A rigid body is in equilibrium when the forces on it add up to zero and the moments about any point add up to zero. In 2D that gives three equations, which is enough to find three unknown support reactions once the free body diagram is drawn.
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The equilibrium equations
A particle only needs its forces to balance. A rigid body can also rotate, so its moments must balance too. For a body loaded in one plane, the conditions are ΣFx = 0, ΣFy = 0 and ΣM_O = 0, where O is any point you choose.
In 3D there are six: three force sums and three moment sums, one about each axis. Every equation comes from the free body diagram, so draw it first and write the equations from it, not from the original sketch.
Choosing where to take moments
You can take moments about any point, so choose one where two unknown forces cross. Those forces have zero moment arm and drop out, leaving one unknown in the equation. Pins are usually the best choice.
You may swap a force equation for a second moment equation, as long as the line through the two moment points is not perpendicular to the remaining force direction, or use three moment equations about points that are not on one line. Solving with one set and checking with another is the quickest way to catch an arithmetic error.
Distributed loads, couples and determinacy
Replace a distributed load with its resultant: a force equal to the area under the load curve, acting through that area’s centroid. A uniform load w over a length L becomes w·L at L/2.
A couple moment adds to the moment equation but not to the force equations, and its effect is the same about every point.
A body is statically determinate when the unknowns equal the independent equations. More unknowns means it is indeterminate, and you need deformations from mechanics of materials to solve it. Fewer, or reactions that all pass through one point or run parallel, means it is improperly constrained and can move.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
2D rigid-body equilibrium
ΣFx = 0, ΣFy = 0, ΣM_O = 0
O = any point
Hibbeler, Engineering Mechanics: Statics
3D rigid-body equilibrium
ΣF = 0, ΣM_O = 0 (six scalar equations)
vector sums of forces and moments
Beer & Johnston, Vector Mechanics for Engineers: Statics
Resultant of a uniform distributed load
F_R = w·L, acting at L/2
w = load per length (N/m), L = loaded length (m)
Meriam & Kraige, Engineering Mechanics: Statics
Worked examples
Example: A pin-and-roller beam with an angled load and a distributed load
Given
Beam AB is 4 m long, with a pin at A (x = 0) and a roller at B (x = 4 m) on a horizontal surface. Beam weight is negligible.
A 6 kN force acts at x = 1.5 m, pointing down and to the right at 60° below the horizontal.
A uniform load of 2 kN/m acts downward from x = 2 m to x = 4 m.
Find
The reactions A_x, A_y and B_y.
Solution
FBD: A_x and A_y at the pin, B_y (vertical) at the roller, the 6 kN force split into components, and the distributed load replaced by its resultant.
Components of the 6 kN force:P_x = 6 cos 60° = 3.000 kN (right), P_y = 6 sin 60° = 5.196 kN (down)
Resultant of the distributed load:F_R = 2 kN/m × 2 m = 4 kN, acting at x = 2 + 1 = 3 m
Moments about A, counterclockwise positive, removes A_x and A_y:B_y(4) − 5.196(1.5) − 4(3) = 0 → B_y = 19.794 / 4 = 4.949 kN
Sum forces in x:A_x + 3.000 = 0 → A_x = −3.000 kN
Sum forces in y:A_y + 4.949 − 5.196 − 4 = 0 → A_y = 4.248 kN
Check with moments about B:−4.248(4) + 5.196(2.5) + 4(1) = −16.99 + 12.99 + 4.00 = 0 ✓
Answer: A_x = 3.00 kN acting to the left (the negative sign reverses the assumed direction), A_y = 4.25 kN up, B_y = 4.95 kN up.
Example: A cantilever bracket with a couple
Given
A horizontal cantilever is 2 m long and fixed to a wall at A.
It carries a uniform downward load of 1.5 kN/m along its full length, a 2 kN downward point load at its free end, and a 1.2 kN·m clockwise couple at its free end.
Find
The reactions at the fixed support: A_x, A_y and the moment M_A.
Solution
FBD: a fixed support gives A_x, A_y and a couple moment M_A (assume counterclockwise).
Resultant of the distributed load:F_R = 1.5 kN/m × 2 m = 3 kN, acting 1 m from A
Sum forces in x. No horizontal loads act:A_x = 0
Sum forces in y:A_y − 3 − 2 = 0 → A_y = 5 kN
Sum moments about A, counterclockwise positive. The clockwise couple enters as negative, and it has no moment arm:M_A − 3(1) − 2(2) − 1.2 = 0 → M_A = 8.2 kN·m
Answer: A_x = 0, A_y = 5 kN up, M_A = 8.2 kN·m counterclockwise. The couple changed the moment reaction but not the force reactions.
Common mistakes
Taking moments about a point with several unknowns acting away from it. Pick the point where the most unknowns cross. One unknown per equation means no simultaneous equations.
Putting a distributed load’s resultant at the end of the load. The resultant acts at the centroid of the load area: the middle of a uniform load, one third of the way from the tall end of a triangular one.
Multiplying a couple moment by a distance. A couple is already a moment. Add it to the moment sum as it is, with the right sign.
Mixing sign conventions between equations. State the positive directions once, at the top of the solution, and keep them for every equation.
Calling a negative reaction an error. It only means the assumed direction was wrong. Report the magnitude with the corrected direction.
In industry
Frequently asked questions
Why can I take moments about any point?
When the forces balance, the total moment is the same about every point. So if it is zero about one point, it is zero about all of them, and you can pick whichever point is most convenient.
Can I use three moment equations instead of two force equations?
Yes, as long as the three points are not on one straight line. Two moment equations plus one force equation also work if the line joining the two points is not perpendicular to that force direction.
What makes a problem statically indeterminate?
More unknown reactions than independent equilibrium equations, such as a beam on three supports. Equilibrium alone cannot split the load between the supports; you need the beam’s deflection to do it.
Does the size of the body matter in statics?
Only through dimensions and moment arms. Treating the body as rigid means its deformation is too small to change the geometry, which holds for most structures and machine parts under service loads.