Physics for ME · Module 2 of 16
Vectors and Coordinate Systems in Physical Problems
Math Module 3 built the vector toolkit. Here it meets physics: displacements, velocities, and forces that add, project, and combine in real situations.
Readiness check
From Math Module 3. Tick only what you can do closed-notes.
- Resolve any vector into components with correct signs.
- Recombine components into magnitude and direction.
- Build a unit vector between two points.
- Use the dot product for angles and projections.
- Choose and declare a coordinate system before computing.
The core idea
Physical vectors add tip to tail, and the right coordinate frame makes a hard problem easy.
vtotal = v₁ + v₂ (componentwise)Relative motion is vector addition: the velocity you observe is the object's velocity plus the carrier's. Choosing axes along and across the natural directions of a problem (a river, a slope, a track) splits one tangled question into two clean ones.
The skills, taught in order
Vectors carry both size and direction, so they keep their own arithmetic. These seven skills are that arithmetic put to physical work, and they are exactly what Statics and Dynamics assume you already own.
2.1 Displacement, velocity, and force as vectors
A vector needs a magnitude and a direction; a scalar (mass, time, energy, temperature) needs only a magnitude. Displacement, velocity, acceleration, and force are vectors, which is why none of them can be handled by ordinary arithmetic alone.
2.2 Components in physical frames
Pick axes, then split each vector into the parts that lie along them. A force F at angle θ above the x-axis becomes (F cos θ, F sin θ). On a slope, axes taken along and across the surface are almost always the right choice.
Micro-example. A 900 N weight on a 25° ramp, in ramp-aligned axes, has 900 sin 25° = 380 N along the slope and 900 cos 25° = 816 N into it.
2.3 Vector addition in real problems
Add vectors tip to tail, or add their components separately. The two methods always agree; components win the moment real numbers are involved.
2.4 Relative velocity
What you observe is the object's velocity within its carrier plus the carrier's velocity in your frame. Track the frames with subscripts so they cancel correctly.
Micro-example. vboat/ground = vboat/water + vwater/ground: the inner "water" labels cancel, leaving boat over ground.
2.5 Projections and the dot product
The dot product a·b = |a||b| cos θ extracts the part of one vector that lies along another. It answers "how much of this force acts along the motion?" (which is work) and "what is the angle between these directions?"
Micro-example. A 50 N force at 60° to a 4 m displacement does work F d cos θ = 50 × 4 × cos 60° = 100 J: only the along-component counts.
2.6 Position vectors and unit vectors
The direction from point A to point B is the unit vector u = (B − A) / |B − A|. Multiplying a magnitude by a unit vector lays that magnitude along a known line, which is how a cable's tension becomes three force components.
Micro-example. From (0, 0, 0) to (3, 4, 0): the displacement is (3, 4, 0), length 5, unit vector (0.6, 0.8, 0). Worked Example 2 does this in three dimensions.
2.7 Choosing coordinate systems
The frame is a free choice, and a good one halves the algebra. Align the axes with the natural directions of the problem (the flow, the slope, the track, the cable) so the fewest vectors need resolving.
| Operation | Formula | What it gives |
|---|---|---|
| Magnitude | |v| = √(vx² + vy² + vz²) | the size of the vector |
| Unit vector | u = v / |v| | direction only, length 1 |
| Addition | (ax + bx, ay + by, …) | the combined (resultant) vector |
| Dot product | a·b = |a||b| cos θ | a projection, work, or the angle |
| Scaling | T·u | a magnitude placed along a direction |
Engineering connection: the essential bridge into Statics and Dynamics. It is the physical twin of Math Module 3: vector language becomes force and motion language here.
Worked example 1: the ferry crossing
A ferry heads straight across an 80 m wide river at 4 m/s relative to the water. The river flows at 3 m/s. Find the ferry's actual speed and direction, the crossing time, and how far downstream it lands.
- ProblemFind the resultant velocity, crossing time, and drift in Figure 1.
- Given / findvboat/water = 4 m/s across; vwater = 3 m/s downstream; width 80 m.
- AssumptionsUniform flow, constant heading, still start ignored.
- ModelAxes across (y) and downstream (x). The observed velocity is the vector sum: components never interfere.
- Equationsv = (3, 4) m/s t = width / vy
- Solve|v| = √(3² + 4²) = 5 m/s, at θ = tan⁻¹(3/4) = 36.9° downstream of straight across. Crossing: t = 80/4 = 20 s. Drift: 3 × 20 = 60 m.
- CheckThe 3-4-5 triangle confirms the magnitude. Independence check: the crossing time used only the across component; the flow cannot slow the crossing, only displace it.
- ConclusionComponents solved two questions independently that look tangled together. The same decomposition runs every conveyor-loading, crosswind-landing, and moving-platform problem in Dynamics.
Worked example 2: a guy cable in three dimensions
A guy cable supports a vertical mast. It runs from the mast top B = (0, 0, 6) m down to a ground anchor A = (4, 3, 0) m and carries a tension of 500 N. Resolve that tension into its x, y, and z components, which is what every 3D support analysis in Statics begins by doing.
- ProblemResolve the 500 N cable tension from B = (0, 0, 6) m to A = (4, 3, 0) m into x, y, z components.
- Given / findT = 500 N acting along B toward A. Find Fx, Fy, Fz.
- AssumptionsStraight, taut cable; the tension acts along the line from B toward A.
- ModelBuild the position vector from B to A, turn it into a unit vector, and scale it by the 500 N tension.
- Equationsr = A − B = (4, 3, −6) m u = r / |r| F = T·u
- Solve|r| = √(4² + 3² + 6²) = √61 = 7.81 m. u = (0.512, 0.384, −0.768). F = 500u = (256, 192, −384) N.
- Check|F| = √(256² + 192² + 384²) = 500 N, recovering the tension exactly. The signs read correctly: the cable pulls the mast top toward +x, +y, and downward (−z).
- ConclusionA magnitude times a unit vector is the entire recipe for placing a force in space. Statics Module 2 resolves every 3D support and cable this way; its only addition is summing several of them at a joint.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| The flow "slows the crossing" | Crossing time computed from the 5 m/s resultant | "Which component moves the boat across?" | Perpendicular components are independent. Crossing uses 4 m/s; drift uses 3 m/s. |
| Magnitudes added head-on | 4 + 3 = 7 m/s reported | "Are these velocities collinear?" | Tip to tail or components. Only parallel vectors add by magnitude. |
| Frames silently mixed | Boat-relative and ground-relative speeds in one equation | "Relative to what is each velocity measured?" | Label every velocity with its frame: vboat/water + vwater/ground = vboat/ground. |
| Axes changed mid-problem | Signs flip between equations | "Did I keep the frame I declared in step one?" | Draw the axes on the sketch and never move them within a solution. |
Practice ladder
A drone flies 300 m east, then 400 m north. Find the displacement magnitude and bearing.
Show answer
500 m (3-4-5), at tan⁻¹(300/400) = 36.9° east of north.
For the ferry example: at what angle upstream must the pilot head to land directly across, and how long does that crossing take?
Show answer
The upstream component must cancel 3 m/s: sin θ = 3/4, θ = 48.6° upstream. Across component = √(16 − 9) = 2.65 m/s, so t = 80/2.65 = 30.2 s. Landing straight costs ten extra seconds.
A 900 N crate sits on a 25° ramp. Using ramp-aligned axes, find the components of its weight along and normal to the surface, and state which one friction must resist.
Show answer
Along: 900 sin 25° = 380 N (friction resists this). Normal: 900 cos 25° = 816 N. Check: √(380² + 816²) = 900 N. The slope-aligned frame made gravity the only vector needing decomposition.
Watch a crosswind landing video or observe a conveyor transfer. Identify the three velocities and their frames, estimate magnitudes, and draw the vector triangle that the pilot or machine is solving.
What good work looks like
Three labeled velocities with frames (craft/air, air/ground, craft/ground), a closed vector triangle with estimated numbers, and one sentence on what limit (crab angle, belt speed) the geometry imposes.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Produce a half-page "Frames Card": the relative-velocity chain rule with your own notation, the ferry example worked in both headings (straight across, and aimed upstream), and one rule you will follow about declaring axes.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the relative-velocity chain rule with frame subscripts.
vA/C = vA/B + vB/C: inner subscripts must match and cancel.
2. Why are perpendicular components independent?
Each axis's motion is driven only by that axis's component; projections onto perpendicular directions share nothing.
3. What makes a coordinate frame "good" for a problem?
It aligns axes with the natural directions (slope, flow, track) so the fewest vectors need decomposition.
4. On a slope of angle θ, what are the weight components?
W sin θ along the slope, W cos θ normal to it: the pair Statics and friction problems reuse constantly.
5. When may you add speeds directly?
Only when the velocities are along one line. Otherwise components or tip-to-tail addition.