A supporting surface pushes back perpendicular to itself (normal force), a rope or cable pulls along its length (tension), and on a slope you split weight into parallel and perpendicular parts.

Key terms

Weight (w) — the force of gravity on an object: w = mg, always straight down.
Normal force (N) — the support force a surface exerts, perpendicular ("normal") to the surface. Surfaces bend slightly and push back like a stiff spring.
Tension (T) — the pull carried by a flexible connector (rope, cable, chain, tendon). It always acts along the connector, and it can only pull, never push.
Real force — has a physical cause (gravity, a push, a rope).
Fictitious force — only appears because the observer's frame is accelerating or rotating (e.g. the apparent sideways push on a merry-go-round).
Inertial frame — a frame that isn't accelerating or rotating, where Newton's laws hold as written. Earth is close enough for almost all problems.

Normal force rules

Flat surface, nothing else pushing or pulling vertically: N = mg.
On an incline: N = mg cosθ, which is less than the weight.
N can be greater than mg (pushing off when you jump, a car on a banked curve).
Don't confuse N (normal force, a vector) with N (newton, the unit). A normal force is measured in newtons.

Inclines: resolving weight

Tilt the axes so one runs along the slope and one runs perpendicular to it. Then weight splits into two parts:

\displaystyle w_{\parallel} = mg\sin\theta

\displaystyle w_{\perp} = mg\cos\theta

w⊥ is balanced by N, so there's no motion into or off the slope.
w∥ pulls the object down the slope; friction f points up the slope, opposing motion.
Memory check: as θ → 0 (flat), sin → 0 and cos → 1, so all the weight presses into the surface. That's how you tell which is sin and which is cos.

Acceleration down an incline:

\displaystyle a = \frac{mg\sin\theta - f}{m}

With no friction this becomes a = g sinθ, the same for any mass.

Tension rules

A rope holding a hanging object at rest: T = w = mg.
Tension is the same everywhere along an ideal (massless, frictionless) rope.
Pulleys, tendons and brake cables change the direction of tension, not its size.
A rope sagging at angle θ below horizontal, with a weight or push F⊥ at the middle:

\displaystyle T = \frac{F_{\perp}}{2\sin\theta}

Small sag angle means huge tension. A rope can never be perfectly straight under load because that would need infinite tension. This is how you can pull a car out of mud by pushing sideways on a chain tied to a tree.

Worked examples

Problem	Given	Setup	Answer
Skier, no friction (Ex. 4.5a)	m = 60.0 kg, θ = 25°	a = g sin25° = 9.80 × 0.4226	4.14 m/s²
Skier, with friction (Ex. 4.5b)	f = 45.0 N	a = (mg sin25° − f) / m = (248.5 − 45.0) / 60.0	3.39 m/s²
Skier, normal force	same	N = mg cos25° = 60 × 9.8 × 0.906	532 N
Hanging mass	m = 5.00 kg	T = mg	49.0 N
Tightrope walker (Ex. 4.6)	m = 70.0 kg, sag 5.0°	T = mg / (2 sin5.0°) = 686 / (2 × 0.0872)	3,900 N (about 6× his weight)

Problem-solving recipe

Draw a free-body diagram: one dot, every force as an arrow.
Pick axes that line up with the motion (along/perpendicular to a slope; horizontal/vertical for ropes).
Break any force not on an axis into components with sin and cos.
Write ΣF = ma separately for each axis. If nothing moves along an axis, ΣF = 0 there.
Solve, then sanity-check units and size.

Common mistakes

Assuming N = mg on a slope (it's mg cosθ).
Swapping sin and cos on an incline.
Forgetting tension acts on both sides of a sagging rope (the 2 in 2 sinθ).
Mixing up the normal force N with the newton unit N.
Leaving the calculator in radians instead of degrees.