Systems and What Counts as an Internal Force

Every force problem starts with a choice: what's inside the box? That choice is the system boundary, and it decides everything about which forces matter. If two blocks connected by a rope both sit inside your box, the rope pulls on each block with equal and opposite tension. Those two tensions are an internal pair — they act within the system, so they cancel when you add up the net external force. Only forces that cross the boundary from outside, like an applied push or friction from the ground, show up in the free-body diagram of the combined system.

Try a 4 kg block pulled by 30 N, trailing a 6 kg block behind it on frictionless ice, connected by a rope. Treat both blocks as one 10 kg system: the rope tension disappears from the diagram, and only the 30 N force, gravity, and the normal force remain. Net force equals 30 N, so acceleration is 3 m/s^2. Now shrink the boundary around just the 6 kg block. Suddenly the rope tension becomes external — it's the only horizontal force crossing into that smaller system, and it must supply exactly 18 N to give that block the same 3 m/s^2.

This is why system choice sits at the start of every free-body diagram you'll draw this unit. Pick the wrong boundary and you'll either double-count a force or lose track of one that actually acts on your object. Pick it deliberately, matching the boundary to what the question asks about, and Newton's second and third laws do the rest of the work for you.