ENGR 201 Fall 2026 - Lecture 07: 2D Moment

Overview

Bruno Guidio, PhD, introduces 2D moments in statics, explaining how force location and body size matter when a rigid body can no longer be modeled as a particle. He develops the signed moment calculation, M_O = Fd, through perpendicular-distance diagrams, force components, and multi-force examples, using counterclockwise as positive and clockwise as negative.

Key takeaways

For a force F about point O, moment magnitude is F times the shortest perpendicular distance from O to the force's line of action; the arrow's endpoints do not define the lever arm.
A force whose line of action passes through the reference point has zero moment about that point, even if the force is large.
With the lecture's sign convention, counterclockwise moments are positive and clockwise moments are negative; this sign is independent of the force's Cartesian direction.
For an angled force on a horizontal bar, resolving it into horizontal and vertical components can simplify moment calculations because each component pairs with a clear perpendicular distance.
Moments about multiple forces must be added algebraically, preserving each force's sign; the four-force example totals approximately +56.57 N·m.
Changing the reference point changes lever arms and may change both the magnitude and sign of individual moments, so every moment calculation must specify its point of reference.

Chapters

0:00 Course Schedule and the Shift from Particle Models

Homework 3 was due the previous night; the current assignment is due Tuesday rather than Monday.
The course will cover 2D moments first, 3D moments on Thursday, then a new topic and an exam review before Exam 1 in early October.
Earlier particle problems ignored body size, but rigid-body analysis must account for force locations and the body's dimensions.

2:20 Why Force Location Matters for Rigid Bodies

A person sitting at different positions on a table produces different effects because the load acts at different locations.
A force applied away from a support can create a tendency to rotate the body, unlike a model that treats all forces as acting at one point.
Moment analysis extends force-based engineering problems to bodies whose size and geometry matter.

5:30 Moment as a Force's Tendency to Rotate a Body

Moment is defined as the tendency of a force to rotate a body about a point, such as point O.
A door illustrates the concept: pushing at a location away from its hinges tends to rotate it about the hinge axis.
Moment is a vector quantity with a magnitude and a rotational direction.

8:34 Moment Magnitude and Perpendicular Distance

The magnitude of a moment about O is calculated as M_O = Fd, where F is force magnitude and d is the perpendicular distance.
Distance d is measured at 90 degrees from the force's line of action to point O—not from the arrow's tail, tip, or midpoint.
The same force can produce different moments about different reference points because the perpendicular distance changes.

11:14 Sign Convention: Counterclockwise Positive

Guidio adopts counterclockwise rotation as positive and clockwise rotation as negative.
The sign describes the force's rotational effect about the chosen point, not whether the force points up, down, left, or right.
Moment units combine force and distance, such as newton-meters (N·m), not newtons per meter.

15:39 A 10 N Force at 4 m Produces −40 N·m

For a 10 N vertical force with a 4 m perpendicular offset from O, the moment magnitude is 10 × 4 = 40 N·m.
Because the force tends to rotate the bar clockwise, the signed moment is −40 N·m.
Guidio suggests using a pencil as a bar to physically visualize clockwise and counterclockwise rotation.

21:39 Summing Moments and Recognizing Zero-Lever-Arm Forces

For the initial two-force example, the 10 N force contributes −40 N·m while a 15 N force whose line of action passes through O contributes zero.
The total moment about O is the algebraic sum of each force's moment.
Moving O changes the lever arms and can change a force's rotational sign; in the revised arrangement, the 15 N force contributes a counterclockwise moment.

27:46 Resolve an Angled Force into Moment-Producing Components

An 18 N force at 30° is resolved into horizontal and vertical components to use simple perpendicular distances.
With O on the bar's horizontal level, the 18 sin 30° component acts at a 4 m lever arm and produces −36 N·m.
The 18 cos 30° component's line of action passes through O, so its lever arm and moment are zero.

33:54 Changing the Reference Point Changes the Angled-Force Moment

When O is moved below the bar, both components of the 18 N force have 4 m perpendicular lever arms.
The vertical component, 18 sin 30°, produces −36 N·m, while the horizontal component, 18 cos 30°, produces about +62.35 N·m.
The net moment is approximately +26.35 N·m, demonstrating that a moment depends on the selected reference point.

39:30 Practice with Multiple Forces and Signed Moments

A multi-force bar exercise asks students to calculate each force's moment about O and add the signed results.
Guidio emphasizes identifying each line of action, perpendicular distance, and rotational direction before multiplying force by distance.
The example distinguishes force direction from moment sign: a rightward force can still produce a clockwise, negative moment.

49:20 Use the True Perpendicular Offset, Not Bar Length

In a geometry example with a 5 m bar and a 3–4–5 triangle, the vertical force's perpendicular distance to O is 3 m, not the 5 m bar length.
A 10 N vertical force therefore produces a moment magnitude of 10 × 3 = 30 N·m, with a negative sign for clockwise rotation.
For an inclined bar and force, Guidio notes that direct perpendicular distance or component resolution can be used, but the chosen distances must match the components.

53:57 Set Up a Four-Force Moment Calculation

The final practice setup places forces at intervals of 2 m and asks for the total moment about O.
The forces are A = 10 N, B = 15 N at 45°, C = 20 N at 30°, and D = 25 N.
Each contribution is handled separately as M_O due to A, B, C, or D before summing.

1:02:08 Solve the Four-Force Example: Net Moment +56.57 N·m

Force A contributes +10 × 2 = +20 N·m; for B, the 15 sin 45° component passes through O and contributes zero, while 15 cos 45° produces about +21.21 N·m.
Force C contributes −20 cos 30° × 2 ≈ −34.64 N·m; its 20 sin 30° component has zero lever arm.
Force D contributes +25 × 2 = +50 N·m, giving a total of about +56.57 N·m counterclockwise.
Guidio closes by noting that the same moment method applies to engineering structures and biomedical or human-body problems.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Bruno Guidio, PhD.