Resistance to acceleration dependent upon mass or weight?

[TumblingDice]

You are mistaken. My statement follows trivially from Newton's second law.

Could you then please explain, what would the observable difference be? Using our simple 'units' like earlier, we apply a continuous force of three units in an upward direction to a mass 'm' in 1g gravity. This amount of force causes the object to accelerate upwards at 3 ft/sec^2.

Next we divide the mass in half and place one half (m/2) in 2g gravity. We now apply a continuous 'three unit' force in an upwards direction. At what rate will the object accelerate upwards?

 

[voko]

Both objects have equal weight $P$. We apply equal force $F$ to them. The resultant force is $F - P$. Evidently, it is the same for both objects. By Newton's second law, $m_{1g} a_{1g} = F - P$ and $m_{2g} a_{2g} = F - P$, where $m$ is the mass and $a$ is the acceleration, and the 1g and 2g subscripts denote the 1g and 2g environments. Obviously, $m_{1g} a_{1g} = m_{2g} a_{2g}$. But $m_{1g} = 2 m_{2g}$, so $a_{2g} = 2 a_{1g}$.

The resultant acceleration of the body in 2g gravity is twice the acceleration of the body in 1g gravity.

 

[TumblingDice]

The resultant acceleration of the body in 2g gravity is twice the acceleration of the body in 1g gravity.

I provided numbers expecting numbers back, to illustrate observable difference from fancy formulae. Are you saying that the m/2 mass in a 2g environment will accelerate upwards at 6 ft/sec^2, when the m mass in 1g accelerates upwards at 3 ft/sec^2? Equal 'weight' and equal force.

Thank you for your patience, voko!

 

[voko]

OK, I will assume that the 3 unit force applied horizontally to mass $m$ accelerates it to 3 feet per second per second. I will further assume that $m$ is the mass of the object in 1g gravity. For the sake of simplicity I will take $g$ equal to 32 feet per second per second.

The following holds for the body in 1g gravity (Newton's second law): $ma = F - mg$, so $a = \dfrac F m - g = 3 \ ^\text{ft} / _{\text{s}^2} - 32 \ ^\text{ft} / _{\text{s}^2} = -29 \ ^\text{ft} / _{\text{s}^2}$.

In the 2g case: $\dfrac m 2 a = F - \dfrac m 2 (2g)$, so $a = 2 \dfrac F m - 2 g = 2 \times 3 \ ^\text{ft} / _{\text{s}^2} - 2 \times 32 \ ^\text{ft} / _{\text{s}^2} = -58 \ ^\text{ft} / _{\text{s}^2}$.

The minus sign in both cases means that the acceleration is downward.

 

[inertiaforce]

The resultant acceleration of the body in 2g gravity is twice the acceleration of the body in 1g gravity.

are we talking downwards acceleration, upwards acceleration, or horizontal acceleration?

 

[Malverin]

I am aware of the difference between mass and weight. Let me try to clarify my question.

Would the mass, when it weighs 0 pounds in a free fall elevator, have the same resistance to acceleration as it did when it weighed 50 pounds in 1g, or would it have less resistance to acceleration?

Also, would the mass, when it weighs 100 pounds in an upward accelerating elevator with a total gravity of 2g, have the same resistance to acceleration as it did when it weighed 50 pounds in 1g, or would have it more resistance to acceleration?

Because theoretically speaking, the mass should have less resistance to acceleration when it loses its weight, and more resistance to acceleration when it doubles its weight. But I have also heard that inertia, an object's resistance to acceleration, is dependent upon its mass, not its weight. So I'm a little confused.

Acceleration of a mass when you push it against Gravity is

a = (F - m*g)/m

When you push it perpendicular to gravity force (you do no work against gravity), or there is no Gravity

a = F/m

 

[voko]

are we talking downwards acceleration, upwards acceleration, or horizontal acceleration?

We are talking about vertical acceleration in 1g and 2g gravity under the influence of the "3-unit" upward force. That results in downward acceleration in both cases.

With an upward force of a different magnitude, the resultant accelerations could be upward. The 2:1 relationship between their magnitudes would still hold.

 

[CWatters]

Would the mass, when it weighs 0 pounds in a free fall elevator, have the same resistance to acceleration as it did when it weighed 50 pounds in 1g, or would it have less resistance to acceleration?

The same because the mass hasn't changed.

Consider what happens when two objects of different mass hit the ground and are decelerated (negative acceleration). Both are weightless when they hit. The larger mass will cause a bigger impact (eg it's harder to accelerate).

 

[sophiecentaur]

Mentors - please don't penalize me for just being naive or not thinking as well as I should.

They will not penalise you. You are penalising yourself by trying to square Newton's Laws of motion with your intuition. Intuition is very often likely to lead you astray. Try to avoid ever using the word "weight' when you talk of moving something around and stick to the word 'Mass'. When you are considering the effect of gravity, you can then use the phrase 'Weight Force', which is just another Force, acting and is of equal consequence to any other force you may be introducing.
What results from this will be strictly logical and should lead you to the 'right' conclusions..

 

[Dale]

I don't know of the term "resistance to acceleration" having any specific standard meaning, but if I had to give it one then I would say it is $\Sigma F/a$ which, by Newton's second law is equal to m, not mg.

 

[inertiaforce]

The same because the mass hasn't changed.

Consider what happens when two objects of different mass hit the ground and are decelerated (negative acceleration). Both are weightless when they hit. The larger mass will cause a bigger impact (eg it's harder to accelerate).

Very interesting way of putting it. Easy to understand and visualize.

Therefore, if the mass weighed 100 pounds in 2g gravity, it would accelerate the same as it does when it weighs 0 pounds in 0g gravity?

 

[inertiaforce]

Guys, let me ask a new question here to better understand this:

F=ma can be rewritten as ma = F. The "F" in this ma = F equation is the force resisting the acceleration, correct?

F=mg can be rewritten as mg = F. The "F" in this mg = F equation is the weight, correct?

F=ma, and therefore a = F/m

F=mg, and therefore g = F/m.

Therefore, if a = F/m and g = F/m, then both a and g are equal to F/m.

If both a and g are equal to F/m, then a = g.

If a = g, then this means that the "F" in both the equations is equal. If the "F" in both the equations is equal, this means that according to these equations, the force resisting acceleration is equal to the weight.

 

[Bandersnatch]

Mass of a sphere can be written as m=ρV
Mass of a cube can be written as m=ρV
m=m, therefore a cube is a sphere.

Can you see what I'm getting at?

 

[sophiecentaur]

m=m, therefore a cube is a sphere.

Can you see what I'm getting at?

An elephant is a whale?

 

[TumblingDice]

@inertiaforce: Good to see you're looking at the F=ma equation now, too. I hope I can use voko's post #39 as a foundation to show the correct answer in a way that also appeals to intuition.

The key is recognizing that, when applying force to an object in a gravitational field, the force acts against gravity and also against the mass. So if a force is applied upward, it must first overcome the gravitational force before it can accelerate upward. If the force applied is great enough to overcome gravity, any force beyond that now accelerates the mass. This is where the acceleration according to mass begins to make complete sense intuitively.

Consider the original 100lb weight at 1g. Let's calculate how much force needs to be applied to accelerate that object upwards at 3ft/sec^2. We know 1g acceleration is 32ft/sec^2, and mass is 100.

a=32 (at 1g)
m=100

We need to apply a force to achieve acceleration of 35ft/sec^2 upwards to 'net' the 3 ft/sec^2 we're aiming for. That's 32ft/sec^2 to overcome gravity (F=3200), and 3ft/sec^2 more to accelerate upwards (F=300). Make sense? With mass at '100' and acceleration at '35', F=ma produces a total of F=3500.

So 3500 is the upward force required to 'net' an observable acceleration of 3 ft/sec^2 in 1g. Now let's look at applying that same force to half the mass at 2g. Our mass and acceleration now become 50 and 64 ft/sec^2, respectively. So our 'weight' will be measured at 100lb, but mass is actually 50. Let's look at what happens

a=64 (at 2g)
m=50

It will require 3200 units of force to overcome gravity. (50 x 64) The same as before - no surprise because they 'weigh' the same, right? So when we apply the same upwards force as before, 3500, the additional 300 units of force will now be acting on one-half of the mass as before. So 300 / 50 = 6, indicating our object will indeed accelerate upwards at 6ft/sec^2. Twice as rapidly as first example.

How does all of that work for you?

 

[Nugatory]

If a = g, then this means that the "F" in both the equations is equal. If the "F" in both the equations is equal, this means that according to these equations, the force resisting acceleration is equal to the weight.

Although both equations use the symbol $F$ they're using it for two different things. In $F=mg$, $F$ is the gravitational force exerted by the Earth on a mass at the surface of the earth. In $F=ma$ we're talking about the force, no matter what its source, needed to produce an acceleration $a$ on a mass anywhere in the universe. It's true that when $a=g$ both formulas will yield the same numerical result, but it doesn't follow that they're the same thing.

 

[Dale]

F=ma can be rewritten as ma = F.

This is one of my personal "pet peeves". Newtons 2nd law is not: F=ma. Newtons 2nd law is: ∑F=ma. The distinction is important for the reason you showed above. Can you see why?

Also, putting a term on one side or the other doesn't change what it is. Wherever you put it ∑F is the net force causing the acceleration. It is not resisting the acceleration; it is driving the acceleration wherever you put it.

 

[CWatters]

Very interesting way of putting it. Easy to understand and visualize.

Therefore, if the mass weighed 100 pounds in 2g gravity, it would accelerate the same as it does when it weighs 0 pounds in 0g gravity?

Yes, it accelerates like an object with a mass of 50lbs.

Consider a spacecraft (mass 6000kg) docking with the International Space Station (mass 450,000kg). Both are essentially weightless in orbit but they have to be careful to dock very slowly. If they hit each other at say 10-20mph the impact forces and any damage caused would be the same as if it happened on earth. eg similar to a 6000kg truck hitting a 450,000kg object 10-20mph.

An astronaut doing an EVA could hold a 6000kg spacecraft above his head because it's weightless. However if he wanted to give it a push (eg to push it away from the ISS) it would feel like he is trying to push/accelerate a 6000kg object. He might be able to move it slowly because there is no friction but he couldn't accelerate it quickly.

Imagine same 6000kg object mounted on a hovercraft (or similar zero friction device) on earth. Now it weighs 6000kg. If you tried to push it horizontally it would feel like you are pushing a 6000kg mass and would accelerate just as slowly as in space.

Same situation on a planet with 2g. Now it weighs 12,000kg but if you push it horizontally it still feels like you are pushing a 6000kg mass and accelerates just as slowly as on Earth or in space.

 

[Dale]

Hi inertiaforce, I think that this discussion is essentially a semantic discussion. The term "resistance to acceleration" is not a standard term, so you would need to define it.

With electrical resistance, which does have a standard definition, the equation is R=V/I, where V is the voltage driving the current and I is the resulting current.

So, by analogy, in defining "resistance to acceleration" I would define it as ∑F/a, where ∑F is the net force driving the acceleration and a is the resulting acceleration. Then, a simple rearrangement of Newton's 2nd law shows that ∑F/a=m, so I would call m the "resistance to acceleration".

 

[CWatters]

Good explanation Dale.

 

[inertiaforce]

Although both equations use the symbol $F$ they're using it for two different things. In $F=mg$, $F$ is the gravitational force exerted by the Earth on a mass at the surface of the earth. In $F=ma$ we're talking about the force, no matter what its source, needed to produce an acceleration $a$ on a mass anywhere in the universe. It's true that when $a=g$ both formulas will yield the same numerical result, but it doesn't follow that they're the same thing.

Yeah I thought about that myself today. And now I've noticed you've pointed out the same. I am reconsidering my previous post lol.

 

[inertiaforce]

Yes, it accelerates like an object with a mass of 50lbs.

Consider a spacecraft (mass 6000kg) docking with the International Space Station (mass 450,000kg). Both are essentially weightless in orbit but they have to be careful to dock very slowly. If they hit each other at say 10-20mph the impact forces and any damage caused would be the same as if it happened on earth. eg similar to a 6000kg truck hitting a 450,000kg object 10-20mph.

An astronaut doing an EVA could hold a 6000kg spacecraft above his head because it's weightless. However if he wanted to give it a push (eg to push it away from the ISS) it would feel like he is trying to push/accelerate a 6000kg object. He might be able to move it slowly because there is no friction but he couldn't accelerate it quickly.

Imagine same 6000kg object mounted on a hovercraft (or similar zero friction device) on earth. Now it weighs 6000kg. If you tried to push it horizontally it would feel like you are pushing a 6000kg mass and would accelerate just as slowly as in space.

Same situation on a planet with 2g. Now it weighs 12,000kg but if you push it horizontally it still feels like you are pushing a 6000kg mass and accelerates just as slowly as on Earth or in space.

Excellent answer. Well said about the horizontal acceleration.

But what about if you try to push half the mass vertically in 2g gravity? What happens then? Is it the same rate of acceleration as twice the mass in 1g gravity?

 

[inertiaforce]

Hi inertiaforce, I think that this discussion is essentially a semantic discussion. The term "resistance to acceleration" is not a standard term, so you would need to define it.

With electrical resistance, which does have a standard definition, the equation is R=V/I, where V is the voltage driving the current and I is the resulting current.

So, by analogy, in defining "resistance to acceleration" I would define it as ∑F/a, where ∑F is the net force driving the acceleration and a is the resulting acceleration. Then, a simple rearrangement of Newton's 2nd law shows that ∑F/a=m, so I would call m the "resistance to acceleration".

what I mean by resistance to acceleration is inertia.

 

[inertiaforce]

@inertiaforce: Good to see you're looking at the F=ma equation now, too. I hope I can use voko's post #39 as a foundation to show the correct answer in a way that also appeals to intuition.

The key is recognizing that, when applying force to an object in a gravitational field, the force acts against gravity and also against the mass. So if a force is applied upward, it must first overcome the gravitational force before it can accelerate upward. If the force applied is great enough to overcome gravity, any force beyond that now accelerates the mass. This is where the acceleration according to mass begins to make complete sense intuitively.

Consider the original 100lb weight at 1g. Let's calculate how much force needs to be applied to accelerate that object upwards at 3ft/sec^2. We know 1g acceleration is 32ft/sec^2, and mass is 100.

a=32 (at 1g)
m=100

We need to apply a force to achieve acceleration of 35ft/sec^2 upwards to 'net' the 3 ft/sec^2 we're aiming for. That's 32ft/sec^2 to overcome gravity (F=3200), and 3ft/sec^2 more to accelerate upwards (F=300). Make sense? With mass at '100' and acceleration at '35', F=ma produces a total of F=3500.

So 3500 is the upward force required to 'net' an observable acceleration of 3 ft/sec^2 in 1g. Now let's look at applying that same force to half the mass at 2g. Our mass and acceleration now become 50 and 64 ft/sec^2, respectively. So our 'weight' will be measured at 100lb, but mass is actually 50. Let's look at what happens

a=64 (at 2g)
m=50

It will require 3200 units of force to overcome gravity. (50 x 64) The same as before - no surprise because they 'weigh' the same, right? So when we apply the same upwards force as before, 3500, the additional 300 units of force will now be acting on one-half of the mass as before. So 300 / 50 = 6, indicating our object will indeed accelerate upwards at 6ft/sec^2. Twice as rapidly as first example.

How does all of that work for you?

This is interesting. So it appears then that the rate of acceleration of an object remains the same, even in the vertical direction, regardless of the gravity force acting on it. The rate of acceleration of the object then, is based on its mass only, not its weight. It's weight results from its mass. A "heavier" object is only harder to move here on Earth because it has more mass, not because it has more weight. The reason it has more weight is BECAUSE it has more mass. Therefore, it appears that mass is the determining factor here, not weight. Interesting analysis tumblingdice.

 

[TumblingDice]

This is interesting. So it appears then that the rate of acceleration of an object remains the same, even in the vertical direction, regardless of the gravity force acting on it. The rate of acceleration of the object then, is based on its mass only, not its weight. It's weight results from its mass. A "heavier" object is only harder to move here on Earth because it has more mass, not because it has more weight. The reason it has more weight is BECAUSE it has more mass. Therefore, it appears that mass is the determining factor here, not weight.

I've been learning on this thread, and perhaps not done learning still. Thank you for making the OP!

There have been at multiple scenarios discussed on this thread. One with the 'same' mass and another with a different mass, and each of them with different gravity, elevators, etc...

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.

 

[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.