Work as Dot Product

Calculus 2 · Axiom Academy

One relationship — W = F · d = |F| |d| cos θ — decides how much of a push actually moves something. Only the part along the motion counts. Put it in your hands. You're dragging a loaded sled 8 m across flat ground with a rope. Pull 50 N hard — but the rope's angle decides how much of that pull does real work. Only the part along the motion does work Drag the rope angle and watch the force split. The red arrow is your full pull; its shadow on the ground is the part that actually moves the sled. That shadow — the parallel component — is what feeds W = |F| |d| cos θ. Flip it: what force does the job take? Now fix the goal — 300 J of work over the same 8 m — and ask the rope angle to tell you the force. Same relationship, rearranged: F = W ÷ (d cos θ). The part of your pull along the ground — its projection onto d — is locked at 37.5 N (that's 300 J ÷ 8 m). The steeper you pull, the longer the whole arrow has to be to keep that same shadow. How much of your pull is wasted? Here's the real worker's question: of your 50 N, how much actually drags the sled, and how much just lifts it? Drag the angle and watch your effort split — the useful part (F cos θ) versus the wasted part (F sin θ).

This is the written version of the interactive lesson above. See the full Calculus 2 course.