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For every action, there is an equal and opposite reaction.

Look at this free body diagram from an exercise where a man pushes a box up an inclined platform:

enter image description here

I don't see all those "equal and opposite" reactions here. If the man pushes on the box, the box pushes on the man as well, and if the platform pushes on the box, the box also pushes on the platform, so why aren't those forces drawn here? The box also pushes on the air (P) and the air on the box but if those forces were equal wouldn't the box not move? I mean, if A pushes on B with force F and B pushes back on A with force -F, how come anything moves? Is it because the have different masses? Would that make this "opposite and equal reaction" not equal then? I'm confused.

Thomas Fritsch
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Segmentation fault
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  • Note that for free body diagrams, you only draw the forces acting on that body. That is why they are free body diagrams. You technically aren't supposed to draw anything else but the body, and you certainly don't include forces that act on other bodies that the diagram is not focused on. – BioPhysicist Mar 15 '20 at 15:20

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The key point is that

  • Newton's 2nd law applies on one object (or system).
  • Newton's 3rd law applies "between" two objects (or systems).

The action/reaction force pair do not cancel out because they don't act on the same object (or system).

You are right that the box exerts a force on the ramp $F_\text{box on ramp}$ and that the ramp reacts with an equal-but-opposite force on the box $F_\text{ramp on box}$:

$$F_\text{box on ramp}=-F_\text{ramp on box}$$

Both only one of them acts on the box. Only $F_\text{ramp on box}$ acts on the box. When you set up Newton's 2nd law, you only include forces that act on the box, because only those forces will have any influence on the acceleration of the box. All other forces - including forces that the box itself causes on something else - are irrelevant for the motion of the box.

Steeven
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  • But if the formula is $F = ma$ isn't the force of the box on the platform different from the force of the platform on the box? Or is that what you were saying about this formula only applying on one object? – Segmentation fault Mar 15 '20 at 14:54
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    @SilenceOnTheWire The formula isn't $F=ma$. It is $\sum F=ma$. Don't forget the sum-symbol (no single force causes acceleration; it is the sum of all forces on the object that does). – Steeven Mar 15 '20 at 15:06
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    @SilenceOnTheWire Sure, the force that the box exerts on the ramp is the same as the force the ramp exerts on the box (which is the normal force). This is what Newton's 3rd law says. But it is only the ramp's force on the box (the normal force) which should be included in Newton's 2nd law $\sum F=ma$, if $a$ here is the acceleration of the box. Whatever the box does to the ramp should not be included here. – Steeven Mar 15 '20 at 15:07