Equilibrium of a particle
Nothing about a body at rest says the forces on it are small. A bracket can be pulled by ten kilonewtons from each side and never twitch. Equilibrium is not the absence of force. It is cancellation, and cancellation has to be proved one component at a time.
Statics is the study of things that do not move, which sounds like the study of nothing at all until you notice that most of engineering is exactly that: the bridge that stays up, the bracket that does not tear off the wall, the cable that holds. The simplest case, and the one every harder case is built out of, is a body loaded so that every force on it passes through a single point. Then there is nothing to rotate, and the whole of equilibrium collapses into one equation.
The condition for equilibrium
Newton’s first law says a body at rest stays at rest, and a body moving at constant velocity keeps moving that way, so long as the resultant force on it is zero. Turn that around and you have the whole subject:
It is one vector equation, which means it is really two scalar equations in a plane and three in space. Note what it does not say. It does not say the forces are small, or that there are few of them, or that they are gentle. It says only that they destroy each other.
The word particle here has nothing to do with size. A forty-tonne crate is a particle if all the forces acting on it are concurrent: if their lines of action meet at a common point. That is what the idealisation actually buys you: concurrent forces cannot spin the body about the point they share, so there is no moment equation to write and force balance is the entire story. Once the forces stop meeting at a point, you are doing rigid-body equilibrium instead, and you need moments as well.
There is a picture hiding in that equation. Adding vectors head to tail lays them out as a chain; if the chain sums to zero, the last arrowhead lands exactly on the first tail. The force polygon closes. For three forces this is a closed triangle, and it can be solved with the law of sines rather than components at all: a result usually quoted as Lami’s theorem, where each \(\alpha_i\) is the angle between the other two forces:
It is worth carrying because it turns a two-equation solve into a single ratio. It is also worth remembering that it applies to three concurrent forces and nothing else.
Three cables pull on one ring.
The free-body diagram
\(\sum \mathbf{F} = \mathbf{0}\) is not the hard part. The hard part is knowing which forces belong in the sum, and that is settled by a drawing, not by algebra. The free-body diagram is the single most valuable habit in statics, and almost every wrong answer in a first course can be traced back to one that was skipped or drawn carelessly.
The recipe is short:
Decide what the particle is and sketch it on its own, away from everything it touches. Cut every connection to the outside world: each cable, each spring, each contacting surface. At every cut, draw the force that connection was transmitting. Add the weight, acting through the centre of gravity, and any load applied directly. Then label each arrow: a number for what you know, a symbol for what you do not.
The one rule that catches people out is this: draw only the forces exerted on the particle, never the forces the particle exerts on other things. Those third-law partners are real, but they act on other bodies and belong in other diagrams. Pull one into your own and you have counted it twice, usually with the wrong sign.
Where you do not know which way a force points, pick a sense, draw it, and stay consistent. The algebra will correct you. A negative answer means the true sense is opposite to the one you assumed, and that is information, not a mistake, with one important exception, which the next section explains.
A ring, two cords, and a hanging load.
Springs, cables, and pulleys
Cutting a connection is only useful if you know what force that connection can carry. Three of them cover most first-course problems.
A linear elastic spring resists being deformed in proportion to how far it has been deformed. Write its undeformed length as \(\ell_0\) and its current length as \(\ell\):
The stiffness \(k\) is in newtons per metre. Positive \(s\) means the spring is stretched and pulls its ends together; negative \(s\) means it is compressed and pushes them apart. The trap is the reference: \(s\) is measured from the unstretched length, not from wherever the spring happened to sit before the load arrived. Problems are written to punish that assumption.
A cable, cord, or rope is idealised as weightless and inextensible. It can only pull, and it can only pull along its own length. The force it applies is directed away from the particle, straight down the cable. This is why a negative tension is not like other negative answers. A cable cannot push, so a negative tension does not mean the arrow flips; it means the cable has gone slack and the configuration you assumed never existed. The arithmetic is fine. The model is wrong.
A frictionless pulley changes the direction of a cable without changing its tension. The pulley can only push on the cable radially, through the pin, so it exerts no moment about the pin and the two sides must balance. One unknown tension therefore covers the entire cord no matter how many times it wraps, which is exactly what makes pulley systems tractable.
Unloaded, the spring sits at its natural length.
Coplanar systems
When every force lies in one plane, the vector condition splits into two independent scalar conditions:
Two equations, so at most two unknowns can be found. Count them before you start: two unknown cable tensions is solvable, three is not, not without more physics than force balance alone provides.
Resolving each force into components is the arithmetic from force vectors, and the signs come off the diagram rather than out of a formula. Read them from the arrow you drew: an arrow pointing left contributes a negative \(x\) term whatever symbol sits on it.
One choice is worth making deliberately. Nothing requires \(x\) to be horizontal. If you align one axis with an unknown force, that force disappears from the perpendicular equation entirely, and the other unknown drops out in a single line instead of a simultaneous solve. On an inclined plane, align with the slope. In a two-cable problem, align with one of the cables. The equations are equivalent either way; the work is not.
Two cords and a weight, all in one plane.
Three-dimensional systems
In space, the same vector statement becomes three scalar equations. Writing the resultant in components,
and a vector is zero only when every component is, so each bracket vanishes on its own:
You will almost never be handed the angles in three dimensions. You will be handed coordinates, and the direction of each cable comes from its own geometry: the position vector between its endpoints, divided by its own length:
The payoff is that each cable still contributes exactly one unknown, the scalar tension \(T_i\), no matter how awkward its direction. Build every unit vector first, in a list, before you write a single equilibrium equation. The overwhelming majority of three-dimensional errors are sign slips in the position vectors, and they are far easier to catch in a tidy list than buried inside a summation.
Then count again. Three equations support three unknown tensions, so a load held by three non-coplanar cables is statically determinate and the geometry alone gives the answer. A fourth cable does not make the rig safer to analyse: it makes it indeterminate, and you can no longer find the tensions from statics at all. You need how much each cable stretches. That distinction is the boundary between this subject and mechanics of materials, and it is worth recognising the moment you finish counting.