What belongs on a free-body diagram
A free-body diagram shows one object, drawn as a dot or small box, with an arrow for every force acting on it and nothing else. Arrows start at the object, point in the direction each force acts, and their relative lengths should reflect relative magnitudes - with no vertical acceleration, the up and down arrows are drawn the same length.
Only real forces belong, and each has an identifiable source: gravity from the Earth, the normal force from a surface, tension from a rope, friction, an applied push or pull, a spring, or drag from a fluid. If you cannot name what exerts it, it does not go on the diagram.
Three intruders cost students the most: anything the object exerts on something else, since the two halves of a third-law pair act on different objects; the quantity ma, which is the result of the forces rather than one of them; and a separate 'centripetal force', which double-counts the real forces already pointing toward the center.
- One object, one dot, one arrow per force acting on it.
- Every force needs a namable source: Earth, surface, rope, spring, fluid, or hand.
- Never draw both halves of a third-law pair on the same diagram.
- No ma arrow, no separate centripetal force, no velocity arrow.
A repeatable procedure
Run the same sequence every time so it survives exam pressure. Choose the object first - for multi-object problems, decide deliberately whether to treat the objects separately or as one system, since that determines which forces are internal. Then draw and label every force on it: F_g, F_N, F_T, F_f.
Next pick axes and say which direction is positive. On an incline, tilt the axes so one lies along the surface; that single choice is what makes incline problems tractable, because the acceleration then lies entirely along one axis. Decompose any off-axis force into components, then write Newton's second law separately for each axis: the sum of the forces along an axis equals m times the acceleration along it, and is zero when there is no acceleration along it. Only then substitute numbers.
- Choose the object or system first; internal forces vanish for a combined system.
- Tilt the axes on an incline so acceleration lies along one axis.
- Write the second law per axis, symbolically, before plugging in numbers.
Worked example: a block on an incline
A 2.0 kg block sits on a ramp inclined at 30 degrees. Two forces act: gravity straight down, mg = (2.0)(9.8) = 19.6 N, and the normal force perpendicular to the surface. Tilt the axes so positive x points down the slope. Only gravity needs decomposing, into mg sin(theta) along the slope and mg cos(theta) into the surface.
Along the slope, mg sin(30 degrees) = (19.6)(0.50) = 9.8 N. Perpendicular to it, mg cos(30 degrees) = (19.6)(0.866) = 17.0 N. There is no acceleration perpendicular to the surface, so the normal force balances that component exactly: F_N = 17.0 N. Notice F_N is not mg here - an error that propagates straight into any friction calculation.
If the ramp is frictionless, the net force is 9.8 N along the slope, so a = 9.8 / 2.0 = 4.9 meters per second squared - which equals g sin(theta), independent of mass. Now add kinetic friction with a coefficient of 0.20 as the block slides down. Friction opposes the sliding, acting up the slope with magnitude f = (0.20)(17.0) = 3.4 N, so the net force is 9.8 - 3.4 = 6.4 N and a = 6.4 / 2.0 = 3.2 meters per second squared, still down the slope.
- Gravity's component along an incline is mg sin(theta); perpendicular is mg cos(theta).
- On an incline the normal force equals mg cos(theta), not mg.
- A frictionless incline gives a = g sin(theta), independent of mass.
Worked example: standing in an elevator
A 60 kg person stands on a scale in an elevator. Two forces act: gravity downward, mg = (60)(9.8) = 588 N, and the normal force from the scale upward. The scale reads the normal force, so the question is what F_N must be. Take upward as positive: F_N - mg = ma.
Accelerating upward at 2.0 meters per second squared gives F_N = m(g + a) = (60)(11.8) = 708 N, and the person feels heavier. With the acceleration 2.0 meters per second squared downward, a is negative in this convention, so F_N = (60)(9.8 - 2.0) = (60)(7.8) = 468 N. At constant velocity - up, down, or at rest - a = 0 and F_N = mg = 588 N. In free fall, a equals g downward, so F_N = m(g - g) = 0 and the scale reads zero: apparent weightlessness, straight out of one diagram and one equation. The reading depends on acceleration, never on velocity.
- State the sign convention before writing F_N - mg = ma.
- Accelerating upward raises the normal force; accelerating downward lowers it.
- Constant velocity means zero net force, so F_N = mg regardless of speed.
- In free fall the normal force is zero - that is apparent weightlessness.
Friction, tension, and connected objects
Static and kinetic friction differ, and the exam tests that difference. Kinetic friction, acting when surfaces slide, has magnitude mu_k times F_N. Static friction is whatever value prevents sliding, up to a maximum of mu_s times F_N. So a block sitting still on a rough incline has static friction equal to the along-slope component of gravity; the maximum applies only at the verge of slipping.
For connected objects, draw a separate diagram for each. An ideal rope over an ideal pulley has one tension throughout, and the objects share the same magnitude of acceleration, giving two equations for two unknowns. Or treat the pair as one system, so tension is internal and cancels, then return to a single object for the tension. For circular motion the diagram is unchanged - draw the real forces, then set the net force toward the center equal to m times v squared over r.
- Kinetic friction is mu_k times F_N; static friction is variable up to mu_s times F_N.
- Connected objects share the acceleration magnitude and, for an ideal rope, the tension.
- For circular motion, set the net inward force equal to m v squared over r.
How to practice until it is automatic
Drawing diagrams is a motor skill, so drill it in isolation: take a stack of forces problems and draw only the diagrams, no algebra, checking each against a worked solution. On free response the diagram often carries points of its own - arrows from a single dot, descriptive labels, no extraneous arrows.
When you review a miss, ask which stage failed: a missing or invented force, a wrong axis choice, a sine-cosine swap, or a sign error. Swaps deserve a routine check, since as the incline angle approaches zero the along-slope component should vanish - true of sine, false of cosine. Every AP Physics 1 question in ScoreMint has a per-choice explanation, and the free-response prompts use rubric-style self-scoring.
- Drill diagrams alone, without solving, for far more repetitions per hour.
- Sanity-check decompositions at extreme angles to catch sine-cosine swaps.
- Sort misses into missing force, wrong axes, bad decomposition, or sign error.
