Statics

Two conditions, and why one is not enough

A body is in static equilibrium when it has no linear and no angular acceleration. Newton's second law turns the first half into a force statement: the vector sum of all external forces on the body must be zero. The rotational half is a separate statement: the vector sum of all external moments must also be zero.

Students routinely treat the second as bookkeeping that follows from the first. It does not. Picture a wrench on a frictionless table. Push the top of the handle right with 5.00 N and the bottom left with 5.00 N, the two lines of action 0.400 m apart. The force sum is exactly zero, so the center of mass never accelerates, but the moment sum is 5.00 N x 0.400 m = 2.00 N.m, so the wrench spins in place. Zero net force buys a stationary center of mass and nothing else.

In components, three dimensions give three force and three moment equations. In a plane the out-of-plane force equation and two of the moment equations carry no information, leaving three: x-forces, y-forces, and moments about the z-axis. Three is a hard ceiling, and most strategy in statics is about spending those three well. Note also that equilibrium is not the same as rest: a body translating at constant velocity satisfies both conditions, which is why statics covers a conveyor at steady speed or an aircraft in unaccelerated level flight.

The moment of a force about a point

A moment measures how strongly a force tends to rotate a body about a chosen point. In the scalar form used constantly in planar work, its magnitude is the force magnitude times the perpendicular distance from the point to the force's line of action. That distance is the moment arm, and perpendicular is doing all the work: not the distance to the point of application, but the shortest distance to the whole infinite line the force acts along. The unit is the newton meter, N.m, which NIST lists as the SI unit for moment of force.

One consequence is worth memorizing. If the line of action passes through the point you are summing about, the moment arm is zero and the force contributes nothing to that equation. This is exact, not an approximation, and it is the lever that makes moment equations useful.

Sign convention in two dimensions is a choice you make and then keep; the usual one takes counter-clockwise as positive, corresponding to a moment vector out of the page.

In three dimensions the moment is genuinely a vector, M = r x F, where r runs from the moment center to any point on the force's line of action. Any point on that line gives the same answer, which is the vector restatement of the perpendicular-distance rule. Direction follows the right-hand rule, and the magnitude reduces to force times perpendicular distance because the cross product already carries the sine of the angle between r and F.

Sum moments about any point, then choose the point that helps

For a body in equilibrium the moment sum is zero about every point in space. Shifting the moment center changes the total by the shift vector crossed into the total force, and in equilibrium the total force is already zero, so the correction vanishes. Equilibrium is what makes the moment center arbitrary.

That freedom is the most useful tactic in statics. Because a force through your chosen point contributes nothing, pick the point that kills the unknowns you do not want: summing about a pin support removes both of its components at once and leaves one equation in one unknown. Where two unknown lines of action intersect is usually the right point, even if no support sits there.

The same freedom gives a free check: solve using one moment center, then re-sum about a different point. Equilibrium demands zero.

A couple is the limiting case. Two parallel forces, equal, opposite, and not sharing a line of action, give a zero resultant force and a moment equal to one force times the perpendicular distance between the lines. That moment is the same about every point, so a couple can be moved anywhere on a rigid body without changing its external effect, which is why applied torques enter equilibrium equations with no location attached. Relocating a couple does change the internal force distribution, though, even when the reactions are untouched. More generally, any force system on a rigid body reduces to a statically equivalent resultant: a single force through a chosen point plus a couple.

Counting: determinate, indeterminate, and improperly constrained

Before solving anything, count. In two dimensions you have three equations against however many unknown reaction components the supports introduce. If the unknowns number three or fewer and the supports are properly arranged, statics alone finds them and the problem is statically determinate. A pin at one end contributing two components plus a roller contributing one gives exactly three, which is why the simply supported beam is the standard example.

Replace the roller with a second pin and you have four unknowns against three equations: statically indeterminate to the first degree. The structure is real and its forces are definite; statics simply cannot name them, because infinitely many reaction combinations satisfy equilibrium, and equilibrium is all statics knows.

Deformation resolves it. Allowing the structure to stretch and bend adds compatibility, which requires the deformed parts still to fit together, and constitutive relations tying force to deformation through material stiffness. Those carry material properties, so the answer depends on what the structure is made of and how large its cross sections are. That is why indeterminate analysis belongs to mechanics of materials, and why a steel tie and an aluminum tie in the same over-constrained frame do not share the load equally.

Counting is necessary, not sufficient. If all reaction lines of action meet at one point, nothing resists rotation about it; if all reaction components are parallel, nothing resists translation across them. Both are improperly constrained: the count looks right, the equations turn out not to be independent, and the body is unstable. This kind of stability depends on support geometry, not on the loading.

Two-force and three-force members

Two shortcuts pay for themselves repeatedly. A two-force member is a body loaded at exactly two points with no applied couples. Force equilibrium makes the two loads equal and opposite; moment equilibrium then makes them collinear, because equal and opposite forces on separate lines form a couple, and a couple has no moment-free position. So the force in a two-force member acts along the line joining its load points. If the member is straight that means pure axial tension or compression, and one unknown magnitude replaces two unknown components.

A three-force member carries forces at exactly three points. If those forces are not parallel, their lines of action must all pass through one common point. Take moments about where two of the lines cross: those two contribute nothing there, so the third must contribute nothing either, meaning its line passes through the same intersection. A simple beam carrying one load and two reactions is the everyday example; parallel forces are the degenerate alternative.

The payoff is directional information for free: knowing a reaction's direction before writing any equation often collapses a three-unknown problem to one unknown.

Trusses: assumptions first, methods second

A truss is a framework of slender members joined at their ends, and the classical analysis works only because of a stack of idealizations. Every joint is modeled as a frictionless pin, so no joint transmits a moment. Every load and reaction is applied at a joint, never partway along a member. Consequently every member is a two-force member carrying purely axial force, and self-weight is neglected or split onto the two end joints. Real connections are not frictionless pins, so the model reports axial forces well and is blind to whatever secondary bending the real joint introduces.

The method of joints isolates one pin at a time. Forces at a pin are concurrent, so the moment equation gives no information and only two force equations are available per joint. That caps you at two unknowns per joint, so start where unknowns are few, usually at a support once the reactions are known, and walk outward. Convention assumes every unknown member force is tensile; a negative answer means compression.

The method of sections cuts through the truss and treats a whole portion as one rigid body, exposing only the members the cut crosses, so it reaches a mid-span member without solving every joint first. Each cut member is one unknown, so a planar cut through more than three members leaves you short of equations.

Zero-force members carry no load in the case analyzed. Two patterns find them: at an unloaded joint connecting exactly two non-collinear members, both are zero-force; at an unloaded joint where three members meet and two are collinear, the odd one out is zero-force. The member is still not useless - it may brace a compression chord against buckling or be loaded under a different load case. The same two-equations-per-joint fact gives a determinacy count: a planar truss with j joints supplies 2j equations against m unknown member forces plus r unknown reaction components, so m + r = 2j signals static determinacy and more unknowns than that means indeterminate.

Distributed loads, centroids, and center of gravity

Real loads are spread out. Snow on a roof, water against a wall, and a beam's own weight act as an intensity, in newtons per meter for a line load, rather than at a point. For rigid-body equilibrium, replace a distributed load with one equivalent force whose magnitude equals the area under the load-intensity diagram and whose line of action passes through the centroid of that area.

Two cases cover most problems. A uniform intensity w over length L has resultant wL at the midpoint. A triangular load rising from zero to peak w over length L has resultant wL/2 acting one third of the length from the tall end, the centroid of a triangle. Watch units: kN/m times m gives kN.

The centroid is a weighted average of position taken with a purely geometric weight such as area or volume; the center of mass uses mass as the weight, and the center of gravity uses weight. NASA describes the center of gravity as the average location of an object's weight, notes that weight always acts through it, that a freely rotating body rotates about it, and that it lies on any plane of symmetry when mass is uniform.

The three coincide when the body is homogeneous and the gravitational field uniform, because density and gravitational acceleration then factor out of the weighted average and cancel. Both conditions can fail: a plastic housing with a steel insert has its center of mass away from its centroid, and a satellite large enough to feel a gravity gradient has a center of gravity slightly offset from its center of mass.

Friction in equilibrium: impending motion, sliding, and tipping

Friction enters statics as an inequality rather than an equation, which is what makes it awkward. On a dry surface, static friction takes whatever magnitude and direction are needed to prevent sliding, up to a ceiling of the static coefficient times the normal force. Below that ceiling friction is an unknown you solve for like any other. At the ceiling motion is impending, the inequality becomes an equality, and you gain one usable equation. Beyond it the body slides and the kinetic model takes over, with a coefficient that is usually smaller. The coefficient model is an approximate empirical description, not a law, and coefficients for nominally identical surfaces scatter with finish, contamination, and humidity, so treat any tabulated value as an estimate.

Pushing a crate raises a second question: slide or tip? Solve both and compare. The force that starts sliding is the static coefficient times the normal force. The force that starts tipping comes from moment equilibrium about the edge the body would rotate over, using the fact that at impending tipping the normal force has migrated as far as it can and acts right at that edge. Whichever threshold is lower happens first, and the comparison is as much geometric as frictional: a tall narrow object pushed high tips at a force a short wide one would shrug off.

Every model here has edges. Rigid bodies do not deform, real joints are not frictionless pins, and equilibrium written on the undeformed geometry stops being valid once a structure deflects appreciably. Naming the assumption you are standing on is part of the answer.

This lesson is educational material. It is not engineering design guidance, and nothing here should be used to size, check, or approve a real structure, lifting device, or machine. Real design must be performed by a licensed engineer working to the governing code and standards.