Q:

prove F=Ma

- Anonymous

- Anonymous

A:

It would be difficult to prove F=ma for at least two reasons.

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.

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)*

Q:

Is there any reasoning as to why F=ma? Intuitively it makes sense but is it all just experimental?

- Bob (age 19)

Seattle, WA

- Bob (age 19)

Seattle, WA

A:

As you know, it's not quite true that F=ma, but let's for now just think about your deep question for the case of slow-moving objects, where F=ma works pretty well. I hope you'll forgive a long-winded, fuzzy answer, since I don't know any good short answer.

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.

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)*

Q:

f=ma doesn't seem to match real life observations. If you pushed hard on a wall and it still won't move then a=0, and f=ma is also 0 showing that no force is exerted on the wall, which is not true. Another example, if you place a heavy object on a surface of high friction such as rubber then the force needed to cause it to accelerate is is much higher than the force needed to give it similar acceleration on a slippery surface such as ice, yet according to Newton's law f=ma the forces should be the same! It seems that f=ma is nothing more than a special case of the object (to be accelerated) can move on the same type of surface. It seems like this is what Mr. Newton did while experimenting on accelerating objects before he came up with this law, but since he was the head of the British Royal Society of science, no one dared question him. It doesn't have to take an Einstein to realize that, but a willing to be disattached from prejudices.I hope no one gets offended by my questioning as if I attacked their religious beliefs. Historically speaking, humans held unshakable beliefs as facts for hundreds of years before they finally tossed them out as falsehoods, and I believe that there are still lots of falsehoods that will be tossed out of science facts in the future. Thank you.

- Hazem (age 55)

Phoenix, AZ, USA

- Hazem (age 55)

Phoenix, AZ, USA

A:

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)*

Q:

I completely agree to the above explanation given by Mr. Mike W. All I want to ask is why F=ma?? I mean F=ma states that mass and acceleration has an equal effect on force. Intuitively and even logically I understand that force applied to generate a specific acceleration in an object depends on mass of the object. Let me put my point through a thought experiment. Consider a ball that weighs 500 grams and another that weighs 1 kg. Now make your friend drop them from the balcony of the first floor(so that it doesn't gives air resistance enough time to change their accelerations by a lot. So when the balls fall down they will be having almost similar acceleration) and you stand down to catch them. Now from the 1 kg ball you will experience a greater impact on your hands than with the 500 grams. This impact can be considered as the reaction of the force you applied on the ball to give it an acceleration of -g m/s2(-g because you brought it to rest). So the forces you applied on 2 balls having different masses gave the same acceleration and the one with the greater mass had the greater force applied on it. Hence F is directly proportional to mass. But is it?? I just said that both F are different and the one with the greater mass had the greater force applied on it but greater by what extent?? What if F is proportional to m^2 or m^3?? I understand the force is not to quantify naturally(how can calculate its absolute value, you need comparisons by setting a basic unit. ) but when you give an equation it should be justified?? F=ma is a very loose equation that can only be understood by experiments. Can anyone suggest me one where I can conclude that F is proportional to m and not to m^2 or m^3 and also F is proportional to a and not to a^2 or a^3??

- Akshat Khatri (age 20)

461001

- Akshat Khatri (age 20)

461001

A:

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)*

Q:

Then what is the right equation to find net force??

- Vansh (age 14)

Delhi, India

- Vansh (age 14)

Delhi, India

A:

I don't usually think of **F**=m**a** 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**=d**p**/dt, as we said earlier in the thread. **p** =m**v**/(1-(v^{2}/c^{2}))^{1/2}. ("m" here means the rest mass.) When v/c << 1, **F**=d**p**/dt becomes very close to **F**=md**v**/dt=m**a**.

Mike W.

*(published on 03/04/2018)*