Metatransformer report
Forces and free-body diagrams
Foundation · 45 minutes · Turn a physical assembly into a force balance before choosing a bracket, axle, or actuator.
Text report: images and diagrams are omitted, and links to project files are inactive. Keep the exported project alongside this PDF for wiring diagrams, models, and linked files.
Learning outcomes
- Draw a free-body diagram for one isolated part and solve a two-dimensional static force balance.
- Use the result to place supports and estimate an actuator or structural load.
Prerequisites
- algebra-trig
- vectors-frames
Concepts
A shelf, camera arm, or toy-car axle can look motionless while several forces act on it. The useful first move is to choose one object, mentally remove everything touching it, and replace every contact with a force or moment. That sketch is a free-body diagram. Gravity acts through the center of mass; a cable pulls along itself; a smooth support pushes normal to its surface; a pin can usually react in two directions. Do not draw forces that the isolated object exerts on something else.
Static equilibrium means acceleration is zero, not that forces are absent. In a Cartesian frame, the formal conditions are ΣF_x = 0, ΣF_y = 0, and ΣM_O = 0 about any chosen point O. Weight is W = mg, in newtons, and a force's moment magnitude is M = Fd_perpendicular, in newton-metres. Choosing the moment point at an unknown support often removes that unknown from the moment equation. For accelerating objects, replace zero with ma: ΣF = ma.
Suppose a small screen is mounted on a horizontal arm. Its weight and the arm's weight both create moments at the wall; the bracket reactions close the force balance. This calculation drives a real build decision: moving the screen inward can reduce bracket moment more effectively than merely using a thicker decorative cover.
A common failure is to balance force magnitudes while ignoring direction or lever arm. Equal upward and downward forces can still form a nonzero couple, and a force passing through the pivot creates no moment about that pivot. Another mistake is to add a mysterious “centripetal force”; that phrase names the net inward force required for curved motion, not an extra interaction. Always label the body, frame, application point, and units before writing equations.
Worked example
Problem
A 1.2 kg camera sits 0.30 m from a wall pivot on a massless horizontal arm. A vertical cable attaches 0.40 m from the pivot. Find the cable tension and pivot's vertical reaction using g = 9.81 m/s².
Steps
- Convert mass to weight: W = 1.2 kg × 9.81 m/s² = 11.77 N downward.
- Take moments about the pivot: T(0.40 m) - 11.77 N(0.30 m) = 0, so T = 8.83 N upward.
- Balance vertical forces: R_y + 8.83 N - 11.77 N = 0, so R_y = 2.94 N upward.
Result
The cable carries 8.83 N and the pivot supplies 2.94 N upward.
Sanity check
The cable attaches farther from the pivot than the load, so its force should be smaller than the 11.77 N weight; 8.83 N is consistent.
Check your understanding
A downward force on a lever is moved from 50 mm to 100 mm from its pivot without changing magnitude. What happens to its moment about the pivot?
A. It halves B. It stays unchanged C. It doubles D. It quadruples
Your answer: ______________________________
Reasoning:
Answer key
ANSWER KEY — check only after attempting the question.
C. It doubles
Moment is force times perpendicular distance, so doubling the distance doubles the moment. Choosing unchanged confuses force balance with moment balance; quadrupling incorrectly squares distance; halving reverses the lever-arm relationship.
Hands-on lab
Goal
Map support reactions on a tabletop beam and diagnose an omitted load.
Equipment
- 30–50 cm ruler
- Two kitchen scales or spring scales
- Known small masses
- Tape measure
Procedure
- Rest the ruler on two scales, zero if possible, and record support spacing and the ruler's own baseline readings.
- Place a known mass at three measured positions and predict both support forces from force and moment balance before reading the scales.
- Record actual readings separately from predicted values and compare their sum with total weight.
- Deliberate fault: hide a small taped mass under one end; use the changed total and moment distribution to infer which side contains the unmodelled load.
What to observe
- Expected: moving the mass toward one support raises that support's share and lowers the other's.
- Measure: actual support readings, positions, and residuals; friction, scale zero error, and ruler weight may prevent exact agreement.
Transfer challenge
Draw a free-body diagram for a desk-robot sensor mast and decide where a brace would most reduce base moment.
Cautions
- Keep the beam and masses low over the table so a slip cannot damage feet or equipment.
Evidence record
Learner: ____________________ Date: ____________________
Evidence artifact or file: ____________________________________________
| Item | Record |
|---|---|
| Prediction or expected result | |
| Actual observation or measurement | |
| Conditions and tools used | |
| Difference from expectation | |
| Next decision or follow-up |
Notes: