2D Equilibrium: ΣF=0, ΣM=0
The single highest-leverage skill in statics — turn any rigid body into three simple equations.
Every bridge, shelf bracket, and crane arm that isn't moving is obeying the same three equations. Learn to write them correctly and you can find the forces holding up almost any rigid structure — before you ever pick up a wrench or pour concrete.
From force balance to three equations
Newton's first law says an object at rest stays at rest unless a net force acts on it. For a rigid body — something that doesn't bend or deform for the purposes of this analysis — "not accelerating" means two things must both be true at once: the forces must balance, and the turning effects of those forces must balance. Miss either one and the body would either slide away or spin.
In two dimensions, "forces balance" splits cleanly into two independent directions (commonly x and y), and "turning effects balance" is one more equation about rotation. That gives exactly three independent scalar equations — no more, no fewer — to describe equilibrium of a single rigid body in a plane.
No matter how many forces act on a 2D rigid body, you only ever get three independent equilibrium equations from it. If your problem has more than three unknown reaction components, the structure is statically indeterminate — you'll need Mechanics of Materials tools (later in this course) to solve it, not statics alone.
Reading a support: how many unknowns does it add?
Before you can write equilibrium equations, you need to know what each support contributes as an unknown force. This is where most beginners lose points — not in the algebra, but in misreading the supports.
- Roller (shown as a circle on a surface): resists motion in only one direction, perpendicular to the rolling surface. 1 unknown.
- Pin / hinge (shown as a triangle): resists motion in any direction in the plane, but allows rotation. 2 unknowns (usually written as horizontal and vertical components).
- Fixed support (shown built into a wall): resists motion in any direction and resists rotation. 3 unknowns (two force components plus a reaction moment).
The beam above has a pin (2 unknowns: A_x, A_y) and a roller (1 unknown: B_y) — exactly 3 unknowns for exactly 3 equations. That's not a coincidence for a properly supported, statically determinate structure.
ΣM=0 can be taken about any point in the plane — even a point with nothing physically there. The smart move is to sum moments about a point where an unknown force passes through, because that force then has zero moment arm and drops out of the equation entirely.
In the beam above, summing moments about point A makes both A_x and A_y vanish from that equation (they pass straight through A), leaving an equation with only B_y as the unknown — solved in one line. Then ΣFy=0 gives A_y directly, and ΣFx=0 gives A_x directly (which is zero here, since the only applied load is vertical).
- Draw the FBD: unknowns are A_x, A_y (pin) and B_y (roller). The only applied load is vertical, so A_x = 0 immediately from ΣFx = 0.
- Sum moments about A (this eliminates both A_x and A_y from the equation): ΣM_A = 0 → −(12 kN)(2 m) + B_y(6 m) = 0.
- Solve for B_y: B_y = (12 × 2) / 6 = 4 kN (upward).
- Sum vertical forces: ΣFy = 0 → A_y + B_y − 12 kN = 0 → A_y = 12 − 4 = 8 kN (upward).
- Sanity check: the load is closer to A (2 m) than to B (4 m), so A should carry the larger share of the load. 8 kN > 4 kN — checks out.
- Sum moments about A: −(20)(4) + B_y(10) = 0.
- B_y = 80 / 10 = 8 kN.
- ΣFy = 0: A_y + B_y − 20 = 0.
- From the previous question, B_y = 8 kN, so A_y = 20 − 8 = 12 kN.
Check your understanding
- A rigid body in equilibrium gives exactly three independent 2D equations: ΣFx=0, ΣFy=0, ΣM=0.
- Read supports carefully: a roller gives 1 unknown, a pin gives 2, a fixed support gives 3.
- Sum moments about a point where an unknown force acts — that force then drops out of the equation.
- A properly (statically determinate) supported structure has exactly as many unknowns as equations.
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