K basically this is F = Ma right?

Isn't this law impractical for any situation outside of a vacuum? If I was pushing a toy car with a constant force 10N in order to make it accelerate, resistive forces would build up to oppose the resultant force and give it uniform motion. Therefore I would have to apply more force (20N) to get it to accelerate again. But resistive forces keep building up, so it's like I need an insane amount of force because the acceleration is never going to be constant.

It's always going to keep decreasing unless I add more force.

Freefall I'm falling at 9.81ms-2. Resistive forces build up, so my 9.81m-2 is not constant anymore and decreases to 0 until I reach terminal velocity. The only way I can picture a constant acceleration is in a vacuum where there are no resistive forces and I can apply a force and get a permanent/constant acceleration out of it

So is f=ma wrong? Or am I deluded? Where have I gone wrong if I am?

Why does nobody want to answer my question? What have I done?

Well, it takes some time for me to wake up and look at your question :rolleyes:

Joke apart;

F=ma is valid

F is the net force on the object. If any resistive forces are concerned, then the net force will decrease, decreasing the acceleration, hence to constant velocity.

But doesn't f=ma cater for a constant acceleration? You just said that if any resistive forces are concerned (which is mostly the case on earth) then net force will decrease and therefore acceleration will decrease.

Newtons 2nd law is not: dF=Mda where 'd' is delta. Is it not meant a uniformly accelerating object is acted on by a resultant force? This is never the case this is why I opened this thread. :/

Hello ED. The F and a in F=ma are actually vectors - meaning they have both magnitude and direction - and the equation should properly be written as \small \sum \vec{F}= m \vec{a}, meaning that you have to add up all the forces acting on the object to find the total magnitide and direction of \small \vec{F} . So F=ma leads to a result of constant acceleration ONLY if the sum of the forces acting on the object is constant.

In the example of the falling object, the force of gravity acting on it in the downward direction is resisted by a force of air resistance acting on it in the upward direction. The magnitiude of the air resistance increases as the object's velocity increases. At first air resistance is nil, and the object falls with an aceleration that is m\vec{a} =\small \vec{Weight} \large = m\vec{g}, or \vec{a} = \vec{g}. But as the velocity increases you need to consider the upward force of air resistance:


\sum \vec{F} = \vec{Weight} - \vec{F_{air}}


As long as \vec{F_{air}} is less than the object's weight, it continues to accelerate as it falls, although at a smaller rate than before. At some point the magnitude of the air resistance equals the magnitude of the object's weight, so that \small \sum \vec{F} = 0, and the object no longer continues to accelerate. At that point it has reached its "terminal veocity."

Hope this explanation helps.

Thank you, that explained it.