Statics
That something does not move does not mean there are no forces: it means they cancel. Two vector equations hold up bridges, shelves and the support of an antenna.
01The two conditions of equilibrium
A body not moving does not mean there are no forces: it means they cancel. And for an extended body it is not enough for the forces to cancel, because a pair of equal and opposite forces applied at different points makes it rotate. Two conditions are needed:
M = F·d, where d is the perpendicular distance from the pivot point to the line of action of the force —the lever arm—. That is why a long wrench loosens what a short one cannot: same force, longer arm.
If ΣF = 0, the total moment has the same value about any point. That freedom is the central tool: you choose as the center the point where the unknowns you do not care about intersect, and they disappear from the equation.
02Constraints and reactions
Each support prevents certain motions, and for every motion prevented an unknown reaction appears:
| Support | Prevents | Unknowns |
|---|---|---|
| Roller support | Displacement perpendicular to the surface | 1 |
| Pin support (hinge) | Both displacements | 2 |
| Fixed support (clamped end) | Both displacements and rotation | 3 |
If there are as many unknowns as equations, the problem is statically determinate and is solved with statics alone. If there are more unknowns than equations it is statically indeterminate, and the deformation of the material has to be considered: that is already strength of materials.
Move the supports and the load. The reactions come from the two equilibrium conditions, and if one comes out negative the bar tips over.
03Center of gravity
It is the point where the whole weight can be considered to act. For a homogeneous body it coincides with the geometric center; for a composite one it is calculated as a weighted average:
A body resting on a surface tips over when the vertical through its center of gravity falls outside the base of support. That is why a tall, narrow cabinet is unstable, why a rack is anchored to the wall and why a bench instrument has its transformer at the bottom: lowering the center of gravity and widening the base are the only two ways to improve stability.
04Beams and trusses
A truss is a structure of bars pinned at their ends and loaded only at the joints. Under those assumptions, each bar works only in tension or compression, never in bending, and this makes it possible to build light and very stiff structures: bridges, cranes and guyed antenna towers.
- Method of joints: set ΣFx = 0 and ΣFy = 0 at each joint.
- Method of sections: cut the structure and balance one of the parts. It is convenient when only a few bars matter.
In a beam, on the other hand, transverse loads produce shear force and bending moment, which vary along its length. Sizing it means making sure the material holds up at the most heavily stressed section.
05Statics for an electronics engineer
- Antenna support. Calculating the tensions in the guy wires and the compression of the mast under wind load is a statics problem. A mistake there takes the whole link down with it.
- Mounting cabinets and racks. The wall anchorage is sized from the moment produced by the weight at a distance.
- Securing boards and heavy components. A large transformer on a circuit board generates moments that fatigue the solder joints if it is not fastened mechanically.
- Load cells. An electronic scale measures deformation and, through it, force: its design starts from the equilibrium of moments in the beam that supports it.
- Cable trays. They are calculated as continuous beams, with their distributed load and their supports.
06In the lab
Rest a board on two kitchen scales and place a weight at different positions. Verify that the sum of the two readings is constant and that the distribution follows the law of moments.
Tilt an empty cabinet and then the same one with a mass at the base. Measure the angle at which it tips and relate it to the position of the center of gravity.
Instrument a cantilever beam with strain gauges in a Wheatstone bridge. Verify that the output is proportional to the bending moment, that is, to the weight times the distance.
07Common mistakes
- Using the distance to the force instead of the perpendicular distance to its line of action.
- Forgetting the self-weight of the bar or the structure.
- Taking moments about a poorly chosen point and ending up with two unknowns in one equation, when choosing a support would have eliminated one.
- Confusing mass with weight when entering the data: scales give kilograms, the equations require newtons.
- Accepting a negative reaction as valid at a support that can only push. It means that the structure tips over or that an anchor is needed.
08Self-assessment
A weightless 4 m bar, supported at its ends, carries 600 N at 1 m from the left end. What is each reaction?
Moments about A: RB·4 = 600·1, so RB = 150 N and RA = 450 N. The load is shared in inverse proportion to the distances.
Why is it convenient to take moments about a support?
Because the reaction at that support has zero lever arm and disappears from the equation, which is left with a single unknown.
When does a body resting on a surface tip over?
When the vertical through its center of gravity falls outside the base of support.
How many unknowns does a fixed support contribute in the plane?
Three: two force components and a moment.
Why do the bars of an ideal truss not work in bending?
Because they are pinned at their ends and loaded only at the joints: each bar receives forces only at its two ends, and to be in equilibrium they must be aligned with its axis.
A 30 kg cabinet hangs with its center of gravity 25 cm from the wall. What moment does the anchor carry?
M = 30·9.8·0.25 ≈ 73.5 N·m, shared among the screws according to their vertical spacing.
09Further reading
- Ferdinand P. Beer and E. Russell Johnston. Vector Mechanics for Engineers: Statics. 10th ed., McGraw-Hill, 2013 (Spanish edition, “Mecánica vectorial para ingenieros: estática”). The reference text: constraints, trusses and centers of gravity, with problems on real structures.
- Raymond A. Serway and John W. Jewett. Physics for Scientists and Engineers with Modern Physics, Volume 1. 9th ed. (Spanish edition, “Física para ciencias e ingeniería”), Cengage Learning, 2014. Chapter 12 treats the equilibrium of the rigid body with a basic-physics approach.
- Russell C. Hibbeler. Engineering Mechanics: Statics. 12th ed., Pearson, 2010 (Spanish edition, “Ingeniería mecánica: estática”). Very graphic, with step-by-step free-body diagrams.