Question

determine the magnitude and direction of the resultant of the following displacements: 5m east, 8m 45 degrees North of east, 3m 60 degrees south of west, and 7m south.

Answer 1

Task. Determine the magnitude and direction of the resultant of the following displacements:

a) $5\mathrm{m}$ east,

b) 8m 45 degrees north of east,

c) $3\mathrm{m}60$ degrees south of west,

d) $7\mathrm{m}$ south.

Solution.

a) Consider the following figure:

The displacement $5\mathrm{m}$ east is a displacement into the positive direction of $x$ -axis by 5 meters. Hence the displacement vector has coordinates $v = (5,0)$ . Its magnitude $|v| = \sqrt{5^2 + 0^2} = 5$ and its angle with $x$ -axis is $0{}^\circ$ .

b) The displacement 8m 45 degrees north of east is shown in the following figure:

The displacement vector $v$ has magnitude 8 and its angle with $x$ -axis is equal to $45{}^{\circ}$ . Hence $v$ has the following coordinates:

$$v = (8 \cos 4 5 {}^ {\circ}, 8 \sin 4 5 {}^ {\circ}) = \left(8 * \frac {1}{\sqrt {2}}, 8 * \frac {1}{\sqrt {2}}\right) = (4 \sqrt {2}, 4 \sqrt {2}).$$

c) The displacement $3\mathrm{m}60$ degrees south of west is shown in the following figure:

The displacement vector $v$ has magnitude 6 and its angle with $x$ -axis is equal to $180 + 30 = 210{}^{\circ}$ . Hence $v$ has the following coordinates:

$$v = (6 \cos 2 1 0 {}^ {\circ}, 6 \sin 2 1 0 {}^ {\circ}) = (- 6 \cos 3 0 {}^ {\circ}, - 6 \sin 3 0 {}^ {\circ}) = \left(- 6 * \frac {\sqrt {3}}{2}, - 6 * \frac {1}{2}\right) \left(- 3 \sqrt {3}, - 3\right).$$

d) The displacement $7\mathrm{m}$ south is shown in the following figure:

The displacement vector $v$ has magnitude 7 and it is directed down along $y$ -axis. Hence $v$ has the following coordinates:

$v = (0, - 7)$