M3c Vectors and components

In this chapter

• Tell scalars (a size) from vectors (a size and a direction).
• Add vectors tip to tail, and find the resultant's size and direction.
• Stretch, flip and subtract vectors.
• Split a vector into components, and put it back together.
• Add several vectors by adding their components.
• Resolve a weight on a slope into parts along and into the slope.

Scalars and vectors

Checkpoint: Scalars and vectors (1 of 2)

Checkpoint: Scalars and vectors (2 of 2)

Adding vectors

Adding vectors

Adding vectors

Adding vectors

The order doesn't matter; drawing both from one point, the resultant is the diagonal of the parallelogram (Newton, 1687).

Multiplying by a number stretches a vector (\(2\mathbf{a}\) is twice as long, the same way); \(-\mathbf{a}\) points the opposite way; and \(\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b})\).

Adding vectors

Adding vectors

Checkpoint: Adding vectors (1 of 3)

Checkpoint: Adding vectors (2 of 3)

Checkpoint: Adding vectors (3 of 3)

Components

Components

Components

Components

Components

Components

Checkpoint: Components (1 of 3)

Checkpoint: Components (2 of 3)

Checkpoint: Components (3 of 3)

Adding with components

Adding with components

Adding with components

Checkpoint: Adding with components (1 of 2)

Checkpoint: Adding with components (2 of 2)

Forces on a slope

Forces on a slope

Forces on a slope

Forces on a slope

(The angle between the weight and the line into the slope equals the slope's angle.) Checks: on flat ground, \(\sin 0^\circ = 0\); on a vertical wall, \(\sin 90^\circ = 1\).

Forces on a slope

Forces on a slope

Checkpoint: Forces on a slope (1 of 2)

Checkpoint: Forces on a slope (2 of 2)

Vector mistakes

Checkpoint: Vector mistakes (1 of 2)

Checkpoint: Vector mistakes (2 of 2)

Key terms

Scalar / vector

A quantity with a size only / with a size and a direction.

Magnitude

The size of a vector, \(|\mathbf{v}|\).

Resultant

The single vector equal to the sum of several (tip to tail, start to finish).

Components

The parts of a vector along two directions at right angles, e.g. \(v\cos\theta\) and \(v\sin\theta\).

Key terms

Resolving

Splitting a force into components along chosen directions (e.g. along and into a slope).

Relative velocity

Velocity measured from something that is itself moving; velocities add as vectors.

Newton's parallelogram (1687)

Simon Stevin's wreath (Leiden, 1586)

Gibbs and Heaviside (1880s)

The problem

Problem. (Illustrative numbers.) A plane must fly 600 km due north at an airspeed of 250 m/s, through a wind from the west of 30 m/s. (a) If it points due north, where does it end up? (b) Which way should it point, and how long does the trip take?

Solution

1. Pointing north: ground velocity \(= (30\ \text{east}, 250\ \text{north})\): \(\sqrt{250^2 + 30^2} \approx 252\) m/s at \(\tan^{-1}(30/250) \approx 6.8^\circ\) east of north. Going 600 km north takes \(600\,000/250 = 2400\) s (40 min), and the wind carries it \(30 \times 2400 = 72\) km east.
2. Aiming: point at angle \(\varphi\) west of north so the westward component cancels the wind: \(250\sin\varphi = 30\), \(\sin\varphi = 0.12\), \(\varphi \approx 6.9^\circ\) (heading about 353°).

Solution

3. Speed and time: northward speed \(250\cos 6.9^\circ \approx 248.2\) m/s; time \(600\,000/248.2 \approx 2417\) s ≈ 40.3 min, about 17 s longer than in still air.
4. Checks: no wind gives \(\varphi = 0\); a wind as fast as the plane gives \(\varphi = 90^\circ\) and no progress.

★ Warm-up

1. A force of 80 N acts at 35° above the horizontal. Find its vertical component, in newtons. Then find the horizontal one too.

   Your answer NCheck

   Solution

   \(80\sin 35^\circ \approx 45.9\) N up; \(80\cos 35^\circ \approx 65.5\) N across.

Problem 2 (★ Warm-up)

2. A vector has components \((-12, 5)\). Find its magnitude. Then find its direction, measured anticlockwise from the positive \(x\) axis (sketch it first).

   Your answer Check

   Solution

   \(\sqrt{144 + 25} = 13\). It points left and up, so the angle is \(180^\circ - \tan^{-1}(5/12) \approx 157.4^\circ\), not the calculator's −22.6°.

Problem 3 (★ Warm-up)

3. Add 7 m east and 24 m north. How far is the end from the start, in metres? Then which way is it?

   Your answer mCheck

   Solution

   \(\sqrt{49 + 576} = 25\) m, at \(\tan^{-1}(24/7) \approx 73.7^\circ\) north of east.

★★ Standard

4. Three displacements. A walker goes 5 km east, then 4 km north, then 6 km south-west (45° south of west). How far is she from the start, in kilometres? Then which way is she from the start?

   Your answer kmCheck

   Solution

   Leg 3: \((6\cos 225^\circ, 6\sin 225^\circ) \approx (-4.24, -4.24)\). Total \((5 + 0 - 4.24,\ 0 + 4 - 4.24) \approx (0.76, -0.24)\) km: about 0.80 km, 17.8° south of east. Almost back where she started.

Problem 5 (★★ Standard)

5. Crossing a river. A river is 80 m wide and flows at 1.5 m/s. A boat moves at 4.0 m/s through the water. (a) Heading straight across, how far downstream does it land, in metres? Also find its speed over the ground and the time. (b) At what angle upstream should it head to land directly opposite, and how long does that take?

   Hint

   In (b), the upstream component of the boat's velocity must cancel the current: \(4.0\sin\varphi = 1.5\).

   Your answer mCheck

   Solution

   (a) \(\sqrt{4.0^2 + 1.5^2} \approx 4.27\) m/s at \(\tan^{-1}(1.5/4.0) \approx 20.6^\circ\) downstream; \(80/4.0 = 20\) s; \(1.5 \times 20 = 30\) m downstream. (b) \(\sin\varphi = 0.375\), \(\varphi \approx 22.0^\circ\) upstream; across speed \(4.0\cos 22.0^\circ \approx 3.71\) m/s; time \(80/3.71 \approx 21.6\) s.

Problem 6 (★★ Standard)

6. A rope pulls a sledge with 120 N at 25° above the horizontal. Find the horizontal component, in newtons. Then find the vertical one, and check them.

   Your answer NCheck

   Solution

   \(120\cos 25^\circ \approx 108.8\) N; \(120\sin 25^\circ \approx 50.7\) N. Check: \(\sqrt{108.8^2 + 50.7^2} \approx 120\) ✓.

Problem 7 (★★ Standard)

7. A cable. A cable pulls with 500 N at 25° from the vertical. What is the vertical component of the pull, in newtons?

   Hint

   The angle is from the vertical, so the vertical part gets the cosine.

   Your answer NCheck

   Solution

   \(500\cos 25^\circ \approx 453.2\) N up. (The horizontal part is \(500\sin 25^\circ \approx 211.3\) N.)

Problem 8 (★★ Standard)

8. Ground speed. A plane points due north at 220 m/s through a wind of 40 m/s from the west. How fast does it move over the ground, in metres per second?

   Your answer m/sCheck

   Solution

   The two velocities are at right angles: \(\sqrt{220^2 + 40^2} \approx 223.6\) m/s, about 10.3° east of north.

★★★ Stretch

9. Forces on a slope. A 20 kg crate rests on a smooth (frictionless) 30° slope. Take \(g = 9.81\ \text{m/s}^2\). (a) Find the component of its weight along the slope, in newtons, and the component into the slope. (b) What force, parallel to the slope, holds it still? (c) If instead you push horizontally, how hard must you push?

   Hint

   In (c), only the part of your horizontal push along the slope holds the crate: \(H\cos 30^\circ\).

   Your answer NCheck

   Solution

   (a) \(W = 196.2\) N: along \(196.2\sin 30^\circ = 98.1\) N; into \(196.2\cos 30^\circ \approx 169.9\) N. (b) 98.1 N up the slope. (c) \(H\cos 30^\circ = W\sin 30^\circ \Rightarrow H = W\tan 30^\circ \approx 113.3\) N: more than (b), because part of a horizontal push presses into the slope.

Problem 10 (★★★ Stretch)

10. Two ropes. Two ropes pull a post, each with 50 N, one at 20° either side of north. What is the resultant, in newtons?

    Your answer NCheck

    Solution

    The east–west parts cancel; the north parts add: \(2 \times 50\cos 20^\circ \approx 94.0\) N due north.

Problem 11 (★★★ Stretch)

11. Half the weight. On a smooth slope, at what angle does the part of the weight along the slope equal half the whole weight, in degrees?

    Hint

    The part along the slope is \(W\sin\theta\). Set it equal to \(\tfrac12 W\).

    Your answer °Check

    Solution

    \(W\sin\theta = \tfrac12 W\), so \(\sin\theta = 0.5\) and \(\theta = 30^\circ\). The weight itself cancels, so it works for any mass.

Problem 12 (★★★ Stretch)

12. Extension: two forces at an angle. Forces of 8 N and 6 N pull from the same point, 60° apart. What is the size of the resultant, in newtons?

    Hint

    Put the 8 N along the \(x\) axis, split the 6 N into components, and add the \(x\) parts and the \(y\) parts.

    Your answer NCheck

    Solution

    \(x = 8 + 6\cos 60^\circ = 11\), \(y = 6\sin 60^\circ \approx 5.196\); \(\sqrt{121 + 27} = \sqrt{148} \approx 12.17\) N, at \(\tan^{-1}(5.196/11) \approx 25.3^\circ\) from the 8 N force.

Answer these out loud, from memory. Then try again in a few days.

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Check yourself

Summary