Newton’s Laws & Free-Body Diagrams

You’ll be able to

State Newton’s three laws and apply the second law in the form ΣF = ma = m dv/dt
Draw free-body diagrams and resolve forces into perpendicular components
Solve for acceleration, normal force, and friction on a single object

Force is the cause of acceleration

Newton’s first law says an object keeps its velocity — at rest or moving straight at constant speed — unless a net force acts. Newton’s second law makes that quantitative: the net force sets the acceleration, ΣF = ma. Written the deeper way it is ΣF = dp/dt, the rate of change of momentum; when mass is constant this reduces to m dv/dt = ma. Newton’s third law adds that forces come in equal-and-opposite pairs acting on different objects.

Newton’s second law

ΣF = ma = m dv/dt

The net (vector) force equals mass times acceleration. Applied one axis at a time: ΣFₓ = maₓ and ΣF_y = ma_y.

The free-body diagram

Before any equation, isolate one object and draw every force acting on it as an arrow: weight mg (down), the normal force N (perpendicular to the surface), tension T (along strings), friction f (along the surface, opposing sliding), and any applied push or pull. Then pick axes — often tilted to line up with the motion — and write ΣF = ma separately along each axis. The diagram, not intuition, tells you which components add and which cancel.

Friction force

f_k = μ_k N f_s ≤ μ_s N

Kinetic friction has fixed magnitude μ_k N once sliding. Static friction adjusts up to a maximum μ_s N to prevent sliding; it equals whatever is needed below that cap.

Worked example

A 5 kg block slides down a frictionless incline tilted at 37° (sin 37° = 0.6, cos 37° = 0.8). Using g = 10 m/s², find its acceleration along the incline and the normal force from the surface.

1.Tilt the axes: let x point down the slope and y perpendicular to it. Gravity mg = 50 N splits into a component mg sin θ down the slope and mg cos θ into the surface.
2.Along the slope there is no friction, so ΣFₓ = mg sin θ = ma → a = g sin θ = 10 × 0.6 = 6 m/s².
3.Perpendicular to the slope the block does not accelerate, so N − mg cos θ = 0.
4.Solve for the normal force: N = mg cos θ = 5 × 10 × 0.8 = 40 N.

Answer: a = g sin θ = 6 m/s² down the incline; N = mg cos θ = 40 N

Checkpoint

A 4 kg block on a frictionless horizontal surface is pushed by a single horizontal force of 12 N. What is its acceleration?

Weight is not the same as normal force

A frequent error is to set N = mg automatically. That is only true on a level surface with no vertical push. On an incline the normal force is mg cos θ; in an elevator accelerating upward it is m(g + a); if you press down on the object it grows. Always get N from the perpendicular equation ΣF⊥ = 0 (or = ma⊥), never by assumption.

Checkpoint

A block rests on a frictionless incline of angle θ and is released. What is the magnitude of its acceleration along the incline?

On the exam

On the AP exam, draw the free-body diagram first and tilt your axes to match the motion. Write ΣF = ma along each axis separately. Getting the components of weight right — mg sin θ along a slope, mg cos θ into it — is where most points are won or lost.

Answer the 2 checkpoints as you read.