## Adding Vectors Using The Component Method

by

Betty Roombos

A vector of the form 6 units 30^{0} N of E may be
broken down into its north and east components and then be combined with other
components of the same type. Since 6 units is the hypotenuse of a right
triangle, the north and east components are the legs of this right triangle.
The first direction listed is always found using the hypotenuse and the sine of
the angle and the second direction listed is always found using the hypotenuse
and the cosine of the angle. With the above information,
let's add the following displacement vectors to find the resultant: 4 km 25^{0} E of S
and 7 km 35^{0} W of N.

The east component of the first displacement is 4(sin 25^{0})
= 1.69 km E

The south component of the first displacement is 4(cos 25^{0})
= 3.63 km S

The west component of the second displacement is 7(sin 35^{0})
= 4.02 km W

The north component of the second displacement is 7(cos 35^{0})
= 5.73 km N

Now the components in the same or opposite directions may
be combined. To be consistent let's make east and north the positive
directions. Then 1.69 km E - 4.02 km W = -2.33 and since the answer is
negative, this means that the direction is west or 2.33 km W.
Similarly, -3.63 km S + 5.73 km N = +2.10 km, and since the answer is positive,
the direction is north or 2.10 km N.

2.33 km W and 2.10 km N are now the legs of a right
triangle, and the hypotenuse of this right triangle is the sum of the original
displacements. Therefore, we may use the Pythagorean theorem to find the
length of the hypotenuse and we may use tangent to find the angle.

R = [ (2.33 km)^{2} + (2.10 km)^{2}]^{1/2}
= 3.14 km

Then the tangent of the angle = either 2.33 km W / 2.10 km
N or 2.10 km N / 2.33 km W and the corresponding angles are 48^{0}
and 42.0^{0 }respectively.

The final answer is then either 3.14 km 48^{0} W
of N or 3.14 km 42^{0} N or W. Notice that the direction listed in
the numerator is written first and that the direction in the denominator is
listed second.

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