2D Equilibrium: ΣF=0, ΣM=0

The single highest-leverage skill in statics — turn any rigid body into three simple equations.

StaticsMechanical Engineering Year 1Free preview
⏱️ About 18 min

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.

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The big idea: If a rigid body is not accelerating, the forces on it must cancel out in every direction, and the turning effects (moments) about any point must also cancel out. That's it — two physical ideas, expressed as three scalar equations in 2D.
🎯 By the end, you'll be able to
  • Write the three 2D equilibrium equations (ΣFx=0, ΣFy=0, ΣM=0) for a rigid body
  • Draw a correct free-body diagram showing every external force and reaction
  • Choose a smart point to sum moments about, to solve for one unknown at a time
  • Solve for unknown support reactions on a simply supported beam
📎 Helpful to know first
  • Free-Body Diagrams — The Master Skill
  • Moments & Couples

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.

\[ \sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum M_O = 0 \]
The three 2D equilibrium equations. M_O is the sum of moments about any chosen point O.
🔑 Only three equations — ever

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.

Free-body diagram of a simply supported beam with a pin at A, a roller at B, and a downward point load P at midspan A (pin) B (roller) P A_x A_y B_y L/2 L/2

A simply supported beam with a pin support at A carrying reactions A_x and A_y, a roller support at B carrying reaction B_y, and a downward point load P applied at midspan, distance L/2 from each support.

A simply supported beam: a pin at A can push in any direction (two unknown reaction components, A_x and A_y); a roller at B can only push perpendicular to the surface it rests on (one unknown, B_y).

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.

✨ Choosing where to sum moments is the whole trick

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).

🎮 Interactive: build a free-body diagram and solve it LIVE
Predict first: Before you solve it: which support do you think carries more of the load — the one closer to P, or the one farther away?

An interactive simulator showing a simply supported beam where the user can move a point load and see the pin and roller reaction forces update in real time, computed from the three equilibrium equations.

Drag the load along the beam and watch the reactions update live. Notice how the support closer to the load always carries more of it.
📝 Worked example: A 6 m simply supported beam has a pin support at A (left end) and a roller support at B (right end). A downward point load of 12 kN is applied 2 m from A. Find the reactions at A and B.
  1. 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.
  2. 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.
  3. Solve for B_y: B_y = (12 × 2) / 6 = 4 kN (upward).
  4. Sum vertical forces: ΣFy = 0 → A_y + B_y − 12 kN = 0 → A_y = 12 − 4 = 8 kN (upward).
  5. 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.
✓ A_x = 0, A_y = 8 kN ↑, B_y = 4 kN ↑
✏️ Practice: A 10 m simply supported beam (pin at A, roller at B) carries a single downward point load of 20 kN located 4 m from A. What is the reaction B_y, in kN?
kN
Solution
  1. Sum moments about A: −(20)(4) + B_y(10) = 0.
  2. B_y = 80 / 10 = 8 kN.
✏️ Practice: Using the same beam as above (10 m span, pin at A, roller at B, 20 kN load 4 m from A), what is the reaction A_y, in kN?
kN
Solution
  1. ΣFy = 0: A_y + B_y − 20 = 0.
  2. From the previous question, B_y = 8 kN, so A_y = 20 − 8 = 12 kN.

Check your understanding

1. How many independent equilibrium equations exist for a single rigid body in 2D?
Two force-balance equations (x and y) plus one moment equation — three total, no more.
2. A roller support contributes how many unknown reaction components?
A roller resists motion in only one direction (perpendicular to its surface), so it contributes exactly one unknown.
3. Why is it usually smart to sum moments about a support with an unknown force?
A force passing through the point you're summing moments about has a moment arm of zero, so it contributes nothing to that equation — leaving fewer unknowns to solve for.
✅ Key takeaways
  • 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.
➡️ 2D equilibrium handles most planar problems, but real structures often need a third dimension — next, extend the same three ideas into 3D equilibrium.
Want to test yourself on this? Try the Mechanical Aptitude test →
🎓 Go deeper: external courses & trusted references