Resistance to acceleration dependent upon mass or weight?

[inertiaforce]

Yes, it would. But why did you begin with, "However,"? What you just described is what we were wrestling with earlier. That it would indeed, be more difficult to accelerate identical mass in the 2g elevator scenario than in 1g. The subtlety is, *only* when you attempt to accelerate it upwards, and that's because of the 2g's.

So to recap, an object with a given mass will always accelerate the same when the direction of the acceleration is *perpendicular* to any existing force(s). It will however, accelerate differently when you apply the same force in a direction against or in concert with other forces that act on the mass.

Oh ok I see what you are saying. You are talking about identical masses. I thought you were referring to differing masses. My mistake. I agree with you. Identical masses would have different rates of upward acceleration in different gravity fields, but the same rates of horizontal acceleration in differing gravity fields. Good point. That is a good way of visualizing this as a thought experiment.

 

[voko]

Therefore, would the 2g gravity making the object harder to accelerate in the upward direction, be offset by half the mass making the object easier to accelerate, and the two cancel each other out so that the object accelerates the same as the 100 pound object in 1g gravity?

For bodies of equal weights, applying an equal upward force produces an equal resultant force (the difference between the applied force and the weight).

But, because the bodies have different masses, their resultant accelerations must be different. In fact, the resultant accelerations are inversely proportional to their masses.

 

[TumblingDice]

That is a good way of visualizing this as a thought experiment.

Here's a thought experiment I'll offer to address the original post, is it mass or weight that resists acceleration:

In our elevator we have a unique "hover-scale". It's a plate with a propulsion jet underneath, and a couple of digital readouts. With the elevator stationary, we put a mass on our plate and turn on the engine. The hover-scale applies just the amount of force to make the plate hover in front of you. We push a 'calibrate' button, the scale assumes 1g, and the readouts display "1g" and also "20lb". So our hover scale has established a 'weight' based on 1g, AND the scale and mass together do not experience any friction except for the air all around. We 'lock' the propulsion and you do acceleration tests in all directions by applying measured forces.

Now we begin accelerating the elevator upward. Our calibrated scale readouts indicate increasing weight and g-force. When the readouts display 2g and 40lb, you repeat the acceleration tests. Tests in all directions should match the results in the stationary elevator at 1g gravity and 20lb weight.

So, yes, you could argue we've made the mass weight-less, but the experiment has allowed us to separately measure 'weight' - we can see what it 'weighs' , and observe that the mass continues to have the same reaction to applied forces.

 

[TumblingDice]

For bodies of equal weights, applying an equal upward force produces an equal resultant force (the difference between the applied force and the weight).

But, because the bodies have different masses, their resultant accelerations must be different. In fact, the resultant accelerations are inversely proportional to their masses.

If you re-read the quote you replied to, what you wrote isn't quite clear. If you apply an equal upward force to a mass in 1g and the same force to half that mass in 2g (what inertiaforce wrote) the acceleration will be the same.

 

[voko]

If you re-read the quote you replied to, what you wrote isn't quite clear. If you apply an equal upward force to a mass in 1g and the same force to half that mass in 2g (what inertiaforce wrote) the acceleration will be the same.

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

 

[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.