5.12 Trigonometry in context: vectors, kinematics and forces — Practise

Work through these on your own. Draw the triangle for each question, mark the angle you are given, and decide whether you need a component, a resultant or a direction. Give lengths and forces to three significant figures.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A force of $50$ N acts at $30^\circ$ above the horizontal. Find its horizontal and vertical components.
Two forces of $5$ N and $9$ N act at $90^\circ$ to each other. Find the magnitude of the resultant.
Two forces of $6$ N and $8$ N act at $60^\circ$ to each other. Find the magnitude of the resultant.
Two forces of $4$ N and $7$ N act at $50^\circ$ to each other. Find the magnitude of the resultant, to three significant figures.
A swimmer heads due east at $1.2$ m s $^{-1}$ in a current of $0.5$ m s $^{-1}$ due south. Find the size of the resultant velocity.
A block of weight $120$ N is on a slope of $20^\circ$ . Find the component of the weight down the slope.
A boat moves $4$ m s $^{-1}$ due east with a current of $3$ m s $^{-1}$ due north. Find the angle of the resultant velocity north of east, in degrees to three significant figures.
A force of $100$ N acts at $65^\circ$ above the horizontal. Find its vertical component, to three significant figures.
Two perpendicular displacements of $3$ m and $4$ m act from the same point. Find the magnitude of the resultant displacement.
A block of weight $200$ N is on a slope of $25^\circ$ . Find the component of the weight into the slope, to three significant figures.

Answers and concise solutions

1. The two components

Answer: Horizontal $= 50\cos 30^\circ = 25\sqrt{3} \approx 43.3$ N. Vertical $= 50\sin 30^\circ = 25$ N.

2. The resultant

Answer: The vectors are perpendicular, so $R = \sqrt{5^2 + 9^2} = \sqrt{106} \approx 10.3$ N.

3. The resultant

Answer: $R^2 = 36 + 64 + 2 \times 6 \times 8 \times \cos 60^\circ = 100 + 48 = 148$ . So $R = \sqrt{148} = 2\sqrt{37} \approx 12.2$ N.

4. The resultant

Answer: $R^2 = 16 + 49 + 56\cos 50^\circ \approx 65 + 36.0 = 101.0$ , so $R \approx 10.05$ N, which is $10.0$ N to three significant figures.

5. The resultant speed

Answer: The velocities are perpendicular, so $v = \sqrt{1.44 + 0.25} = \sqrt{1.69} = 1.3$ m s $^{-1}$ .

6. The component down the slope

Answer: The component down the slope is $120\sin 20^\circ \approx 41.0$ N. The component into the slope is $120\cos 20^\circ \approx 112.8$ N.

7. The direction

Answer: The angle satisfies $\tan\theta = \dfrac{3}{4}$ , so $\theta \approx 36.9^\circ$ north of east. The size of the resultant is $5$ m s $^{-1}$ .

8. The vertical component

Answer: Vertical component $= 100\sin 65^\circ \approx 90.6$ N.

9. The resultant

Answer: For perpendicular displacements, $R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ m.

10. The component into the slope

Answer: The component into the slope is $200\cos 25^\circ \approx 181$ N, to three significant figures. The component down the slope is $200\sin 25^\circ \approx 84.5$ N.