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Engineering & Technology · Forces, Energy & Simple Engineering Systems

Forces, moments, supports and simple equilibrium

A narrow board rests on two supports. Put an object closer to the left end and the left support carries more of its weight. The total support force still balances the weight, but equal sharing is no longer the right prediction.

Useful preparation is Measurement, Units & Engineering Evidence and Engineering Sketches, Models & Technical Communication. Be comfortable with multiplication, division, dimensions and reading a simple graph. The lessons introduce each force, motion and energy relationship without calculus. Supplied numerical values, device records, limits and SVG models are original constructed teaching cases, not equipment specifications, safe loads or instructions to operate machinery, electrical or heating equipment.

Back to the chapter

After reading, you can

Draw forces on one selected body and check both translation and turning balance in a simple static model.

01

A force diagram needs a receiver

A force is a directed push or pull, measured in newtons, N. Weight is the gravitational force acting on an object; mass is measured in kilograms and describes a different quantity. Begin by choosing the body whose motion or support you want to understand. An arrow on its force diagram represents a force acting on that body, with an identified source.

For the board, include forces from the load, the supports and Earth if the board’s own weight matters. The board also pushes on its supports, but those forces act on the supports and belong in their diagrams. Equal-and-opposite interaction pairs do not cancel each other within a diagram of only one of the bodies.

02

Equal total forces can still turn a body

For a body initially at rest to remain in static equilibrium in this rigid-body model, the net external force and the net external moment must both be zero. Balance forces in each relevant direction. A moment, also called torque, describes a force’s turning effect about a chosen point or axis. Its magnitude is force multiplied by the perpendicular distance from that point to the force’s line of action.

In our beam sketch all forces are vertical, so their moment arms about the left support are horizontal distances. The 60 N load at 0.4 m tends to turn the beam clockwise. The right reaction at 1.2 m tends to turn it counterclockwise. The left reaction passes through the chosen point and has zero moment about it. Choosing that point simplifies the arithmetic; it does not remove the left support force.

Both support force and turning effect must balanceA weightless rigid beam spans 1.2 m, with a 60 N load 0.4 m from its left support. Upward support reactions are 40 N at left and 20 N at right. Horizontal positions and force-arrow lengths use consistent scales.60 N down40 N up20 N up0.4 m1.2 m between supportsBeam weight neglected
Left support at 0 m, load at 0.4 m, right support at 1.2 m. Downward load 60 N; upward reactions 40 N left and 20 N right. The total upward force is 60 N. About the left support, clockwise moment is 60×0.4=24 N·m and counterclockwise moment is 20×1.2=24 N·m. Beam weight is neglected.
03

Equilibrium is one part of a design assessment

Moment is measured in N·m. Although force times distance also appears later in work calculations, moment is a turning quantity, not an energy amount. Keep the name and direction with the unit. A slanted force requires its perpendicular moment arm; simply multiplying by the full length of a handle can give the wrong result.

The beam model treats the board as rigid, neglects its weight and assumes the stated supports provide the required reactions. It predicts reaction forces under those assumptions. It does not establish that the board, fasteners or supports resist deformation, buckling, sliding or failure. A balanced model can describe forces in a design that is still unsuitable.

Find the unequal support reactions

The support spacing is 1.2 m and the downward load is 60 N at 0.4 m from the left support. Its clockwise moment about the left support is 60 × 0.4 = 24 N·m.

For turning balance, the right upward reaction multiplied by 1.2 m must equal 24 N·m. Therefore the right reaction is 24 ÷ 1.2 = 20 N. Vertical force balance then needs left reaction + 20 = 60 N, so the left reaction is 40 N.

Check both conditions: 40 + 20 − 60 = 0 N, and 20 × 1.2 − 60 × 0.4 = 0 N·m about the left support. Equal 30 N reactions would satisfy the first equation but fail the second. The load’s position is why equal sharing was wrong.

Try the reasoning

  1. Why are the board’s downward forces on its supports absent from a force diagram of the board?

    Show the answer and reasoning

    They act on the supports. The corresponding upward forces exerted by the supports act on the board and belong in its diagram.

  2. Can equal upward and downward totals alone establish static equilibrium?

    Show the answer and reasoning

    No. An unbalanced turning effect can remain. Both net force and net moment must be checked in the rigid-body model.

  3. If a force’s line of action passes through the chosen moment point, what moment does that force have about it?

    Show the answer and reasoning

    Zero, because its perpendicular moment arm is zero. The force can still matter in the force-balance equation.

What to carry forward

Select the body, name the forces acting on it and check both net force and net moment before interpreting a support model.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. OpenStax2 sources
    1. University Physics Volume 1 — 12.1 Conditions for Static EquilibriumOpenStaxTeaching resourceopenstax.orgRead source
    2. University Physics Volume 1 — 10.6 TorqueOpenStaxTeaching resourceopenstax.orgRead source