1
$\begingroup$

Newton third law of motion says that "To every action, there is always an equal and opposite reaction". The vector study tells us that if two vectors are of same nature and equal magnitude but opposite direction, say $\vec{A}$ and $\vec{B}$, and they acts then the resultant is given by,

$\vec{A}$ $=$ $\vec{-B}$

$=$$>$ $\vec{A}$ $+$ $\vec{B}$ $=$ $0$

My question is, if I carry object upward then its reactive Force must be acting against my Force and according to Newton's third law of motion it should be equal to my force and according to above equation both must cancel each other then how I am able to do work on it by moving it upward?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is a common misconception. When you apply a force upward on the object, the "reaction" force in Newton's 3rd Law is NOT the force of gravity down on the object; they do not have to be equal, and as you said, cannot be equal if you are to accelerate the object upwards.

It is just a confusing coincidence that the force of gravity kind of looks like a reaction force, but it's not. One key giveaway that you have not identified a "action/reaction pair" is that both of the forces are on the same object - the force of your hand on the object, and the force of gravity on the object. Those can't possibly be a "3rd law pair".

The more accurate way of saying Newton's 3rd Law is this:

"If object A puts a force on object B, object B puts an equal and opposite force on object A, and the forces are the same type, and occur at the same time".

In your situation, those two forces are part of two separate action/reaction pairs;

  • the reaction force to you pushing up on the object is the object pushing down on you
  • the force of gravity on the object (from the Earth) is a reaction force to the force of gravity on the Earth from the object.

You can put an arbitrarily large amount of force on the object, and the force of gravity opposing you on the object will remain the same. On the other hand, the reaction force from the object pressing on you will become equally arbitrarily large. Luckily, you plus whatever you're standing on (probably the Earth) is way more massive than the object is, so the reaction force produces a negligible acceleration on you & the Earth, but you produce a large acceleration for the object (F = m a).

$\endgroup$
6
  • 1
    $\begingroup$ That's a pretty different question, since you removed the reference to gravity - modifying the question makes it harder for future readers to understand what's going on. In any case, the last paragraph of my response briefly addresses the new question. $\endgroup$
    – Brionius
    Commented Feb 22, 2015 at 18:53
  • 1
    $\begingroup$ Now you've deleted the comment I was responding to. $\endgroup$
    – Brionius
    Commented Feb 22, 2015 at 18:53
  • $\begingroup$ Sorry, I read your last paragraph after posting my comment and found some of my answer there thereby I removed my comment. $\endgroup$
    – user73555
    Commented Feb 22, 2015 at 18:55
  • $\begingroup$ That's fine - it's usually best to just leave the question and comments in place, unless there's some important reason to make changes. Changing questions and removing comments makes it hard for someone who wasn't watching in real time understand, and this site is supposed to be a resource for future readers too, not just the asker. So, not a big deal, but it's better not to do that. $\endgroup$
    – Brionius
    Commented Feb 22, 2015 at 18:58
  • $\begingroup$ Yes, I can understand. This site don't allow to pay thank therefore I can not pay it so to you otherwise I would do it as I needed the answer immediately and you gave. $\endgroup$
    – user73555
    Commented Feb 22, 2015 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.