Foundations — Normal force — reaction force, not always = mg

Before you can understand "the normal force is not always $mg$", you must be fluent in a small set of symbols and pictures. The parent note assumes all of them. We build each from zero, in an order where every idea rests on the one before it.

1. Mass — the symbol $m$

The picture: imagine a shopping trolley. An empty one ($m$ small) is easy to shove; a full one ($m$ large) resists your shove. That resistance-to-being-pushed is mass.

Why the topic needs it: every normal-force equation multiplies $m$ by something. A $20\text{kg}$ box and a $2\text{kg}$ box on the same floor get very different pushes — the surface must support their different masses.

2. Force — the arrow $\vec{F}$

The little arrow on top, $\vec{F}$, is a reminder: "this quantity has a direction." A plain $F$ means we only care about its size (its length).

Why the arrow matters: the whole topic is about directions — forces point one way, gravity another. If we forgot direction and only tracked sizes, we could never say "this force points out of the surface" or "gravity points down."

3. Gravitational field strength — the symbol $g$

The picture: think of $g$ as the "strength dial" of the planet you stand on. On Earth the dial reads about $10$; on the Moon it reads about $1.6$ (a weaker pull). The dial doesn't care what object you drop — only where you are.

Why the topic needs it: $g$ is the bridge between an object's mass and its weight. On its own it is just a number; multiplied by mass it becomes a downward force. The parent note also compares $g$ against an elevator's acceleration $a$ (next section), so $g$ must stand as its own symbol before we combine it.

4. Gravity and weight — the symbol $mg$

The picture: an apple hanging from a string — the string is taut because gravity tugs the apple down. That downward tug is the weight arrow $mg$.

Why $mg$ and not just "gravity"? Because it is a number we can compute. A $5\text{kg}$ block has weight $5\times 10=50\text{N}$. The parent note constantly compares the surface's push against this $50\text{N}$ to decide whether that push is bigger, smaller, or equal.

5. Perpendicular — the symbol $\perp$ and the word "normal"

The picture: stand a pencil straight up on a table. The pencil points normal to (straight out of) the table. Now tilt the table into a ramp — the pencil, still pointing out of the surface, now leans. "Normal" always means "straight out of the surface," whatever way the surface faces.

Why the topic needs it: the surface's push is defined by its direction — it always points $\perp$ to the surface. On flat ground that's straight up; on a ramp it's tilted. Everything hinges on knowing what "perpendicular to the surface" looks like.

6. The tilt angle $\theta$, and the fractions $\cos \theta$, $\sin \theta$

The picture: shine a torch straight down on a leaning stick. The shadow length on the ground is the stick's length times $\cos \theta$ — the more it leans, the shorter that shadow. That "shrink factor" is exactly $\cos \theta$. Its partner $\sin \theta$ measures the sideways part.

Anchor values you can trust:

• $\theta =0^{\circ }$: $\cos 0^{\circ }=1$ (nothing lost — the whole arrow still points ahead), $\sin 0^{\circ }=0$.
• $\theta =30^{\circ }$: $\cos 30^{\circ }\approx 0.866$, $\sin 30^{\circ }=0.5$.
• $\theta =90^{\circ }$: $\cos 90^{\circ }=0$ (nothing left pointing ahead), $\sin 90^{\circ }=1$.

Why the topic needs it: on a ramp we must know what fraction of gravity presses into the surface. That fraction is $\cos \theta$. The deeper "which angle gives which fraction" machinery lives in Inclined Plane Problems; here you only need that $\theta$ is the ramp's tilt and $\cos \theta$ shrinks with steepness.

7. Splitting an arrow into components

Why we split forces: on a ramp, gravity points straight down but the surface faces a tilted direction. To ask "how hard must the surface push?" we only care about the part of gravity aimed into the surface. Splitting lets us isolate exactly that part.

Sanity across all tilts:

• Flat ground, $\theta =0^{\circ }$: $\cos 0^{\circ }=1$, so the perpendicular part is the full $mg$. Makes sense — all of gravity presses into a flat floor.
• Vertical wall, $\theta =90^{\circ }$: $\cos 90^{\circ }=0$, so none of gravity presses into the wall. Makes sense — gravity slides straight down a wall, pressing on nothing.
• In between: the steeper the slope, the smaller the perpendicular slice.

8. Adding forces — the symbol $\sum F$

The picture: three people pushing a stuck car. To know if it moves, you don't track each shove separately — you add the arrows into one net arrow. That net arrow is $\sum F$.

Why the topic needs it: the surface's push is never found alone. We add all forces in the perpendicular direction (gravity's slice, any extra push, and the surface's push itself) and set the total equal to what Newton's law demands. Without $\sum$ we couldn't write that balance.

9. Acceleration — the symbol $a$

The picture: press the gas pedal — you're pushed back into the seat; that's positive $a$. Slam the brakes — you lurch forward; that's $a$ in the other direction. Cruise at steady speed — $a=0$.

Why the topic needs it: the parent note's headline is "the surface's push $\ne mg$ when there is vertical acceleration." In an elevator, $a\ne 0$ in the vertical direction, and that is exactly why the floor's push changes. The symbol $a_{\perp }$ means "the acceleration in the perpendicular direction."

10. Newton's 2nd law — the engine $\sum F=ma$

The picture: a bigger net shove ($\sum F$ up) gives a bigger acceleration ($a$ up). A heavier object ($m$ up) accelerates less for the same shove. That trade-off is the equation.

Why this is THE tool: the parent note says the surface's push "is only $mg$ in one special case." The reason is that it is always found by plugging into $\sum F_{\perp }=m{a}_{\perp }$ and solving. Change $a_{\perp }$ or add a force, and the solved push changes. The full engine is unpacked in Newton's Second Law.

11. The normal force — the symbol $N$, and why $N\ge 0$

Now every ingredient is on the table, we can name the star of the topic.

The picture: press your hand flat on a table — the table pushes back on your palm. Now try to pull your palm upward off the table: the table cannot grab and hold it, it just stops pushing. A surface has no "glue"; the moment it would need to pull, it lets go and $N$ drops to $0$ — never below.

Why this edge case matters: it is exactly why free fall gives weightlessness ($N=0$) in the parent note, and why $N$ can shrink to zero but never go negative. Any answer with a negative $N$ is a signal that the object flew off the surface.

12. Reaction / constraint force — the idea behind $N$

The picture: sit on a sofa cushion — it squishes until its springiness pushes back just enough to hold you. Add a second person — it squishes more and pushes back harder. The cushion's push adjusts. That is exactly what the normal force does (via atomic bonds, per the parent note).

Why the topic needs it: this is the heart of "not always $mg$." Because $N$ is a response, it can be $mg+F$, $m(g+a)$, $mg\cos \theta$, or even $0$ — whatever $\sum F_{\perp }=m{a}_{\perp }$ requires (staying $\ge 0$). The distinction between action–reaction pairs (which live in Newton's Third Law) is also built on this idea.

How these foundations feed the topic

Read it top to bottom: mass and field strength build weight; arrows, perpendicularity and the tilt angle let us split and add forces; acceleration plus that sum give Newton's 2nd law; the constraint idea (with $N\ge 0$) plus that law produce the normal force — and only then can we see why it need not equal $mg$.

Equipment checklist

Cover the right side and see if you can answer each before revealing.

Connections

• Newton's Second Law — the engine $\sum F=ma$ that solves for $N$
• Newton's Third Law — action–reaction pairs vs. same-body forces
• Free Body Diagrams — the tool that draws all the arrows you just learned
• Inclined Plane Problems — where splitting into components gives $mg\cos \theta$
• Apparent Weight & Elevators — where $a\ne 0$ makes $N\ne mg$
• Friction — the force built on top of $N$ ($f=\mu N$)
• Parent: Normal Force