Resistance to acceleration dependent upon mass or weight?

[inertiaforce]

Although all that you wrote (above quote) sounds accurate to me, I'm not as sure about the connections you may be making in tying it all together.

My post was an effort to reply to your OP #1 after digesting the good stuff from this thread, and to provide an intuitive way to understand why acceleration is dependent on mass, not weight. Not sure we're both there yet.

I'm not as sure about the connections I may be making in tying it all together either lol.

I agree that I'm not sure we're both there yet either. But I do believe we have made some significant progress in considering the problem and analyzing it.

 

[Simon Bridge]

@inertiaforce (OP) and TumblingDice:
You are both overthinking matters, and you keep confusing common uses of terms like force and weight for the use in physics.

Strictly keeping to Newton's Mechanics (Einstein adds some extra wrincles that won't help your ideas here):

Newton's law is properly: "The vector sum of all the forces on an object produces a vector acceleration."
We can write this down as: $$\sum_n \vec F_n=m\vec a$$ The force of gravity - by definition: the weight - adds to the left hand side of this equation.
The inertia - which is the same as the mass - adds to the right hand side of this equation.
The force of gravity happens to be proportional to the mass - which means that a mass term will appear on both sides of the equation - this can confuse things.

Caution: The equation is best handled in an inertial reference frame - i.e. when the system you are looking at is not inside an accelerating box. That would include the elevator examples that have been throwing you off track.
Some confusion has also been generated by using the same variable name to refer to different things - so I am going to be careful to label my variables for different situations. I can make typos though...

I can feel the "but"'s rising - just hold on for a bit, I'll give you a bunch of examples that should illustrate some of the situations that have come up in this discussion. I've numbered them for easy reference.1. A force F is applied to a mass m in weightless conditions.
the weight is $W_1=0$ ... because it is weightless.
the acceleration is: $$a_1=F/m$$ ...in the same direction as the applied force F.

2. The same magnitude force F is applied, upwards, to a mass m, in 1g gravity.
the weight is $W_2=mg$
the acceleration (in the direction of F) is $$a_2=\frac{F}{m}-g$$Notice that the mass still responds the same way to the force as in scenario 1 - the mass m offers the same amount of "resistance to acceleration" as before - it's just that the sum of the forces is different.

The mass is responding differently to the applied force F. There is nothing odd about this - there is another force present. If someone else was pushing on it, providing the extra force, you wouldn't find it at all strange that it behaved differently.

3. The same magnitude force is applied, upwards, to a mass m/2, in 2g gravity.
the weight is $W_3=mg=W_2$
the acceleration (in the direction of F) is $$a_3=2\left(\frac{F}{m}-g \right)=2a_2$$ ... the weight is the same, but the effect of the same applied force is double the acceleration.

In this case the sum of the forces has remained the same as in 2 but the mass has halved.
Therefore - twice the result.

4. the mass is stationary in an elevator that is accelerating at a rate of 1g in 1g gravity.
the weight is $W_4=mg$ ... because that is the force due gravity.
the force of gravity points down.
the elevator is also pushing on the mass with a force $F_p=2mg$ upwards
the sum of forces comes to: $$\left[ \sum_n F_n = F_p - W = 2mg-mg = mg\right] = ma_4$$ ... giving the acceleration: $a_4=+g$ ... which is to say that the mass is accelerating at the same rate as the elevator."Weight" is a technical term in physics that has a slightly different meaning to the common use. It is important not to confuse them.
In physics, weight is defined as the force due to gravity.
In common use, your weight is how hard you press into the ground - or how hard someone has to struggle to hold you up. This is a different thing.

In the elevator example, you can see that how hard you press into the floor is not the same thing as the force of gravity on you.

Which leads to the last scenario:

5. Apply force F upwards on mass m in an elevator that is accelerating upwards at 1g in a 0g gravitational field. i.e. weightless.

the weight is $W_5=0$
the acceleration of the mass is $$a_5=\frac{F}{m}$$ ... in the direction of the applied force F.

BUT: the floor of the elevator is also accelerating ... so the distance between the floor and the object is changing at a different rate than we'd expect if the elevator were not accelerating. That rate is $$a_f=\frac{F}{m}-g$$ ... which looks the same as if the elevator was stationary in 1g of gravity.

We call this a "simulated" gravity giving rise to an "apparent" weight.
The weight is only apparent because there is no gravity and weight is defined as the force due to gravity. The acceleration of the elevator gives the appearance of gravity from the POV of anyone riding in the elevator.

I want to add just one more situation that leads to a lot of confusion.

6. mass m sitting on the floor in an elevator that is accelerating at 1g downwards in 1g gravity.
the weight is $W_6=mg$
the acceleration is $$a=g$$ ... downwards.

BUT the floor is also accelerating at the same rate downwards, so the distance between the floor and the object remains the same.

Notes:
the force of the floor on the mass is zero ... so the "apparent weight" is zero.
this situation is called "free fall" and is often casually referred to as "weightless".
notice that it is not the same thing as zero gravity.

Concluding remarks
That should cover all the bases.
You will now feel a great urge to write out a post immediately - please resist that urge and give these things a chance to settle in first. Please use the correct physics jargon I and others have been trying to teach you - the jargon exists specifically to make sense of this sort of confusion so it is not just a matter of being pedantic for the sake of it. Please also try to apply Newton's laws, as demonstrated above, to any further scenarios you come up with before you write about them here: this is so you will be better able to articulate your ideas.

Thank you and have fun ;)

 

[Dale]

what I mean by resistance to acceleration is inertia.

Inertia refers either to mass (usually) or momentum (occasionally), never weight. In the context of resistance to acceleration I think mass is a better choice than momentum. In either case, it is not weight.