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I've been thinking about Newton's third law lately because I couldn't understand a few things and I think I actually answered my own question. Could someone confirm if my reasoning is right or show me my mistakes?

So the question is: If force and reaction force are always the same, and less mass means greater acceleration, how can I actually push heavier objects? Then I got my idea of an answer that I will try to show with a picture:

  1. I exert $F_{\text{action}}$ force on a Box (much heavier than me) and an equal $F_{\text{reaction}}$ is exerted on me in the opposite direction, but $F_{\text{action}}$ is caused by the $F_{\text{muscle}}$ force which is exerted on me and has the same direction as $F_{\text{action}}$

  2. The box changes position but I don't change position myself because $F_{\text{muscle}}$ still acts and prevents it.

  3. Step 1 is repeated and that way I can push much heavier objects.

(I am sorry that box looks little different in every step. It was hand-drawn - the box in steps 2 and 3 is meant to be in the same position)

enter image description here

So can someone confirm if that reasoning is right or wrong? Also do the $F_{\text{muscle}}$ and $F_{\text{action}}$ force have the same value in the 1st step?

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No, that's not right, for a couple of reasons.

Firstly, your Fmuscle is a force that you are exerting on other objects. That force doesn't count with regard to what happens to you. Think of it this way: Can you lift yourself by your own bootstraps? (That's a dated term nowadays. Can you lift yourself off the ground by pulling on your shoelaces?) The answer is no. That isn't an external force, and neither is your Fmuscle.

Secondly, you haven't drawn all of the forces you are exerting on other objects and hence that other objects are exerting on you. Always find all of the forces when you are drawing a free body diagram!

You won't be able to push the box if you are standing on ice and the box is not. Your feet will slip out from under you. This is the external force you are missing. You are using your leg muscles to push against the ground via your feet. This force has both downward and backward components. The reaction to this is that the ground pushes you up and forward. The upward force counters gravity (another force that you missed). It's the forward force that is crucial in enabling you to push the box.

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  • $\begingroup$ Put another way: if you are not moving there must be zero net force on you. In this case friction with the ground is the force missing in your diagram. Coefficient of friction between you and the floor must be greater than between the object and the floor (if you are pushing horizontally) or you are pushing up against the object (lowering the normal force between the object and the ground) $\endgroup$
    – Floris
    Commented Aug 13, 2014 at 1:49
  • $\begingroup$ So when I push someone in a way that he will move few meters and I won't move at all is it because of my legs muscles pressing the ground? $\endgroup$
    – lijki
    Commented Aug 13, 2014 at 16:31
  • $\begingroup$ @lijki - Yes. If you aren't moving the net force on you is zero. Note very well: The individual forces acting you are not Newton third law pairs. Newton's third law applies to the same force acting equally but oppositely on two different bodies. When a single body has external forces that happen to be equal but opposite, those are two distinct external forces. (continued) $\endgroup$
    – David Hammen
    Commented Aug 13, 2014 at 17:03
  • $\begingroup$ Consider a book at rest on a table. The forces on the book are gravity pulling it downward and the normal force pushing it upward. Even though these are equal but opposite, these are not third law force pairs. The third law counter parts to these forces are the book pulling the Earth upward gravitationally and the book pushing the table downward with the normal force. (Aside: Those forces on the book actually aren't equal but opposite. The Earth is rotating. There has to be a net residual force on the book to keep it rotating with the Earth.) $\endgroup$
    – David Hammen
    Commented Aug 13, 2014 at 17:04

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