If you were on a platform in space that accelerated at 9.8m/s^2, this would obviously have the same effect as would earth's gravity. The acceleration vector of this plate would point upward, but why is the gravitation acceleration vector on earth always depicted downwards? obviously the gravitational force is downward, but if i accelerated an object to the left, i would feel a force opposite to the direction of acceleration. Maybe i am confusing acting and reacting forces, if there really is any distinction.
On earth, the gravitational acceleration vector points downward because the acceleration due to gravity is downward. On your accelerating platform, the effective acceleration vector would also point downward (assuming the platform is accelerating upward)--it's the apparent acceleration of a "falling" object with respect to the platform that counts. As far as what force you feel if you push an object to the left, you are feeling the contact force of the object on you, which points to the right. (Yes, you are confusing "action/reaction" pairs.)
But wouldnt the actual direction of acceleration be upward? if im in an accelerating train and let go of an object, it will move towards the back of the train, yet the acceleration of the train is forward. so the force acting on the object, and acceleration vector are in opposite directions.
The train is accelerating, so there's a force on it. As long as you hold the object--exerting a force on it--the object accelerates as well. Let the object go, and it no longer accelerates: there's no more force on it. So, yes, the object appears to accelerate (backwards!) with respect to the train, but there's no horizontal force acting on it so it's not really accelerating. If you were to view this scene from the train platform, it would be obvious that as soon as you released the object it maintained whatever horizontal speed it had at the moment you released it. The train, however, continued to accelerate.
Thanks, but due to me possibly not being able to completely articulate what my problem is, that didnt really help. It is a fully thorough answer to what i asked, and i fully understood that explanation already. However, i think i can use this to narrow in on what my real problem is. There is no force acting on the object once it leaves my hand, just like there is no force acting on an object in free-fall. but on earth, the direction of the motion of the object (with respect to me) is the same as g, however, the g of the train is opposite to the motion of the object. What is different between these 2 scenarios?
The only force acting on the ball is my hand before i let go of it, same as if i was holding it on the earth. the direction of the force i have to apply to the ball on the train is the same as the train's a (forward), but the direction of the force i have to apply to the ball on the earth is opposite to the direction of g (upward).
There most certainly is a force acting on the object: gravity! In both cases the accelerated object accelerates in the direction of the force: A dropped object accelerates down; A train accelerates forward. The difference is that in one case your frame of reference is an inertial one (the unaccelerating earth) which does not accelerate but the dropped object does; in the other case your frame of reference is accelerating (the train), but the dropped object does not accelerate horizontally (with respect to an inertial frame). Of course, as seen from the accelerating train, the object appears to accelerate backwards.
Maybe another perspective: back on the accelerating train, but consider gravity as well. i am holding a ball in my hand. on a free body diagram of the ball, we draw the forces acting on the ball. the force due to gravity till point down, and the force from my hand will point up. the force due to the train's acceleration will point backward, and the force from my hand will point forward. the directions of the forces are inconsistent with the directions of the accelerations. maybe ive just re-stated my previous post, but maybe it will help to understand what exactly im asking. Edit: I think i just need to play around with this in my head a while longer and mess with frames of reference.
The gravitational acceleration of an object points in the direction that the object is accelerating. You confusing the acceleration of the platform with the acceleration of the object. In an accelerating frame there is also a force acting on it. This kind of force is called an inertial force. In Enstein's general theory of relativity inertial forces are of the exact same nature as the gravitational force. For this reason Einstein stated that one can "produce" a gravitational field by a change in the coordinate system (from an inertial frame to an accelerating frame). Pete
In general relativity, there is no "force of gravity", but just "locally accelerating reference frames" ; that is: frames in which free-falling objects seem to accelerate (have non-zero second derivatives of their coordinates wrt the time variable of the frame). Consider the frame fixed to the surface of the earth as "accelerating upwards". In this frame, free falling particles seem to "accelerate downwards". If your space platform is accelerating upwards (using its engines), free objects inside seem to accelerate downwards wrt to a coordinate frame fixed to the platform. In a train, accelerating forwards (by throttling its engines), free objects seem to accelerate backwards in it. On the surface of the earth (which is "gravitationally accelerating upwards"), free objects seem to accelerate downwards in a frame fixed to the surface of the earth (but not in a free-falling elevator frame, for instance). BTW, I know that saying that "the surface of the earth is accelerating upwards" sounds bizarre. It is just applying the equivalence principle, though.
That is incorrect. This question was asked in the context of Newtonian physics since this is not the special/general theory of relativity. Within Newtonian physics there is a force of gravity. Within Einstein's GR there is also a force of gravity. The force of gravity in GR is a frame dependant quantity unlike the Lorentz force. There are two classes of forces (1) inertial forces (2) non-inertial forces. Both of which Einstein held to be "real" forces. People who work in Newton physics like to refer to inertial forces to "pseudo-forces." But this is not Einstein's view. Pete
The answer is simpler in the context of general relativity, that's why I referred to it. When invoking the equivalence between accelerated observers and gravity, it is probably easier to explain things from the PoV of a theory that takes the equivalence principle as a basis, than with forces and non-inertial frames. Clearly, the OP invokes the equivalence principle, and it is rather strange that he invokes it (how did he get to the equivalence principle in Newtonian physics in the first place) and then wonders what way the inertial force points. Because in Newtonian physics, *a priori* there doesn't need to be equivalence between an upward accelerating platform in space and the force of gravity on earth: it is only after working out the "pseudo force" (with the right sign of course) in the non-inertial platform frame, and compare it with the Newtonian force of gravity on the earth's surface, that one notices that any property of the "falling" object itself (such as its mass), drops out, and that this is hence a property intrinsic to the point in the coordinate system, and not to the "falling object". But in order to see this in the first place, one needs already to have worked out the pseudoforce in the non-inertial frame, with the right sign: hence one cannot wonder about its sign afterwards ! So clearly, the OP took the equivalence principle as a starting point: that, A PRIORI, a frame with constant acceleration g upward will be equivalent with a frame at the surface of the earth (neglecting tidal effects), and then wondered in what direction a force should be applied in order to take into account both equivalent phenomena (falling down on the platform, or falling down on earth). On earth, the OP seemed to understand: it comes from an "attractive force of gravity", but on the platform, he thought that, the platform accelerating upward, the pseudoforce should be upward too, and hence WONDERED about how the equivalence came about. It was therefor, in my opinion, easier to "get rid of the Newtonian force of gravity", and just say that the surface of the earth can be considered "accelerating upward" just as the platform. "Accelerating upwards" is just a property of the metric expressed in the coordinate system "at rest" (fixed to the platform, or fixed to the earth surface), so that the corresponding geodesics "bend down". You can hold this view, but the concept of "force" is a bit silly in GR. Force is "interaction", and the property of freely moving test bodies is that they don't undergo any interaction: they follow geodesics. Now, if you happen to use a coordinate system in which these geodesics are CURVED, then you will say that, wrt your coordinate system, the freely moving test body undergoes accelerations, but it is a far cry to call that forces. This is like saying that a straight line is curved when looking at it in polar coordinates. The "inertial force" is the ABSENSE OF A FORCE WHICH WOULD BE NEEDED IN THE OPPOSITE SENSE TO HOLD THE FREELY MOVING TEST BODY ON A STAIGHT COORDINATE LINE. Just as the centrifugal force in a rotating frame is the opposite of a force needed to keep a body on a constant coordinate line (in this case, constant radius). And just as gravity, at the surface of the earth, is the opposite of the force needed to KEEP A BODY AT REST (or in "uniform motion") WITH RESPECT TO THE SURFACE OF THE EARTH. If it were not for the surface of the earth, we wouldn't call it a force. It's only a "force" when you want to keep something from doing its natural motion, which is falling down. So I would rather say that Einstein didn't instore pseudo-forces as real forces, but realised that gravity is a pseudoforce which appears when we insist on working in a Euclidean space.