Half of every lost mark in mechanics is a sign error, and every sign error is a convention chosen carelessly — or never chosen at all. This primer is the chapter's compass: how to pick axes, how to build a free-body diagram that cannot lie to you, which way work and g point, and when a pseudo-force is honest. Set these once, then trust them.
§01
Choose axes along the motion
Axes are yours to choose — so choose the pair that makes the acceleration live along one of them. On an incline, tilt the axes with the slope: gravity splits into mg sin θ along it and mg cos θ into it, and the messy geometry collapses into two clean lines.
Rule 1 — align with acceleration. Pick +x along the expected motion (or acceleration), +y perpendicular. Then ΣFy = 0 handles the geometry and ΣFx = ma does the physics.
Rule 2 — any choice works, once. "Up the slope positive" and "down the slope positive" both give the right answer — if you keep the choice for the whole problem. Signs flip when you flip mid-solution.
Rule 3 — connected bodies share a sense. For a rope over a pulley, march one positive direction along the rope through the system: down on the heavy side = positive = up on the light side. One consistent thread, no contradictions.
§02
The free-body recipe
A free-body diagram cannot lie to you if you build it the same four steps every time. The whole art is step one: draw only the body, stripped of everything it touches — then let each removed contact leave its force behind.
STEP 1
Isolate the body
Redraw it alone — no incline, no rope, no wall. Everything it touched will reappear as a force in step 2. If two blocks move together, isolating them as one system hides the internal forces you don't need.
STEP 2
Mark every force — and no more
One long-range force (weight mg, always down) plus one force per contact: normal pushes ⊥ from the surface, friction along it, tension pulls along the rope, spring pushes or pulls along its axis. "Force of motion" does not exist — never draw one.
STEP 3
Resolve along your axes
Split each force into components along the axes you chose in §01. On an incline only gravity needs splitting — that is exactly why you tilted the axes.
STEP 4
Write ΣF = ma per axis
One equation per axis, signs read straight off the diagram. If the body doesn't accelerate along an axis, that equation reads ΣF = 0 — usually the one that hands you N or T.
§03
The sign of work — and of g
Work carries the angle between force and displacement: W = Fd cos θ. Pull the handle below through the angles and watch the sign flip — then keep the g-convention card beside it in mind for every projectile you ever throw.
Bench · the angle decides the sign
angle θ{{ wkTheta }}°
W = Fd cos θ
{{ wkVerdict }}
{{ wkNote }}
g has no fixed sign — you give it one. Throw a ball up with "up = +": then a = −g = −9.8 m/s² the whole flight, even at the top. Drop a stone with "down = +": a = +g. Choose per problem, never per half-of-problem.
Friction's work is negative — usually. Kinetic friction opposes sliding, so it drains energy (W < 0) on a sliding block. But static friction on an accelerating car's tyre points forward and does positive work. Read the arrows, not the folklore.
Normal force and tension do no work when they stay perpendicular to the motion — the circular-motion string, the incline's push. θ = 90° is the silent case of the dial.
§04
Pseudo-forces — honest only in an accelerating frame
Solve from the ground and there is no such force. Step inside the accelerating lift or car, and to keep ΣF = ma working you must add one fictitious force: magnitude ma, direction opposite the frame's acceleration. The weighing machine in the lift reads the normal force — pick a motion and read it.
Bench · a 60 kg student on a lift's weighing machine · g = 10 m/s²
scale reads · N = m(g + a)
{{ lift.N }} N · {{ lift.kgf }} kgf
pseudo-force in the lift frame
{{ lift.pseudo }}
{{ lift.note }}
Use it or don't — never both. Either solve from the ground (no pseudo-force, the frame's acceleration appears in ma) or from inside the frame (add −ma, then treat the body as if in equilibrium when it's at rest there). Mixing the two double-counts the acceleration.
Centrifugal force is the same deal. In the rotating frame of a turntable, a stationary rider needs an outward pseudo-force mω²r to balance the books. From the ground there is only the inward centripetal pull. Both ledgers close; pick one.
✎
The pocket card
six lines that close most sign leaks
◦ Axes along the acceleration; keep them for the whole problem.
◦ FBD: one body, mg + one force per contact, nothing else.
◦ W = Fd cos θ — the angle between force and displacement.
◦ g is a magnitude; your axis choice gives it its sign.
◦ Rope through a system: one positive sense, end to end.
◦ Pseudo-force −ma only inside the accelerating frame, never from the ground.