accelleration going uphill at a steady speed (gravity)?

[DavidC]

Hi.

If you're driving up a slope (30 degrees) at a static speed (50 km/h), are you considered to be accelerating, due to fighting gravity?

Just curious.

[godBoy]

[ QUOTE ]
If you're driving up a slope (30 degrees) at a static speed (50 km/h), are you considered to be accelerating, due to fighting gravity?

[/ QUOTE ]
I wouldn't think so - acceleration is an increase in velocity.

[AdamL]

Just face it, it doesn't accelerate. Ever. /images/graemlins/grin.gif

[Silent A]

You're using the horsepower of your car to fight gravity.

power (watts) = force (N) * speed (m/s) [in the direction of the force you're applying]

Where, the force you must apply to go up hill = car_weight * sin(angle) [ignoring road friction and air resistence]

[DavidC]

[ QUOTE ]
You're using the horsepower of your car to fight gravity.

power (watts) = force (N) * speed (m/s) [in the direction of the force you're applying]

Where, the force you must apply to go up hill = car_weight * sin(angle) [ignoring road friction and air resistence]

[/ QUOTE ]

The thing that's confusing me is if you drop a rock off a building it'll accellerate til it hits the ground (not just drop at a steady speed), unless it somehow reaches terminal velocity... not sure how high you have to be to hit that speed, but lol, I guess it depends on your initial altitude due to air pressure being higher the lower you get.

Anyways, the point is that I view gravity (no education in this) as an "acceleratory force" (lol), so that's where I'm coming from when I ask this.

Another way of looking at it is, would you appear to accelerate if you were being observed from a stationary point in space?

[gull]

Net acceleration equals zero if velocity is constant, according to newtonian mechanics.

[ QUOTE ]
Another way of looking at it is, would you appear to accelerate if you were being observed from a stationary point in space?


[/ QUOTE ]

Stationary relative to what?

[DavidC]

[ QUOTE ]
Net acceleration equals zero if velocity is constant, according to newtonian mechanics.

[ QUOTE ]
Another way of looking at it is, would you appear to accelerate if you were being observed from a stationary point in space?


[/ QUOTE ]

Stationary relative to what?

[/ QUOTE ]

The Sun.

edit to say that the relative to what was a good question. But I guess the Sun is my answer there.

[MidGe]

[ QUOTE ]
[ QUOTE ]
Net acceleration equals zero if velocity is constant, according to newtonian mechanics.

[ QUOTE ]
Another way of looking at it is, would you appear to accelerate if you were being observed from a stationary point in space?


[/ QUOTE ]

Stationary relative to what?

[/ QUOTE ]

The Sun.

edit to say that the relative to what was a good question. But I guess the Sun is my answer there.

[/ QUOTE ]

ORLY

[DavidC]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Net acceleration equals zero if velocity is constant, according to newtonian mechanics.

[ QUOTE ]
Another way of looking at it is, would you appear to accelerate if you were being observed from a stationary point in space?


[/ QUOTE ]

Stationary relative to what?

[/ QUOTE ]

The Sun.

edit to say that the relative to what was a good question. But I guess the Sun is my answer there.

[/ QUOTE ]

ORLY

[/ QUOTE ]

lol piss off

[Matt R.]

[ QUOTE ]
at a static speed

[/ QUOTE ]

[ QUOTE ]
are you considered to be accelerating

[/ QUOTE ]

Nope, by definition you aren't accelerating. Your car engine may be supplying a force to "fight gravity", as you put it, but since speed is static the net force must be zero.

[Phil153]

Einstein asked himself this very question, but expanded it to the reference frame of a piece of light. He basically thought: If a car was accelerating up a hill but being pulled back by gravity, was it really accelerating? This (along with the Michelson-Morley experiment, showing the acceleration of light was an absolute) led him to the photoelectric laws of relativity.

So these simple questions can actually show some interesting flaws in our assumptions. The basic answer to yours is that a car going up a hill isn't accelerating relative to the sun, as others have pointed out.

[MusashiStyle]

[ QUOTE ]
Hi.

If you're driving up a slope (30 degrees) at a static speed (50 km/h), are you considered to be accelerating, due to fighting gravity?

Just curious.

[/ QUOTE ]

No. You are only accellerating if your velocity changes.

a = accelleration , v= velocity, t= time

a = dv/dt

in this case dv/dt = 0

thus a = accelleration = 0.

lol.

[Borodog]

One could imagine that there is a component of acceleration due to each force, and that the net acceleration is due to the net force. I don't know that it would be particularly useful, but you could do it.

[MusashiStyle]

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One could imagine that there is a component of acceleration due to each force, and that the net acceleration is due to the net force. I don't know that it would be particularly useful, but you could do it.

[/ QUOTE ]

I don't know if imagining an accelaration of zero would be all that exciting. I think what u mean is that articulating the forces (force diagram) would be interesting.

[Silent A]

[ QUOTE ]
The thing that's confusing me is if you drop a rock off a building it'll accellerate til it hits the ground (not just drop at a steady speed), unless it somehow reaches terminal velocity... not sure how high you have to be to hit that speed, but lol, I guess it depends on your initial altitude due to air pressure being higher the lower you get.

[/ QUOTE ]

As as object falls from a great height it speeds up as it accelerates due to gravity. However, since it's falling through air applies a resistence force. This force is proportional to the object's speed squared so the resistence increases rapidly as it speeds up. Eventually, the resistence will get so high that it's equal to the force of gravity. The net force on the object is now zero, so it falls at a constant speed.

Air pressure changing with altitude isn't really an important factor since you don't need to fall very far to reach terminal velocity, as far as the atmosphere is concerned anyway, it'll be far enough that you don't want to try it out yourself.

Whenever you have two equal forces acting on an object in opposite directions the total force on the object is zero and so it's acceleration is zero. Finally, just because the forces are in balance right now, doesn't mean they they've always been in balance (for example the object falling in air).

[teampursuit]

Actually, assuming the car is on the Earth, it is accelerating relative to the Sun. The Earth is rotating, so any point on its surface is always accelerating in a Sun-fixed frame. Similarly, the Earth is orbiting the Sun, so that motion is also accelerated.

[Borodog]

[ QUOTE ]
Einstein asked himself this very question, but expanded it to the reference frame of a piece of light. He basically thought: If a car was accelerating up a hill but being pulled back by gravity, was it really accelerating? This (along with the Michelson-Morley experiment, showing the acceleration of light was an absolute) led him to the photoelectric laws of relativity.

So these simple questions can actually show some interesting flaws in our assumptions. The basic answer to yours is that a car going up a hill isn't accelerating relative to the sun, as others have pointed out.

[/ QUOTE ]

You don't happen to publish in that little square in the classifieds at the back of Physics Today, do you? Because nothing you just said is remotely correct.

"Photoelectric laws of relativity"? A car going up a hill isn't accelerating relative to the sun?

Lol.

[Phil153]

[ QUOTE ]
You don't happen to publish in that little square in the classifieds at the back of Physics Today, do you? Because nothing you just said is remotely correct.

"Photoelectric laws of relativity"? A car going up a hill isn't accelerating relative to the sun?

Lol.

[/ QUOTE ]
It's called a joke, Mr. Physics assistant. Thanks for outing me.

I don't read physics today but there used to be some pretty good ones in the back of New Scientist.

[Silent A]

Boro, turn up the sensitivity on your sarcasm detector.

edit: too late.

[Borodog]

[ QUOTE ]
[ QUOTE ]
You don't happen to publish in that little square in the classifieds at the back of Physics Today, do you? Because nothing you just said is remotely correct.

"Photoelectric laws of relativity"? A car going up a hill isn't accelerating relative to the sun?

Lol.

[/ QUOTE ]
It's called a joke, Mr. Physics assistant. Thanks for outing me.

I don't read physics today but there used to be some pretty good ones in the back of New Scientist.

[/ QUOTE ]

Mea culpa. It was excellently dry.

[DavidC]

[ QUOTE ]
Einstein asked himself this very question, but expanded it to the reference frame of a piece of light. He basically thought: If a car was accelerating up a hill but being pulled back by gravity, was it really accelerating? This (along with the Michelson-Morley experiment, showing the acceleration of light was an absolute) led him to the photoelectric laws of relativity.

So these simple questions can actually show some interesting flaws in our assumptions. The basic answer to yours is that a car going up a hill isn't accelerating relative to the sun, as others have pointed out.

[/ QUOTE ]

Thanks dude.

My next question will be funnier, maybe. /images/graemlins/smile.gif

edit: been totally leveled

Thanks everyone else too

[AdamL]

fwiw, I think the actual answer is that yes, it is accelerating. Gravity = acceleration. You should ask Chris about this one.

[Silent A]

[ QUOTE ]
fwiw, I think the actual answer is that yes, it is accelerating. Gravity = acceleration. You should ask Chris about this one.

[/ QUOTE ]

No.

Gravity = force

force = mass * acceleration

but you only have acceleration if there's a net force.

Since gravity is never the only force acting on any object, you can never assume that just because gravity is acting on it there must be an acceleration.

[AdamL]

I'm not sure I follow, SA. The car is gaining energy going up the hill, it is not trading momentum for potential but adding it from nought.

So, if the energy in the system (the car) is going up as it climbs the hill, is this not a form of acceleration? Can you differentiate between acceleration and gravity?

I think what you are saying is that gravity is distinct as a thing than acceleration, am I right?

[Matt R.]

gravity != acceleration
gravity is a warping of spacetime

An easier way to see that it's NOT accelerating it to just look at the definition of acceleration
a = dv/dt, or change in velocity divided by the time interval (if it's constant a)
The OP specified that velocity was static, so change in velocity = zero.
Thus acceleration must be zero.

Thinking about forces and gravity in the context of the question is extraneous and irrelevant. You COULD conclude that the sum of all forces is zero, but op wasn't asking that. Since he said velocity = static, we know a = 0.

This is assuming we're in the same inertial frame as the car though. And not on the sun or something /images/graemlins/tongue.gif.

[AdamL]

But the sum of all forces in the car is not zero. He is adding energy to the system by keeping a steady speed up an incline. What is that addition of energy to fight gravity called if not acceleration?

[Matt R.]

Ya, it is zero from Newton's definition of force.

He's adding potential energy to the car via burning of fuel. It's doing work on the car (but it's simply increasing potential energy rather than kinetic due to the incline... energy conservation and all that). The "addition" of energy to the system (assuming fuel is not part of the system) increases the potential energy of the car in this case, but it doesn't increase its kinetic energy since OP specified velocity = constant.

[Silent A]

I agree with what Matt said.

P.S. don't confuse "acceleration due to gravity" with gravity itself.

[AdamL]

Velocity is relative. /images/graemlins/smile.gif What do you call the addition of energy into a system? E=MC2 right? So the mass of the object must be increased as well.

You have to understand that velocity is completely relative, and changes in velocity which are relative don't tell us a lot. But acceleration is measured in G-Forces, not velocity changes. That is the physics standpoint, it is a non-relative function. It's an absolute addition of energy into any system. As fas as I know this is acceleration "in reality" but is not the common understanding of change in velocity.

You are in a space ship with no windows - the classic example - and you jam the rockets. The way you measure acceleration is in G-force, which is a single unchanging value as long as rate of acceleration is constant...

Correct me if I'm wrong.

[Silent A]

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Velocity is relative. /images/graemlins/smile.gif

[/ QUOTE ]

Unfortunately for you, acceleration is not.

[Matt R.]

You're confusing me lol.

E=mc^2 is the rest energy of an object of mass m. If it's moving its energy is given by E^2 = (pc)^2 + (mc^2)^2. When an object has relativistic momentum p = gamma*m*v, its rest mass does not increase. The increase in momentum occurs through the gamma term. Some physicists combine the gamma and mass terms into a "relativistic mass", but it isn't exactly analogous because the same combination doesn't work when you try to do it for kinetic energy. I don't really like thinking of it in terms of increase in mass, but some others do.

Velocity is indeed relative, which means that depending on your frame of reference the measured velocity may change. However, it does NOT change your measured acceleration if you are in an inertial frame (this is why I specified that I wasn't measuring v from, say, the sun). Putting yourself off in space from the surface of the earth complicates things, and I don't think the OP was asking that. But, when doing that, you are no longer in an inertial frame. The earth (and the incline + mass) are now accelerating relative to you, and it becomes a central force problem I believe. Then, gravity is causing the car to accelerate, but it's a centripetal acceleration. Unless you're analyzing the rotational dynamics of the earth or something I don't think this is very useful in regards to the OP.

So, in regards to the OP, we should put us, the observer, in the same rest frame as the incline. In this case, there is a net force of zero acting on the mass. Acceleration can indeed be measured in G-forces if you know the objects mass, but what makes you think the G-force is non-zero? It must be zero due to the constraint velocity = constant. Yes, there is a force due to the gravitational acceleration, but it is offset by the force due to friction of the tires on the incline. Therefore, the g-forces on the car = zero. I think the problem is you aren't thinking in terms of net force... simply having a force on an object does not mean it is accelerating because you can have a force exactly equal in magnitude opposite in direction on the same object; i.e. a zero net force.

In your space ship example, yes, you are accelerating when you jam the rockets. The difference b/w that example and the car + incline is there is no other force opposing the force due to the rocket fuel. With the car, we have gravity, and that offsets the force due to the tires on incline. Also, just to point out, it's also possible the rocket in your example can have zero net force on it as well. Put it next to a black hole, trying to accelerate in the opposite direction. The "g-force" it feels will be in the OPPOSITE direction as force due to the thrust (or, zero, if it's placed at just the perfect distance).

[AdamL]

Physics confusing? Never. /images/graemlins/smile.gif

"Some physicists combine the gamma and mass terms into a relativistic mass"

This might be what I am thinking of. What I am observing is that there needs to be some accounting for the increase in total energy for the car. Without an engine momentum is transferred into the "bank" of energy at the top of the hill, so the trade is even. With an engine, velocity remains constant, and without worrying about the road friction for a moment, energy is going into the car. A car going 50kph at the top of the hill has more energy than the car going 50kph at the bottom of the hill. So what is that depositing of extra energy called in physics? It is acceleration, without respect to relative velocity, but measured absolutely by g-force.

So what do you call it if not acceleration? What is the term for an absolutely measurable injection of energy into a system?

"I don't really like thinking of it in terms of increase in mass, but some others do."

Is it really a matter of preference one way or the other? Relativity says the only acceleration is g-force, which is literally undistinguishable from gravity. Gravity as a warping of space-time = mass = energy. Acceleration is an increase in gravitational force. Car climbs hill. Dave steps on the gas to maintain 50kph, and the *g-force* goes up. When he crests the hill, he lets off the gass and the g-force returns to the "normal" amount of 1.0 here on earth. Which is simply the feeling of accelerating in any direction.

"Acceleration can indeed be measured in G-forces if you know the objects mass"

Mass need not be known, nor acceleration measured, for it to be measured in g-forces as a fact. We don't always know how much we weigh, but we always know that we don't measure our weight in terms of say, distance, or velocity. We measure it in pounds. We measure acceleration in g-force. It is the only way to measure acceleration without a reference system. It's pure gravitational force, there is no other way to talk about it to my knowledge. (?)

"Yes, there is a force due to the gravitational acceleration, but it is offset by the force due to friction of the tires on the incline."

Sure. It is nowhere near that significant to cancel out the g-force of putting the gas on up a hill.

":a force on an object does not mean it is accelerating because you can have a force exactly equal in magnitude opposite in direction on the same object; i.e. a zero net force"

Irrelevant - space ship example again ok? No windows. You burn fuel and you're pinned into your seat. The "energy" from the engines is getting blasted out into space behind your ship. Your "net force" in the locality of your ship is constant while the total actualized energy of your spaceship increases monstrously as the fuel is spent - ie., acceleration, an increase in total energy relative to nothing at all.

"The "g-force" it feels will be in the OPPOSITE direction as force due to the thrust (or, zero, if it's placed at just the perfect distance)."

That doesn't make sense to me either. It shouldn't matter where you are relative to the black hole. If you fire the rockets you're going to get acceleration as the fuel is realized into actual energy for the ship.

Interesting stuff.

[Silent A]

[ QUOTE ]
":a force on an object does not mean it is accelerating because you can have a force exactly equal in magnitude opposite in direction on the same object; i.e. a zero net force"

Irrelevant - space ship example again ok? No windows. You burn fuel and you're pinned into your seat. The "energy" from the engines is getting blasted out into space behind your ship. Your "net force" in the locality of your ship is constant while the total actualized energy of your spaceship increases monstrously as the fuel is spent - ie., acceleration, an increase in total energy relative to nothing at all.

[/ QUOTE ]

OMG, it's not irrelevant. It's totally, 100% relevant. It's bloody everything about the OP's scenario.

In your spaceship example there is no balancing friction or gravitational force, so the ship accelerates.

Also, acceleration is not "an increase in total energy". It's an increase in speed, and only speed - nothing else. Speed is an important aspect of energy, but is only one aspect of many. The two are not synonymous.

[AdamL]

SA,

The point about the space ship scenario is that a single system can accelerate without picking up speed relative to anything else. Acceleration in physics is measured in g-force, not looking out the window to see how much faster you pass the objects nearby. Objects nearby can go any direction they please and change your velocity. Your acceleration within the system (spaceship, car) is a function of the fuel being burned and that energy being added to the object.

[Max Raker]

No you are not accelerating. Just think of a person sitting in a chair. Gravity is acting on you with a force of mg going down but the chair is forcing you up with the exact same force (magnitude) but in the oposite direction. In the car example the engine is providing a force to cancel the component of gravity in the x and y dir.

[Matt R.]

[ QUOTE ]
So what is that depositing of extra energy called in physics? It is acceleration, without respect to relative velocity, but measured absolutely by g-force.

So what do you call it if not acceleration? What is the term for an absolutely measurable injection of energy into a system?


[/ QUOTE ]

It's called potential energy. I mentioned it earlier. It's the same energy added to a book when you pick it up off of the floor. When in a gravitational field, it's proportional to height, so when the car drives up the incline it's gaining potential energy.

A gain in energy does NOT automatically mean it is accelerating.

And Silent is right about the net force thing that you said was irrelevant. It's the most important point of the problem if you're looking at things from a force perspective... which you don't even need to find if the car is accelerating or not.

I also think you are misunderstanding g-forces. I could be wrong on this, because I typically don't work in those units, but I think a g-force is a multiple of gravitational force; i.e. 2 g's on an object of mass m is 2*m*g = 19.6*m. So, it's not a measure of acceleration, but a measure of force. Simply having a g-force acting on an object (ie rocket thrust) doesn't mean it is accelerating if there is an equal force acting the other way.

[ QUOTE ]
Sure. It is nowhere near that significant to cancel out the g-force of putting the gas on up a hill.


[/ QUOTE ]

Also, why do you think this? It depends on how hard you are pushing the pedal. Sure, if you mash the gas pedal down as you go up the incline you will be accelerating, but that's because the velocity is now increasing /images/graemlins/tongue.gif.

[Matt R.]

[ QUOTE ]
The point about the space ship scenario is that a single system can accelerate without picking up speed relative to anything else.

[/ QUOTE ]

Why do you keep saying this? Pick two reference frames. For instance, a frame attached to an astronaut outside the spaceship and one attached to a rocket. If the velocity of the rocket changes relative to the astronaut (it doesn't matter how much faster it is going initially relative to the astronaut), it is accelerating. Same thing if you pick the reference frames to be the ground and the car.

If you make everything else disappear, like removing the windows on a rocket ship or whatever so you don't have a reference, yon can tell if you are accelerating or not if a force is applied to a rocket ship (you can feel it or do a simple experiment). BUT, if an equal force pushes the rocket back, you will *not* be accelerating and you will not feel anything on the inside... assuming no vibrations etc. from the thrust. This is exactly what would be going on with the car + incline if velocity is static. Frictional force pushing car up the hill due to rotation of tires cancels out the gravitational force pushing it back.

I know you're trying to think of all this and justify it in terms of reference frames, but trust me, you're making it far too complicated. Just look at the definition of acceleration, delta v/(delta t), and you can immediately see that a car w/ static velocity is not accelerating. This is true in ANY inertial frame.

[daryn]

[ QUOTE ]

The point about the space ship scenario is that a single system can accelerate without picking up speed relative to anything else.

[/ QUOTE ]

not that this even matters but it does pick up speed relative to the rocket fuel.

[Copernicus]

[ QUOTE ]
[ QUOTE ]
Hi.

If you're driving up a slope (30 degrees) at a static speed (50 km/h), are you considered to be accelerating, due to fighting gravity?

Just curious.

[/ QUOTE ]

No. You are only accellerating if your velocity changes.

a = accelleration , v= velocity, t= time

a = dv/dt

in this case dv/dt = 0

thus a = accelleration = 0.

lol.

[/ QUOTE ]

And the corrolary is F=ma or a=F/m. Since a = 0, F = 0, ie as stated earlier the net forces are 0...the engine is just counteracting gravity.