If the definition of balanced forces is "two opposing forces that are equal" and an object with a net force of zero acting upon it means that the forces are balanced on the object, then it should follow that an object travelling through space at constant velocity should have a net force of zero acting upon it. But there was an initial force acting upon it when it was thrown into space. What is the force that is balancing that initial force to make the net force zero?
How does an object in space travelling at constant velocity have a net force of zero acting upon it?
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$\begingroup$ It is good to remember that F=ma is a law, an extra axiom, in order to pick up from mathematical equatiions describin motions those solutions that fit observations. see this answer of mine physics.stackexchange.com/questions/250035/… $\endgroup$– anna vCommented Oct 14, 2022 at 3:19
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2 Answers
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You are right. If we have an object at rest and then we want it to start moving, we apply a force to the object. While the force is being applied, the object accelerates according to $F=ma$.
Now let's say we stop applying this force. Then there is no longer a net force acting on the object. Therefore, the acceleration is $0$ and the object now moves at a constant velocity.
I'm not sure if I fully understand where you are having difficulty, but it seems to me that you are thinking of objects that can "remember" the "history of forces". So that if we apply a force and then take the force away, we still need to apply an opposite force to undo what the first force did to cause the acceleration to be $0$. This is not the case. Once the first force is gone, the acceleration is then $0$.
However, if we wanted to stop the object and bring it back to rest, then we would need to apply a force opposite to the first force to produce an acceleration in the opposite direction.
Side note, for this answer I am assuming we are in an inertial reference frame and just applying a force to an object. Any discussed motion or velocity is assumed to be viewed from this reference frame. I know I need to specify this or else I might get attacked by the relative motion police (of course, what constitutes an attack is also relative, so maybe let's just all agree to boost to a frame where we don't have to worry about all of this). :)
answered Nov 14, 2018 at 4:24
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$\begingroup$ Thanks. So you are saying that once the object is going at constant velocity, there are no forces acting on the object ( because objects don't need a force to move, but only to accelerate). Is that correct? $\endgroup$– suseCommented Nov 14, 2018 at 4:35
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1$\begingroup$ @suse That is right. No net force means no acceleration. Accelerations don't persist after a force is applied and then "turned off". You are also right about forces aren't needed for an object to move relative to you. If it is already moving it will keep moving. You do need a force to get an object to start moving relative to you if it starts at rest, but once it is moving you can take the force off and it will keep moving unless acted upon by a different force that causes it to stop. $\endgroup$ Commented Nov 14, 2018 at 4:45
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I think I understand your point, and it does make sense.
The way I see it is the following : the object you're throwing in space is firstly acted upon a force (let's call it F1). You're suggesting that since the object is traveling with a constant velocity, then the net force on that object must be 0, which is correct. However, where you might get confused is what kind of force F2 might have cancelled F1 to get a net force of 0, right?
Technically, no force F2 was applied. In fact, you threw the object in space with a force of F1 (the object currently accelerates from a velocity of 0 to velocity V). The object keeps on accelerating as long as F1 is applied.At some point, you're going to stop applying F1, and leave the object alone. The object already has a velocity, so it will keep on moving. Since F1 is no longer applied, no acceleration takes place (F = ma, if F = 0 then a = 0). So then your net force is 0, which explains the constant velocity and the 0 acceleration.
Hope this helps!
