Engineering question: Why does the same downward load produce different support reactions when it moves along a bridge?
Everything needed for the introductory calculations, diagrams and self-checks is provided below.
When you open a door by pushing near the handle, the force acts farther from the hinge and creates more turning effect than the same force applied near the hinge.
For a force perpendicular to a horizontal lever, the moment arm equals the horizontal distance. 30 N × 0.40 m = 12 N·m.
To give the moment a sign, check which way the force tends to rotate the body around the selected point O.
The lever arm is not always the distance along the object. It is the shortest perpendicular distance between the pivot and the infinite straight line of action of the force.
If the force's line of action passes through O, then d⊥ = 0 and its moment about O is zero, even if the force is large.
We will use the conventional sign system: counterclockwise (CCW) is positive, and clockwise (CW) is negative.
Push down to the left of a pivot, or push up to its right: the tendency is counterclockwise.
Push down to the right of a pivot, or push up to its left: the tendency is clockwise.
If a force is not perpendicular to the lever, use the perpendicular force component—or the perpendicular moment arm—to calculate its moment.
Here r is the straight distance from the pivot to where the force acts, and θ is the angle between the position vector and the force vector.
Example: 40 N at 30° below a 0.50 m horizontal lever → |M| = (0.50)(40)(sin30°) = 10 N·m, clockwise, so M = −10 N·m.
For a horizontal lever extending right from the pivot, rᵧ = 0, so M = rₓFᵧ. Horizontal force Fₓ makes no moment about the pivot in this case.
This directly builds on the sine/cosine components in Lesson 2.5.
These two verified lessons cover both the concept of torque/moment and its application to beam support reactions. Use the direct links if YouTube embeds are blocked at school.
Watch and do: Sketch one door or lever example. Label the pivot, force, perpendicular lever arm, and CW/CCW direction.
Watch on YouTube ↗Watch and do: Before the example is solved, draw Aᵧ and Bᵧ. Explain why moments about the left support eliminate its reactions.
Watch on YouTube ↗Always begin with a free-body diagram, choose a pivot, and write the sign convention beside your equations.
Interpretation: doubling the force or lever arm doubles the moment magnitude.
Taking moments about B: ΣMB = +80(3.0) −Aᵧ(4.0) = 240 −240 = 0 N·m.
Why choose a support as pivot? Any force whose line of action passes through that chosen pivot has a zero moment about it, eliminating its unknown magnitude from the equation.
For one downward point load P between a pin at A and roller at B, with span L and load distance a from A. These are equilibrium reactions, not a material strength or deflection calculation.
Use the two models below. Move one slider at a time and make a prediction before checking the calculated answer.
Force applied at the tip of a horizontal lever extending right. Angle θ is measured counterclockwise from +x: −90° means straight down.
Clockwise (negative)
The diagram shows a straight lever and applied force. Arrow length is illustrative; the numbers use F sinθ and rF sinθ.
Slide the vertical load along a 4.0 m simply supported beam. Pin A is at the left, roller B at the right. Support reactions follow static equilibrium.
The load is closer to A, so the left support carries more.
Try: center the load (a = 2.0 m). Why do both reactions become 40 N?
You can investigate turning effects with a ruler or small lever, a fulcrum and teacher-approved small masses. Your drawing and calculations are the main work; any physical test must be instructor-approved.
Safety: No standing on bridges, school fixtures or improvised loaded spans. Use only instructor-approved small weights and supports.
| Type | What to measure or control |
|---|---|
| Independent | Distance from pivot |
| Dependent | Calculated moment magnitude |
| Controls | Same force, lever orientation, pivot and measurement method |
Prediction: Increasing the distance by 50% while force stays constant increases moment magnitude by 50% for a perpendicular force.
Use CCW positive, CW negative. Include moment units and show how you chose the pivot.
A 30 N force acts vertically downward 0.40 m to the right of pivot O. Calculate its signed moment.
M = −(30)(0.40) = −12.0 N·m, clockwise.
A 15 N downward force acts 0.60 m left of pivot O. What is its moment?
Downward to the left produces counterclockwise rotation. M = +(15)(0.60) = +9.0 N·m.
A 50 N force acts 0.40 m to the right of a pivot at 30° above the horizontal. Calculate its moment about the pivot.
M = rF sinθ = (0.40)(50)(sin30°) = +10.0 N·m (CCW).
A 4.0 m simply supported beam carries 80 N downward 1.0 m from A. Find Aᵧ and Bᵧ.
ΣMA = 4Bᵧ−80(1) = 0 ⇒ Bᵧ = 20 N upward. ΣFᵧ = Aᵧ+20−80=0 ⇒ Aᵧ=60 N upward.
A 5.0 m simply supported beam has a 120 N downward load 3.0 m from A. Find the reactions.
ΣMA = 5Bᵧ−120(3) = 0 ⇒ Bᵧ = 72 N upward. Aᵧ = 120−72 = 48 N upward.
Why do statics problems often take moments about a support? What happens if a force's line of action passes through that point?
A force whose line of action passes through the chosen pivot has zero perpendicular moment arm, so its moment about that point is zero. Choosing a pivot at a support can eliminate unknown reaction forces from the moment equation.
Open-access textbooks and references expand the on-page examples, with additional equations and diagrams.
University-level statics chapter describing how to choose a moment center and solve unknown support reactions.
Engineering Statics — Rigid-Body Equilibrium ↗Why translational and rotational equilibrium are both necessary.
Engineering Statics — Equations of Equilibrium ↗Reference diagrams and calculations of pin-and-roller reactions and how they change with load location.
CalcResource — Simply Supported Beam Reactions ↗Moment: turning effect of force about a point.
Moment arm: perpendicular distance to force's line of action.
Moment equilibrium: sum of signed moments equals zero.
Support reaction: force or moment supplied by a restraint.
Complete the POE 2.6 Calculating Moments classwork/homework worksheet or teacher-directed notebook equivalent. Use the on-page resources to learn the method and check your calculations.
Due date: Check the Daily Agenda and follow your teacher's submission instructions.
