I am wondering about a question regarding Newton's Second Law of Motion.

For an object to have a constant velocity, it means the total net force is 0 since there is no acceleration. Does that mean there is no net force at all? From what I think, there is still a net force but it's just 0 because of the forces in all directions cancel out, it doesn't mean there is "no" net force. Thus, I am wondering would it be responsible to say there is a net force, but it is just 0.

  • $\begingroup$ The key is that the word "net" means "resultant", like "after doing all the math and cancel out all possible terms, after working on it until it cannot be simpler". $\endgroup$
    – FGSUZ
    Feb 21, 2019 at 22:50

1 Answer 1


You are just getting caught up in semantics. Saying there is no net force is the same thing as saying that the net force is $0$. Think about your bank account. If you have $0 in it, could you not also say you have no money in it?

With that being said, it's pretty standard to say no net force when the net force is $0$. People will most likely get annoyed with you if you hit them with "well technically there is a net force, but it is $0$." This is similar to if you went up to someone who has no money in their bank account and said to them "Hey don't worry. You technically do have money in your bank account. It's $0 worth of money."

Another issue you can see with what you propose is that the term "net force" becomes completely pointless by itself because you can say there is always a net force acting on an object. This then means you always have to specify "zero net force" and "non-zero net force". Now this technically isn't wrong, but I would say it's pointless. For a point particle, there is no way to distinguish between no forces acting on it at all and the forces acting on it cancelling out, so it makes sense to described these scenarios in the same way by just saying "no net force".

Using your reasoning you could also say that there are always forces acting on something, even if there isn't a force. For an extreme example, as I type this now I could say you are punching me in the head with a force of $0\ \rm N$. I guess you could argue that this is true, but I'm not sure why you would want to.

So, in summary, what you are proposing isn't technically wrong (if others agree on the use of your words), but it is unintuitive and not the common usage of the term "net force".

  • $\begingroup$ Thanks for the reply. Yes, I understand what you mean. It is just there was a question on a multiple choice test and it has two choices. The question is: for an object to have a constant velocity; a) there is a net force acting on it. b) the forces in all direction are equal. Technically speaking, i suppose there are real differences based on your explanation. However, I just think there is a net force but it's just zer so the answer would be A. My teacher would probably think the other way, that's why I am trying to get a better grasp on this concept before i argue. $\endgroup$ Feb 21, 2019 at 8:01
  • $\begingroup$ When we say "a force is acting on an object", we almost always mean a non-zero force is acting, even if not stated explictly. Similarly, your a) should probably be read as a non-zero net force. $\endgroup$
    – BowlOfRed
    Feb 21, 2019 at 8:12
  • $\begingroup$ Is it possible that something got lost in translation for b)? I think it could mean "forces in opposite directions are equal". This would be a correct answer, as opposed to a) and current b) $\endgroup$
    – Jasper
    Feb 21, 2019 at 9:24
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    $\begingroup$ @Andrew Pang “I am trying to get a better grasp on this concept before i argue” I strongly discourage you from doing that. There is no way you should argue this point. It will only make you seem antagonistic and litigious. $\endgroup$
    – Dale
    Feb 21, 2019 at 11:56
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    $\begingroup$ @AndrewPang That question doesn't really make sense as worded. For a), if there were a net force acting on it, that force would cause a change in velocity. The b) one is worded really poorly; because "the forces in all directions are equal" is very unclear. They probably mean opposite directions; but to me, saying "all directions" implies something like a constant pressure acting on all surfaces of the object. The wording on the choices doesn't make much sense as you have it worded there. $\endgroup$
    – JMac
    Feb 21, 2019 at 12:25

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