Equilibrium of rigid bodies

A book resting on a table is easy to explain: the forces on it cancel, so it does not accelerate. A ladder leaning against a wall is a harder question — the forces on it can cancel perfectly and it can still tip or swing, because forces alone say nothing about turning. A rigid, extended body needs a second, independent condition before you can call it truly at rest: the turning effects of every force about any point must also cancel. This article builds that second condition — rotational equilibrium — alongside the familiar force-balance condition, then applies both together to the classic three-force case every ISC student meets: a ladder resting against a smooth wall with its foot on rough ground. You will see why the three forces on such a ladder must pass through one point, why choosing the right pivot for moments turns a two-unknown problem into a one-line equation, and how the same reasoning solves a hinged rod held by a string. By the end you should be able to write both equilibrium conditions for any rigid body under three coplanar forces and solve for an unknown reaction without ever writing more equations than you need.

The core insight

A rigid body is in equilibrium only when the net force on it is zero and the net moment (torque) about every point is zero — force balance alone is not enough, because a body with zero net force can still rotate.

The core idea

Why force balance alone fails. Push on opposite ends of a metre rule with equal and opposite forces, offset from each other. The net force is exactly zero — the rod does not translate — yet it spins, because the two forces form a couple: equal, opposite, non-collinear forces whose moments add rather than cancel. This single example is why a rigid, extended body needs a condition on rotation as well as a condition on force. A particle (or any object treated as a point) only ever needs the first condition; a rigid body always needs both.

The two conditions, stated together. For a rigid body to be in complete (static) equilibrium:

∑F⃗=0and∑τ⃗=0\sum \vec F = 0 \qquad\text{and}\qquad \sum \vec\tau = 0

where: ∑F⃗\sum \vec F is the vector sum of every external force acting on the body (in components, ∑Fx=0\sum F_x = 0 and ∑Fy=0\sum F_y = 0 for coplanar forces), and ∑τ⃗\sum \vec\tau is the vector sum of the moments (torques) of every external force, taken about any one chosen point. The first condition fixes translational equilibrium — the centre of mass does not accelerate. The second fixes rotational equilibrium — the body does not begin to spin. A remarkable and useful fact, proved in ISC mechanics, is that if the net force is already zero, the net torque comes out the same about every choice of pivot; that freedom to pick the most convenient point is what makes rigid-body problems tractable by hand.

The three-coplanar-force case. The commonest ISC problem — the "ladder problem" — restricts attention to a rigid body acted on by exactly three coplanar forces: for instance a ladder with its foot on rough ground and its top against a smooth wall. The three forces are the ladder's weight WW (acting at its centre of gravity), the wall's normal reaction NwN_w (horizontal, since a smooth wall exerts no friction), and the ground's reaction, which splits into a normal component NgN_g (vertical) and a frictional component ff (horizontal, preventing the foot sliding out). A body under exactly three coplanar forces can only be in equilibrium if the three forces are concurrent — their lines of action meet at a single point. If they did not, the two conditions could not both hold: shifting any one force's line away from the common point would create an unbalanced moment about that point even while the vector sum of forces stayed zero. This concurrency is not an extra assumption; it is a consequence of the two equilibrium conditions applied to exactly three forces.

Figure
Forces on a ladder leaning against a smooth wall with its foot on rough ground
W N_w N_g f smooth wall (N_w only) rough ground (N_g and f)
Derivation AThe moment equation for a ladder against a smooth wall

Show that for a uniform ladder of weight WW and length LL, resting at angle θ\theta to the ground with its foot on rough ground and top against a smooth wall, the wall's reaction obeys Nw=12Wcot⁡θN_w = \tfrac{1}{2}W\cot\theta and equals the frictional force at the foot, so that once WW and θ\theta are known, both unknown reactions follow from a single moment equation without ever needing to solve two equations in two unknowns simultaneously.

Step 1. List the forces and use translational equilibrium first. Four forces act: weight WW at the midpoint, ground's normal reaction NgN_g (vertical), friction ff at the foot (horizontal), and the wall's normal reaction NwN_w (horizontal, since the wall is smooth). Resolving vertically, the only vertical forces are WW (down) and NgN_g (up), so Ng=WN_g = W. Resolving horizontally, the only horizontal forces are ff and NwN_w, so f=Nwf = N_w. Translational equilibrium alone already links ff to NwN_w, but neither is yet known in terms of WW and θ\theta.

Step 2. Choose the foot of the ladder as the pivot for moments. This is the strategic step: both NgN_g and ff act exactly at the foot, so their moment arms about that point are zero and they vanish from the moment equation entirely, leaving only WW and NwN_w — an equation in one unknown once WW is given.

Step 3. Write the moment of each remaining force about the foot. The weight WW acts at the midpoint of the ladder, at horizontal distance L2cos⁡θ\tfrac{L}{2}\cos\theta from the foot, giving a moment W×L2cos⁡θW \times \tfrac{L}{2}\cos\theta tending to rotate the ladder clockwise (sliding the foot outward). The wall's reaction NwN_w acts at the top of the ladder, at vertical distance Lsin⁡θL\sin\theta from the foot, giving a moment Nw×Lsin⁡θN_w \times L\sin\theta tending to rotate the ladder counter-clockwise (pushing the foot back in). Rotational equilibrium sets these equal:

Nw×Lsin⁡θ=W×L2cos⁡θN_w \times L\sin\theta = W \times \frac{L}{2}\cos\theta

Step 4. Cancel LL from both sides (it appears in every moment arm here, so it always cancels in this configuration) and solve:

Nw=Wcos⁡θ2sin⁡θ=W2cot⁡θN_w = \frac{W\cos\theta}{2\sin\theta} = \frac{W}{2}\cot\theta

Since f=Nwf = N_w from Step 1, the frictional force needed at the foot is also W2cot⁡θ\tfrac{W}{2}\cot\theta. This single result is the working formula for every ladder-against-a-smooth-wall problem: as θ\theta increases (a steeper ladder), cot⁡θ\cot\theta falls, so less friction is needed to hold the ladder — which matches the everyday sense that a nearly vertical ladder is safer than a shallow, leaning one.

Worked example

A uniform ladder of weight W=200 NW = 200\ \text{N} rests with its foot on rough ground and its top against a smooth vertical wall, making an angle of 60°60° with the ground. Find the normal reaction from the wall, the frictional force at the ground, and, if the ground can supply a coefficient of static friction μs=0.3\mu_s = 0.3, check whether the ladder is safe.

Step 1. Translational equilibrium gives Ng=W=200 NN_g = W = 200\ \text{N} (vertical) and f=Nwf = N_w (horizontal), exactly as in the derivation above.

Step 2. Apply the derived formula with θ=60°\theta = 60°:

Nw=W2cot⁡60°=2002×13=100×0.577≈57.7 NN_w = \frac{W}{2}\cot 60° = \frac{200}{2}\times\frac{1}{\sqrt3} = 100\times 0.577 \approx 57.7\ \text{N}

Step 3. Since f=Nwf = N_w, the friction required is also f≈57.7 Nf \approx 57.7\ \text{N}.

Competitive stretch

Step 4. Check safety: the maximum friction the ground can supply is μsNg=0.3×200=60 N\mu_s N_g = 0.3 \times 200 = 60\ \text{N}. The required friction, 57.7 N57.7\ \text{N}, is just under this limit (57.7<6057.7 < 60), so the ladder is on the point of being safe but has very little margin — a slightly steeper wall angle, a slightly heavier person climbing near the top, or a slightly less rough floor would make it slip. The wall pushes back with Nw≈57.7 NN_w \approx 57.7\ \text{N}, an equal friction of ≈57.7 N\approx 57.7\ \text{N} acts at the foot, and the ladder is safe but only just, since the required friction is close to the 60 N60\ \text{N} limit.

Common traps

  • Assuming zero net force is enough for equilibrium; a rigid body with zero net force can still rotate under a couple, so both conditions must be checked.
  • Taking moments about an arbitrary point instead of a point through which unknown forces pass, which needlessly leaves those unknowns in the equation.
  • Forgetting that a smooth surface supplies a normal reaction only, while a rough surface supplies both a normal reaction and friction, and mixing the two up.
  • Losing track of sign convention when several forces produce moments in different senses — fix clockwise or counter-clockwise as positive and stay consistent throughout.

Ask the tutor: Want to try a ladder problem with a person standing partway up, where the weight's line of action shifts and the moment equation gets an extra term?

Why do we need a rotational equilibrium condition at all if force balance already worked for particles?

A particle has no size, so every force on it effectively passes through the same point and can never create a net turning effect. A rigid body is extended, so two forces can have zero vector sum yet act along different lines, producing a couple that spins the body even though it does not translate. That is why rigid-body equilibrium always needs both ∑F⃗=0\sum \vec F = 0 and ∑τ⃗=0\sum \vec\tau = 0 checked independently.

Does it matter which point I choose to take moments about?

Not for the final answer, but it matters enormously for how much algebra you do. If the net force on the body is already zero, the net torque works out to be the same about every possible pivot, so any correct point gives the right equation eventually. Choosing a point where one or more unknown forces act (like the foot of the ladder, where NgN_g and ff act) makes those unknowns vanish from the moment equation, so you solve for the remaining unknown directly instead of juggling simultaneous equations.

Why must the three forces on the ladder pass through a single point?

Because with exactly three coplanar forces, both equilibrium conditions can only hold together if the three lines of action are concurrent. If two of the forces met at a point but the third's line missed it, taking moments about that meeting point would leave an unbalanced moment from the third force alone, breaking rotational equilibrium even if the force vectors still summed to zero. Concurrency is a direct consequence of the two conditions, not a separate rule to memorise.

Before this

Where this leads