Foundations — Equilibrium of rigid bodies — translational + rotational

This page assumes you know nothing. Every arrow, every letter, every little symbol in the parent note the parent topic gets built here from the ground up, in an order where each idea leans only on the one before it.

0. What is a "rigid body"?

Why does the topic need this word? Because for a rigid body, a single force applied at one end is felt as a twist about the whole object. If the object could bend, the far end wouldn't care. Rigidity is what lets "where you push" matter.

Look at the figure: the same sideways push at the top of the rigid rod makes the whole rod swing about its base. That "swing about a point" is the seed of everything on this page.

1. A point and its position — the vector $\vec{r}$

Before forces, we need to say where things are. Pick a reference point (call it the origin or the pivot). The arrow from that point to some spot is the position vector.

The picture: stand at the pivot, point your finger at where a force acts — your arm is $\vec{r}$. Its length is how far away the point is; its direction is which way you point.

Why the topic needs it: torque asks "how far out, and in which direction, is the force applied?" You cannot answer that without an arrow from the pivot. That arrow is $\vec{r}$.

2. Force — the symbol $\vec{F}$

The picture: an arrow starting at the object, pointing the way you shove it. Weight is a force too — an arrow pointing straight down, drawn from the object's centre of mass.

Why the topic needs it: Newton's laws say forces are what change motion. To keep a body still, we must know every arrow acting on it.

3. Adding vectors and the symbol $\sum$

The picture: lay each force arrow tip-to-tail. Where you end up — the single arrow from the very start to the very end — is $\sum \vec{F}$, the net force.

4. Splitting a force into $F_x$ and $F_y$ — components

Adding slanted arrows is fiddly. The trick: give every arrow a shadow on two perpendicular walls — the horizontal $x$-axis and the vertical $y$-axis.

The picture: shine a light straight down — the floor-shadow is $F_x$. Shine a light from the side — the wall-shadow is $F_y$. Together the two shadows rebuild the arrow (they form a right triangle with $\vec{F}$ as the slanted side).

Why the topic needs it: instead of one hard 2-D balance, we get two easy 1-D balances: $\sum F_x=0\text{and}\sum F_y=0.$ "Nothing drags it sideways" and "nothing drags it up or down."

5. Angle $\theta$, and why $\sin \theta$ shows up

Now the key idea for twisting. Take the position arrow $\vec{r}$ (from pivot to where you push) and the force arrow $\vec{F}$. The angle between them is $\theta$.

Why $\sin$ and not something else? $\sin \theta$ is exactly the fraction of a length that points across (perpendicular) rather than along. When $\theta =90^{\circ }$ the force is fully sideways, $\sin 90^{\circ }=1$: maximum twist. When $\theta =0^{\circ }$ (force along $\vec{r}$), $\sin 0^{\circ }=0$: no twist. $\sin \theta$ is the answer to "what fraction of this push actually turns the body?"

6. Torque — the symbol $\vec{\tau }$ and moment arm $d$

Now we can name the "twisting power."

The picture: extend the force arrow into a full straight line (its line of action). Drop a perpendicular from the pivot onto that line. The length of that perpendicular is $d$. Torque $=F\times d$ — "how hard $\times$ how far off-line."

7. The cross product $\vec{r}\times \vec{F}$ and the sign of a twist

A twist has a direction too: clockwise or counter-clockwise. We package all of this into the cross product.

The picture: curl the fingers of your right hand the way the force tries to spin the body; your thumb points along $\vec{\tau }$. Thumb out of the page = positive.

8. Mass $M$, acceleration ${\vec{a}}_{cm}$, and Newton's engine

Newton's second law for the whole body: $\sum {\vec{F}}_{\text{ext}}=M{\vec{a}}_{cm}.$ "Ext" means external — forces from outside the body; internal pushes between its own atoms cancel in pairs. Set ${\vec{a}}_{cm}=0$ and you get the first equilibrium condition, $\sum \vec{F}=0$.

9. Moment of inertia $I$, angular acceleration $\vec{\alpha }$ — the twisting Newton's law

The rotational twin of Newton's law: $\sum {\vec{\tau }}_{\text{ext}}=I\vec{\alpha }.$ Set $\vec{\alpha }=0$ and you get the second condition, $\sum \vec{\tau }=0$. You don't need to compute $I$ for a static problem — but you must know it exists, because it's what makes "$\sum \tau =0\Rightarrow$ no spin-up" true.

10. Weight, centre of mass, and friction — the players in real problems

• Weight $W=Mg$: one downward force arrow, drawn from the centre of gravity (same as centre of mass in uniform gravity). For a uniform body that's the geometric middle.
• Normal force $N$: a support (floor, wall, pivot) pushes perpendicular to its surface.
• Friction $f$: a sideways grip force, $f\le \mu N$, where $\mu$ is the friction coefficient. It's what stops the ladder sliding.

Why here: the worked examples in the parent (seesaw, ladder, couple) are built entirely out of these arrows plus the two balance conditions.

How it all feeds the topic

Equipment checklist