3.7 — Worked problems — Chapter 3 toolkit in action

Seven worked problems exercising every Chapter 3 tool. Each has a step-by-step solution tagged with the relevant §3.x section.

Problem 1 — adding two forces (§3.4)

Scenario: two ropes pull a load:

Find the total force.

Solution:

Decompose:

Add:

Magnitude: F = \sqrt{143.9^2 + 129.3^2} \approx 193.5\ \mathrm{N}

Angle: \theta = \arctan(129.3/143.9) \approx 41.9°

Problem 2 — a moving river (§3.2)

Scenario: a boat can move at 4 m/s relative to the water. The river flows east at 3 m/s. The pilot wants to head due north. What angle should she aim, and what is the actual ground speed?

Solution:

\vec v_{boat/ground} = \vec v_{boat/water} + \vec v_{water/ground}

For the net velocity to be due north, the boat's x component must cancel the current.

Pilot aims at angle \theta west of north:

\[ \theta = \arcsin(0.75) \approx 48.6°\ \text{west of north} \]

Actual northward speed: v = 4\cos\theta = 4 \cdot 0.661 \approx 2.65\ \mathrm{m/s}

Problem 3 — work on an inclined slope (§3.5)

Scenario: you push a 50 kg crate with 300 N parallel to a 20° slope, moving 20 m up. Work done by your force?

Solution:

Force and displacement are both along the slope, so the angle between them is zero:

\[ W = \vec F \cdot \vec d = F d \cos 0° = 300 \cdot 20 \cdot 1 = 6000\ \mathrm{J} \]

Bonus — work done by gravity?

Gravity \vec F_g = -mg\hat j = -490\hat j\ \mathrm{N}. Displacement \vec d = 20\cos 20° \hat i + 20\sin 20° \hat j = 18.79\hat i + 6.84\hat j\ \mathrm{m}.

\[ W_g = \vec F_g \cdot \vec d = 0 + (-490)(6.84) \approx -3352\ \mathrm{J} \]

Negative (gravity opposes climbing). Net work: 6000 - 3352 = 2648 J → kinetic energy.

Problem 4 — angle between two vectors (§3.5)

Scenario: \vec a = 2\hat i - \hat j + 3\hat k, \vec b = -\hat i + 2\hat j + \hat k. Angle between them?

Solution:

\[ \vec a \cdot \vec b = 2(-1) + (-1)(2) + 3(1) = -1 \]

\[ |a| = \sqrt{14},\quad |b| = \sqrt{6} \]

\[ \cos\phi = \frac{-1}{\sqrt{84}} \approx -0.109 \]

\[ \phi = \arccos(-0.109) \approx 96.3° \]

Problem 5 — hand-computed torque (§3.6)

Scenario: a bolt at (0, 0, 0). Force \vec F = 20\hat j\ \mathrm{N} applied at position \vec r = 0.15\hat i\ \mathrm{m} (a 15 cm wrench along x). Torque?

Solution:

\[ \vec \tau = \vec r \times \vec F = (0.15\hat i) \times (20\hat j) = 3(\hat i \times \hat j) = 3\hat k\ \mathrm{N \cdot m} \]

Three newton-meters in +z direction — counterclockwise from above.

Change of force direction: replace \hat j with \hat i (along the wrench). \hat i \times \hat i = 0 → zero torque. Pulling the wrench doesn't twist the bolt.

Problem 6 — cross product and area (§3.6)

Scenario: a triangle with vertices A = (1, 0, 0), B = (0, 2, 0), C = (0, 0, 3). Its area?

Solution:

Two side vectors from A:

Cross product:

\[ \vec{AB} \times \vec{AC} = 6\hat i + 3\hat j + 2\hat k \]

Magnitude:

\[ |\vec{AB} \times \vec{AC}| = \sqrt{36 + 9 + 4} = 7 \]

Triangle area = half parallelogram area = 7/2 = 3.5 square units.

Problem 7 — unit vector in an arbitrary direction (§3.3)

Scenario: \vec v = 6\hat i - 8\hat j + 10\hat k. Unit vector in this direction?

Solution:

\[ |\vec v| = \sqrt{36 + 64 + 100} = 10\sqrt{2} \approx 14.14 \]

\[ \hat v = \frac{6\hat i - 8\hat j + 10\hat k}{10\sqrt{2}} \]

Decimal:

\[ \hat v \approx 0.424\hat i - 0.566\hat j + 0.707\hat k \]

Test: |\hat v| = \sqrt{0.18 + 0.32 + 0.5} = 1\ ✓

Wrap-up

Strategy for any vector problem:

  1. Pick a coordinate system that simplifies (sometimes tilted axes)
  2. Decompose to components — always start here
  3. Perform operations component-wise (add, dot, cross)
  4. At the end, recover magnitude/angle if needed
  5. Sanity check — units, magnitude, direction, sign

Preview of §3.8

Twenty practice problems without step-by-step solutions — with final answers for self-check.

📚 See also: Halliday Vol 1, Ch 3 — Worked Examples.

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