus fuel? In all three cases the mass stays constant.

In a control mass analysis ("closed system"), you're right, the mass of the rocket+fuel doesn't change. However, a rocket is better considered with a control VOLUME analysis ("open system"). Consider the open system to be the volume occupied by the rocket and all of its contents. The system loses mass as burning fuel is expelled out the back of the rocket. Thus, you can use the momentum form of Newton's second law in this case.

 

[robphy]

Although Newton first described his law in terms of momentum, I fail to see much of a distinction. The only instance where there would be a difference would be if the mass of the object in question changed in value. But barring nuclear physics (which isn't applicable to Newton's second law anyway), I am unaware of such a system.

How about a "leaky car" (which drips water from a hole in the bottom)

 

[JohnDubYa]

The rain actually collides with the van. The water dripping from the car is also a collision, just run in reverse. You don't have an object changing mass, you have an object colliding with another mass.

Take two blocks (call them A and B) attached by a compressed spring. We will assume that a rope holds the two blocks together. Now cut the rope.

Does the mass of the system change? Well, what is the system? Block A? No, its mass doesn't change. What about block B? Its mass doesn't change either.

But let's suppose we use the notion that the system is block A and block B. After you cut the string, what is the mass of the system? According to your argument, the system is now just block B, which has dramatically dropped in value. But here is the fundamental problem: Block A applies an external force to block B.

There is another fundamental problem. If the mass of the system has dropped, how long has it taken to drop? This means you have to answer the question "At what point in time is the system only block B?"

Do you now see the problem with considering the "rocket + remaining fuel" as a system?

 

[JohnDubYa]

by the way, for those that disagree with me, let me ask:

If a car is dropping oil, what happens to its speed over time? (Assume no air resistance.)

Hint: Think about the conservation of momentum in the forward direction. Remember, no air resistance.