Proving F=ma?
Most recent answer: 9/15/2018
- Anonymous
1. Physics does not start from a well-defined set of unshakable axioms and proceed via mathematical logic to proven facts. It is put together out of a combination of observations, provisional generalizations, and mathematical glue.
2. Anyway, F=ma happens to be false.
Even if you were to (falsely) assume that space and time had a simple geometry, the proper version of the equation is F=dp/dt where p is momentum. [dp/dt means the rate of change of momentum in time (t).] That happens to be the version that Newton gave, but it too is not true in the way that Newton meant it. The reason is that p=mv where v is velocity and m is a velocity-dependent mass. Newton didn't know about that intrinsic velocity dependence. F=ma would be true only if m were independent of velocity. This was all worked out precisely by Einstein in 1905.
In more valid general geometry, the terms, including 'a', have no fixed meaning.
Mike W.
(published on 10/22/2007)
Follow-Up #1: Why does F=ma?
- Bob (age 19)
Seattle, WA
The first thing you have to ask about F=ma is what it actually tells us about the world, since until we figure that out we can't begin to say why it should be true or false. Here's the problem. Most objects don't come stamped with an "m" value on them. Space and time aren't laid out with a labeled grid of coordinates, so we aren't just given the "a" values for objects. Worst of all, what's F?
Let's say we ignore the question of how to determine a, by assuming that somehow we have a common-sense set of space-time coordinates that we're happy with. The key step to making some meaning out of F=ma is then Newton's 3d law- conservation of momentum. We can bounce objects off each other, and by measuring their velocities before and after the bounce, figure out the ratios of their m's. Pick one object to call the unit mass, and now we have a set of m's.
Now we get to the hard part. What are the F's? Let's say we see some m with an a. What's to stop us from just inventing an F to make F=ma true? If we could always do that, then F=ma would be untestable and meaningless. So we must insist on some rules about the F's. The third law says that there needs to be an opposite F on something else, and we can insist that the something else is fairly nearby. More generally, we can insist that the rules for when there should be an F shouldn't be too weird or complicated. If we can fit what we see within those rules, then we can say that F=ma is true. Up to a point, that program works. Once you start including electromagnetic fields and fast-moving objects, it gets too awkward and we need a different set of rules, in which F=ma is replaced.
That's a very compressed version of a long discussion. Feel free to follow up.
Mike W.
(published on 06/25/2011)
Follow-Up #2: Net forces and Newton
- Hazem (age 55)
Phoenix, AZ, USA
The F in F = ma is the the total force on an object, often called the "net force." This equation is Newton's second law. (By the way, Newton wasn't made president of the Royal Society until 1703, 16 years after his Principia was published.)
Okay, so we need to consider all the forces on an object for F = ma to be valid. In both of your examples, you only mention one force, where there are actually at least two we need to account for.
In the case of pushing against a wall, there is a so-called normal force equal and opposite to the force you apply, so the acceleration is zero (unless you exceed the breaking strength of the wall). F_total = F_push - F_normal = 0.
In the case of a box being pushed along a floor with friction, there is a friction force opposite the pushing force. The friction force depends on the material (more for rubber, less for ice) and is usually proportional to the weight of the object. F_total = F_push - F_friction = ma (smaller for rubber, larger for ice).
You asked a question which came from curiosity and thinking about physical situations (good). With a little more thinking like that, I bet you would have wondered whether there might be additional forces that would explain your observations. Instead, you cheated yourself of any further understanding.
Rebecca H.
(published on 02/29/2016)
Follow-Up #3: testing F=ma form
- Akshat Khatri (age 20)
461001
Sure, we can suggest such experiments.
First, let's take proportionality to F. Take some mass and push on it with 1 compressed spring, measuring a. Now use two, then 3, etc. Common sense says that F should be additive. So you can check that a is indeed proportional to F.
Now for fixed a you want to check if F is proportional to m. You can put together 1, 2, 3... nearly identical masses. Check how many of those nearly identical compressed springs you need to get the combined mass accelerating at the fixed a.
Mike W.
(published on 10/03/2016)
Follow-Up #4: force and acceleration
- Vansh (age 14)
Delhi, India
I don't usually think of F=ma as "an equation to find the force" although I guess it plays that role in some homework problems. I thnk of it more as a way to predict the acceleration, a, when you have some known forces acting on an object. It turns out not to be quite right. Instead, the rule that makes correct predictions is F=dp/dt, as we said earlier in the thread. p =mv/(1-(v2/c2))1/2. ("m" here means the rest mass.) When v/c << 1, F=dp/dt becomes very close to F=mdv/dt=ma.
Mike W.
(published on 03/04/2018)
Follow-Up #5: F=ma?
- Z. Awan (age 19)
Islamabad, Pakistan
As you can see from the preceding answers, F=ma is known not to be generally true, so it can't be the foundation of our current physics. As the discussion above explains, even before it was known to be false, philosophers had noted that the complications you mention make it hard to check whether it holds.
Mike W.
(published on 09/15/2018)