Rethinking Newton's Equation: Re-writing F=MA with Reduced Mass

[NickMMM]

I am the president of the Physics club at my high school. One of our members came up with a very interesting hypothetical. He asked each of us to re-write the famous Newtonian equation for motion, F=MA(force equals mass times acceleration). Here's the question we are all trying to answer, if there was a hypothetical reduction in inertia or mass, how would one re-write the equation F=MA?

Thanks,

Nicholas

 

[HallsofIvy]

Do you mean a if the mass is reduced as the object moves (such as a automobile consuming gas as it moves or a wagon spilling water as it moves)? In that case the general formula would be F= dp/dt where "p" is the momentum, mv. By the "product rule" of differentiation, F= m dv/dt+ v dm/dt. (If m is a constant, dm/dv = 0 so that reduces to the previous "F= m dv/dt".)

 

[NickMMM]

HallsofIvy,

Thanks so much for the response. To make sure I understand, what does d and t stand for, distance & time? Also m is mass and v, velocity? So hypothetically if a car was traveling 75MPH and the mass or inertial force was reduced couldn't we describe it much more crudely as the following very general equation?

< <
F = (M)A

 

[NickMMM]

the "<" should be over F & M, bad formatting

 

[ZapperZ]

HallsofIvy,

Thanks so much for the response. To make sure I understand, what does d and t stand for, distance & time? Also m is mass and v, velocity? So hypothetically if a car was traveling 75MPH and the mass or inertial force was reduced couldn't we describe it much more crudely as the following very general equation?

< <
F = (M)A

I'm guessing that you haven't had any calculus.

df/dt is a differential with respect to time. It is a derivative of any function "f". However, if you don't know calculus, this is all Greek to you, unfortunately. So I'm not sure to what extent what HallsofIvy has stated would be useful to you.

Zz.

 

[NickMMM]

Is there an simplistic was that this can be generally stated? As I don't know calculus but could it be accurate to say there's another way to express this formula as previously stated?

 

[Nugatory]

Is there an simplistic was that this can be generally stated? As I don't know calculus but could it be accurate to say there's another way to express this formula as previously stated?

You could introduce the concept of momentum (which is defined by ##p=mv##), and then because acceleration is just the change in speed with time, you could look at the right-hand side of ##F=ma## and say that that's just another way of saying "the change in momentum with time assuming the mass is constant".

Now it feels natural enough to generalize ##F=ma## to say that F is equal to the change in momentum with time, whether the change is happening because of a change in speed or a change in mass or a combination of the two. Unfortunately, it's very difficult to express that as an equation without using calculus, which is what HallsofIvy was doing.

 

[jbriggs444]

I am the president of the Physics club at my high school. One of our members came up with a very interesting hypothetical. He asked each of us to re-write the famous Newtonian equation for motion, F=MA(force equals mass times acceleration). Here's the question we are all trying to answer, if there was a hypothetical reduction in inertia or mass, how would one re-write the equation F=MA?

In the hypothetical is the suggestion that inertial masses of all objects are changed. So that all bricks are suddenly accelerated twice as easily, for instance? If so...

You are using the equation f = ma, so it follows that you are using a "coherent" system of units. The units of force, mass and acceleration have a necessary relationship to one another. In such a system of units, a systematic change to all masses must be accompanied by a change to the units of force, mass, and/or acceleration.

If, for instance, you cut all the masses in half, you would also reduce the size of the Newton by half. Even though it would take half as much force to accelerate a brick that used to be 1 kg in mass at a rate of 1 meter per second per second, that half-force would still be one "Newton" in the resulting system of units.

 

[Khashishi]

How, exactly, are you changing the mass? Is it a rocket, and you are burning off the fuel? Or are you envisioning something more mystical?

 

[bhobba]

Interesting question for someone at high school level - especially if you haven't done calculus.

Just as a teaser to what's really going on think about this for a minute. Newtons first law follows from the second which is a definition of force anyway - you can't really argue with a definition - you may be able to state it a different way - but you can't argue with it. The real content of Newtons laws is the third law.

But wait - there is more. Newtons third law is basically conservation of momentum and a very interesting discovery was made by a very great mathematician - Emily Noether - called Noethers Theorem - look it up - it's way cool.

When you have done so it should make you ask an obvious question - post here when it occurs to you and your other club members.

But to fully answer your question - F=MA is a prescription that says - get thee to the forces - they are the important things and the paradigm to analyse problems in mechanics. In that sense it can't really be re-written.

Thanks
Bill

 

[meBigGuy]

If the mass changed but the force was constant you would re-write the equation to express the mass in time. You would then be able to solve for the acceleration as it changed in time. Or you could hold the acceleration constant and solve for the required force profile to match the mass change for constant acceleration. But, the basic F=MA would still hold true. The values for the variables would change over time.

Say the mass started at 1 Kg and increased 1 Kg per second, and the force was 1 Newton. You could write F=1, M=1+t and then solve for A at any t.

dv/dt is the derivitive of velocity with respect to time (the slope of the velocity). If the change in V were constant you could think of it as delta-v/delta-t. That is, change in velocity divided by change in time (which is acceleration. meters/sec/sec). Calculus provides the tools to deal with continuously varying variables. I taught myself about derivitives and Integrals in sophmore year high school, so it isn't beyond you, and I found it really cool. How do you determine the area under a parabola or the parabola's slope at any point.