Equilibrium of a Rigid Body

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Equilibrium of a Rigid Body

TL;DR

A rigid body is in equilibrium when the net force and net torque on it are both zero. You'll learn to set up force and torque equations, pick smart pivot points to kill unknowns, and solve for unknown forces like tensions, normal reactions, and hinge forces. Two equations, one good diagram, done.

1. The Mental Model

A rigid body doesn't just need balanced pushes — it needs balanced twists too. Forces can cancel perfectly and still spin something, like pushing opposite ends of a steering wheel. So equilibrium needs two separate checks: forces sum to zero, and torques sum to zero. No net push, no net spin — that's the whole story.

2. The Core Material

2.1 The Two Conditions

For a rigid body to stay at rest (or move with constant velocity and no rotation), two independent conditions must hold simultaneously:

Translational equilibrium:
$$\vec{F}_{net} = \sum \vec{F}_i = 0$$

In components: $\sum F_x = 0$ and $\sum F_y = 0$ (for a 2D problem).

Rotational equilibrium:
$$\vec{\tau}_{net} = \sum \vec{\tau}_i = 0$$

where torque about a point is $\vec{\tau} = \vec{r} \times \vec{F}$, with magnitude $\tau = rF\sin\theta = F \cdot d$, and $d$ is the perpendicular distance from the pivot to the line of action of the force (the "moment arm").

Both conditions are necessary. A body can satisfy force balance and still rotate (think of a couple — two equal, opposite, non-collinear forces). It can also satisfy torque balance about one point and still accelerate linearly (if there's a net force acting through that exact point). You need both.

Sign convention: pick counterclockwise as positive (or clockwise — just stay consistent within one problem). A force that would spin the body counterclockwise about your chosen pivot gets a positive torque; clockwise gets negative.

2.2 The Trick That Makes Life Easy: Choosing Your Pivot

Here's the part that separates fast, clean solutions from messy ones. The torque equation $\sum \tau = 0$ can be taken about any point — not just the body's center of mass or a physical hinge. This is because if the net force is already zero, the net torque is the same about every point in space. (If you're not satisfied with that claim: torque about a new origin $O'$ differs from torque about $O$ by $\vec{r}_{OO'} \times \vec{F}_{net}$, and that term vanishes when $\vec{F}_{net}=0$.)

So the strategy is: choose the pivot at the location of an unknown force you don't want to deal with. Any force acting at that point has zero moment arm ($d=0$), so it contributes zero torque and drops out of the equation entirely. This often lets you solve for one unknown directly without touching the others.

2.3 Setting Up a Problem — the Standard Procedure

1. Draw a free body diagram. Identify every external force: weight (acting at the center of gravity), normal forces, tensions, friction, applied forces, and reaction forces at supports/hinges.
2. Choose axes. Usually horizontal ($x$) and vertical ($y$).
3. Write $\sum F_x = 0$ and $\sum F_y = 0$.
4. Choose a pivot — ideally where an unknown (often an ugly hinge reaction) acts — and write $\sum \tau = 0$ about it.
5. Solve the system. With 3 independent equations (2 force + 1 torque in 2D), you can solve for up to 3 unknowns.

A key subtlety: a rigid body in 2D has exactly 3 degrees of freedom (translate in $x$, translate in $y$, rotate). That's why you get exactly 3 independent equilibrium equations — not more, even if you write torque equations about five different points. Writing extra torque equations about other points can still be useful (sometimes easier algebra), but they won't give you new information beyond the original 3.

flowchart TD
    A["Draw free body diagram: list all external forces + their points of application"] --> B["Write sum Fx = 0"]
    A --> C["Write sum Fy = 0"]
    A --> D["Choose pivot at an unknown force location"]
    D --> E["Write sum tau = 0 about that pivot (unknown force drops out)"]
    B --> F["Solve system of 3 equations for up to 3 unknowns"]
    C --> F
    E --> F

2.4 Center of Gravity and Stability

The weight of a rigid body acts as if concentrated at its center of gravity (CG) — in a uniform gravitational field, this coincides with the center of mass. For torque calculations, you always place the weight vector $Mg$ acting straight down through the CG.

Stability depends on where the CG sits relative to the base of support. If you tilt a body slightly, it's stable if the vertical line through the CG still falls inside the base (restoring torque brings it back), and unstable if that line falls outside the base (gravity's torque tips it further). This is why a low, wide CG (race cars) is more stable than a high, narrow one (a pencil balanced on its tip).

2.5 Common Setup: The Ladder/Beam Problem

A classic configuration: a rigid beam or ladder resting against a wall and floor, or a plank on two supports, with forces at discrete points. The general pattern is always:

• Supports exert normal forces perpendicular to the contact surface.
• Rough contact surfaces can also exert friction, parallel to the surface, whose direction you must guess and let the algebra confirm (negative result = wrong guessed direction).
• Hinges/pins can exert force in any direction — so you resolve the hinge reaction into unknown components $H_x$ and $H_y$, which is exactly why you want your pivot at the hinge: it eliminates both unknowns from the torque equation in one move.

3. Worked Example

Problem: A uniform ladder of mass $M = 20\ \text{kg}$ and length $L = 4\ \text{m}$ leans against a smooth (frictionless) vertical wall, with its base on rough ground. The ladder makes a $60°$ angle with the ground. Find the normal force from the wall, the normal force from the ground, and the minimum coefficient of friction at the ground needed to keep the ladder from slipping. Take $g = 10\ \text{m/s}^2$.

Step 1 — Free body diagram. Forces acting on the ladder:
- Weight $Mg = 200\ \text{N}$, acting at the midpoint (CG), $L/2 = 2\ \text{m}$ from either end.
- Normal force from the wall, $N_w$, horizontal (wall is smooth — no friction there), acting at the top of the ladder.
- Normal force from the ground, $N_g$, vertical, acting at the base.
- Friction from the ground, $f$, horizontal, acting at the base (it must point toward the wall, opposing the tendency of the base to slide away).

Step 2 — Force equations.

$\sum F_x = 0$: $\quad f - N_w = 0 \implies f = N_w$

$\sum F_y = 0$: $\quad N_g - Mg = 0 \implies N_g = 200\ \text{N}$

Step 3 — Pick the pivot. Choose the pivot at the base of the ladder (where it touches the ground). This is smart because both $N_g$ and $f$ act right there — zero moment arm — so they vanish from the torque equation, leaving only $Mg$ and $N_w$.

Step 4 — Torque equation. Set up geometry: the ladder makes $60°$ with the ground.

• Weight acts at the midpoint, horizontal distance from base = $\frac{L}{2}\cos 60° = 2 \times 0.5 = 1\ \text{m}$. This torque tends to rotate the ladder clockwise (base-down) about the base pivot — call clockwise negative.
• Wall's normal force $N_w$ acts at the top, horizontal, at height $L\sin 60° = 4 \times 0.866 = 3.46\ \text{m}$. Its moment arm (perpendicular distance from pivot to the line of action) is this vertical height, $3.46\ \text{m}$, and it tends to rotate counterclockwise.

$$\sum \tau = 0: \quad N_w (L\sin 60°) - Mg\left(\frac{L}{2}\cos 60°\right) = 0$$

$$N_w (3.46) = 200 \times 1 = 200$$

$$N_w = \frac{200}{3.46} \approx 57.8\ \text{N}$$

Step 5 — Back out friction. From Step 2: $f = N_w \approx 57.8\ \text{N}$.

Step 6 — Minimum coefficient of friction. For no slipping, $f \le \mu N_g$:

$$\mu_{min} = \frac{f}{N_g} = \frac{57.8}{200} \approx 0.29$$

So the wall pushes back with≈57.8 N, the ground pushes up with 200 N, friction supplies ≈57.8 N toward the wall, and the ground needs a coefficient of friction of at least 0.29 to keep the ladder from slipping. Notice how choosing the base as the pivot made two unknowns vanish instantly — that's the whole trick in action.

4. Key Takeaways

4.1 Most Important Concepts

• Two conditions, not one: equilibrium requires both $\sum \vec{F} = 0$ and $\sum \vec{\tau} = 0$; satisfying only one isn't enough.
• Torque is pivot-independent at equilibrium: if net force is zero, net torque is the same about every point, so you're free to choose the most convenient pivot.
• Kill unknowns by placing the pivot on them: any force acting exactly at your chosen pivot contributes zero torque, removing it from that equation.
• Three equations, three unknowns (2D): a rigid body has 3 degrees of freedom, so you get exactly 3 independent equilibrium equations — extra torque equations about other points are useful for algebra, not for new information.
• Weight acts at the center of gravity, which you treat as a single point force $Mg$ downward when writing torque equations.
• Friction direction is often guessed, then confirmed: if your solved value comes out negative, the real direction is opposite to your guess.
• Stability depends on the CG relative to the base of support — not on total weight or size alone.

4.2 Common Misconceptions

• "If forces balance, the object must be in equilibrium." False — a couple (two equal, opposite, offset forces) has zero net force but a nonzero net torque, causing rotation.
• "The pivot for torque must be the center of mass or a physical hinge." False — you can choose any point, including one with no physical significance, purely to simplify the algebra.
• "More torque equations about more points give more information." False — in 2D you only ever have 3 independent equations total; extra torque equations are linear combinations of the first three.
• "A heavier object is automatically more stable." False — stability is about the geometry of the CG relative to the base of support, not mass.

4.3 Compare & Contrast

5. Now Try It

Exam-style question: A uniform horizontal plank of mass $12\ \text{kg}$ and length $3\ \text{m}$ rests on two supports: one at the very left end, and one at $2\ \text{m}$ from the left end. A person of mass $18\ \text{kg}$ stands at the right end of the plank. Taking $g = 10\ \text{m/s}^2$, find the normal reaction forces at both supports.

A top-band answer must: (1) draw a clear free body diagram showing the plank's weight at its midpoint (1.5 m from left), the person's weight at the right end (3 m from left), and both unknown normal reactions; (2) write $\sum F_y = 0$; (3) choose a pivot at one support to eliminate that unknown, write $\sum \tau = 0$, and solve for the other support's force; (4) substitute back to find the first force; (5) sanity-check that both reactions are positive (both supports pushing up, not pulling).