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Vectors

Relative Motion

Resolving Vectors

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Resolving Vectors⁠

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All vectors can be thought of as the hypotenuse of a right angle triangle.

In some cases, the length of one of the legs may be zero, but in most cases this is not true (just like some, but not all, rectangles are squares)

We now have a non-trivial use for our knowledge of right-angle triangle geometry

Example 1.

You move through a displacement of 4.0 km [E] and then 3.0 km [N].

What is the total displacement of your movements?

The overall displacement would connect where your motion began, with where it finished (as shown with the green dotted line below). We know that these two vectors are perpendicular (sometimes called orthogonal) to each other, and so these three sides would form a right angle triangle.

We can employ the use of Pythagorean Theorem, and our Primary Trigonometric Ratios (, , ) to figure out the length of the hypotenuse (magnitude of our displacement), and the internal angles (direction of our displacement).

To determine magnitude:

d=d12+d22 From a2+b2=c2…

d=4.0 km2+3.0 km2

d=5.0 km

We will choose to use the original given magnitudes to determine the direction (to avoid rounding error):

=oppositeadjacent

=3.0 km4.0 km

=-13.0 km4.0 km

= 36.9, so the direction is [E 37º N]

Therefore, the resulting displacement is d=5.0 km [E 37 N]

Likewise, we can reverse this process.

If you were given a total displacement vector, d=5.0 km [E 37 N] and asked to find two perpendicular motions that would produce this displacement you could imagine:

…and you would find that:

d1= 4.0 km [E], because =d1d , and

d2= 3.0 km [N] because =d2d

We could also call these dx, and dy respectively.

This is all we are doing when finding components.

We take the vector we are given, imagine how we could construct that vector out of two perpendicular vectors, and then, using primary trigonometric ratios, we find what the magnitudes of those sides could be.

Conventions:

It is important to build yourself a problem solving environment that reduces cognitive load (allows your brain to think about the thinking and not the annoying, minute details that inevitably result in mistakes)

For this reason, use:

For any vector r, this coordinate system will produce:

x-components when you use =xr

y-components when you use =yr

When the direction is somewhere between:

[E] and [N], the angle will be between 0º and 90º, and

x-component will be positive

y-component will be positive

[N] and [W], the angle will be between 90º and 180º, and

x-component will be negative

y-component will be positive

[W] and [S], the angle will be between 180º and 270º, and

x-component will be negative

y-component will be negative

[S] and [E], the angle will be between 270º and 360º, and

x-component will be positive

y-component will be negative

Resolving Vectors Practice Problems

Level 1

Resolve the following vectors into their horizontal and vertical components.

d = 6.0 x 101 km; θ = [E, 74˚ N]

v = 2.50 x 102 m/s; θ = [W, 32˚ S]

v = 4.0 x 101 cm/s; θ = [N, 62˚ W]

F= 9.00 x 102 N; θ = [N, 57˚ E]

p = 25 Ns; θ = [Right 48˚ Down]

Level 2

Bob is pulling his sister, Alice, on a sled, by applying a tension force with magnitude of 1.10 x 102 N. The angle of the rope is 56° above the horizontal.

What is the magnitude of the component of the tension force acting perpendicular to the ground? (the vertical component)

What is the magnitude of the component of the tension force acting parallel to the ground? (the horizontal component)

Level 3

Add the following vectors by resolving the vectors into their components.

Sketch a diagram to support your answer.

Since the answers are vectors, they should include a magnitude and a direction.

35 m/s [NE] + 65 m/s [N 60˚ W]

1.20 x 102 N [S 20˚ W] + 245 N [S 55˚ E]

4.0 m [N 63˚ W] + 15.0 m [W 26˚ S]

8.0 N [S 15˚ E] + 7.0 N [N 28˚ W] + 12.0 N [S, 55˚ W]

A car changes its velocity from 25 m/s [S 35˚ W] to 35 m/s [W 26˚ N] in 0.50 minutes. Please calculate the acceleration of the car. When doing the vector subtraction you are expected to resolve the vectors into their components.

Level 4

Two ranchers have their hands full because they have lassoed a wild horse. They are in a stand off because nobody is able to move (horse or ranchers, therefore Fun= 0 N). Please calculate the force with which the horse must be pulling to be able to match the ranchers. The ranchers are using forces of 3.40 x 102 N [N 33˚ E] and 226 N [E 12˚ N].

For a holiday, an old Newfoundlander decided to go whaling. The ship he boarded maintained a velocity of 35 km/h [N 34˚ W]. The velocity of the ocean current was 27 km/h [E 62˚ N]. After spotting a whale, the Newfoundlander fired a harpoon at 85 km/h [S 10˚ E]. Please calculate the velocity of the harpoon relative to the "ground". i.e. to the shore or the bottom of the ocean.

Little Red Riding Hood was lost in the woods. She suddenly saw the Wicked Wolf right beside her. She ran 80 m due East, then 40 m due north, then 60 m in a direction 60° West of North. There she stopped. The Wolf then walked 25 m due North and 35 m due West where he stopped. How far was he from Red Riding Hood at that time? (Include magnitude and direction.)

Worked Example
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In 4.4 s, a chickadee flies in a horizontal plane from a fence post (P) to a bush (B) and then to a bird feeder (F), as shown in Figure 18(a).

Find the following:

total distance travelled

average speed

total displacement

average velocity

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The total distance travelled is a scalar quantity.

d=22 m +11 m

d=33 m

t =4.4 s

vav= ?

vav=dt

vav=33 m4.4 s

vav=7.5 m/s

The average speed is 7.5 m/s.

Total displacement could be done by using the sine and cosine laws, however, we could also use components to solve this problem.

Vector

x-component

y-component

d1x=d1

d1x=22 m33

d1x=18.45 m

d1y=d1

d1y=22 m33

d1y=11.98 m

d2x=d2

d2x=11 m–28

d2x=9.71 m

d2y=d2

d2y=11 m–28

d2y=–5.16 m

dx=d1x+d2x

dx=18.45 m+9.71 m

dx=28.16 m

dy=d1y+d2y

dy=11.98 m+-5.16 m

dy=6.82 m

There are no rows in this table

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Now I need to use pythagorean theorem and to find the resultant vector.

d=dx2+dy2

d=28.16 m2+6.82 m2

d=28.97 m

=dydx=6.82 m28.26 m

=-16.82 m28.26 m

=13.6

d=28.97 mE 13.6 N

To determine average velocity, we need to use the displacement and the time interval.

d=28.97 mE 13.6 N

t =4.4 s

vav= ?

vav=dt

vav=28.97 mE 13.6 N4.4 s

vav=6.58 m/s E 13.6 N

The average velocity is 6.58 m/s [E 13.6º N]

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