Richard J. Terwilliger

by

Let’s look at someexamples.

We’ll break this GREEN vector up into different

RED vectors.

Here the GREEN vector is broken down into

two RED vectors

The two RED vectors are

called the of the

GREEN vector.

Or, the of the two RED vectors is

the GREEN vector.

The same GREEN vector can be broken up into

2 different RED vectors.

Or two different vectors!

There are

possibilities!

A little later on we’ll show that two components at right

angles are very helpful!

Here the same GREEN vector is broken up into 3 different

component vectors

Again, the of the three RED vectors is the GREEN vector!

And, the GREEN vector can be broken into RED vectors

Let’s see anotherexample.

This BLUE vector is broken down into several

ORANGE vectors.

The tail of the BLUE vector starts at the same point as the tail

of the first ORANGE vector.

STARTING

POINT

And they both end at the same point

ENDING

POINT

STARTING

POINT

The of the ORANGE vectors is the

BLUE vector.

E

E

Or again, theof the BLUE vector are the ORANGE vectors..

E

So what have we learned so far?

E

A vector can be broken down into any number of components.

The different arrangements of components are unlimited.

Components are one of two or more vectors

having a sum equal to a given vector.

A resultant is the sum of the component vectors.

The process of finding the components of a vector given it’s magnitude and direction is called:

E

E

is an easy way to find the resultant of several vectors.

E

To show how, let’s go back to two at right angles..

90o

E

We’ll sketch in the x-axis soit goes through the TAIL of our original vector.

X-axisTAIL

EX-axis

Y-axis

TAIL

Then we’ll sketch the y-axis so it goes through the TAIL of our original vector, too.

Y-axis

E

In the diagram the red vectoris called the horizontal or X- component.

X-axis

Y-axis

E

The blue vector is called the vertical or Y- component.

X-axis

Y-axis

E

To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively.

X-axis

Y-axis

E

To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively.

X-axis

Y-axis

E

To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively.

X-axis

E

Knowing the original vector’s magnitude and direction we can solve for Vx and Vy using trigonometry.

90o

0o180o

270o

E

Let’s label the original vector with it’s magnitude and

direction.

0o180o

90o

270o

90o

270o

E

Let’s label the original vector with it’s magnitude and

direction.

0o180o

90o

270o

E

To determine the value of the X-component (Vx) we need

to use COSINE.

0o180o

90o

270o

E

Remember COSINE?

0o180o

E0o180o

Cosine = Adjacent Hypotenuse

90o

270o

90o

270o

We need to rearrange the equation:

solving for the adjacent side.

.

Cosine = Adjacent Hypotenuse

E0o180o

90o

270o

Cos = Adj Hyp

E0o180o

Adj = (Hyp) (Cos )therefore

Vx = V cos

90o

270o

E0o180o

Adj = (Hyp) (Cos )therefore

Cos = Adj Hyp

Vx = V cos

90o

270o

E0o180o

Vx = V cos Vx = 36 m/s (cos 42o)

Vx = 26.7 m/s

90o

270o

E0o180o

We can now solve for Vy using Sine.

90o

270o

E0o180o

Sine = Opposite Hypotenuse

90o

270o

We are solving for the side opposite the 42 degree angle, Vy, therefore we’ll rearrange the equation solving for the opposite side.

E0o180o

90o

270o

th

eref

ore

E0o180o

Vy = V sin

Opp = Hyp (Sin )

Sine = Opposite Hypotenuse

90o

270o

th

eref

ore

E0o180o

Vy = V sin

Opp = Hyp (Sin )

Sine = Opposite Hypotenuse

90o

270o

E0o180o

Vy = V sin

Vy = 36 m/s (sin 42o)

Vy = 24.1 m/s

90o

270o

E0o180o

We’ve solved for the horizontal vertical and components

of our original vector!

90o

270o

E0o180o

Let’s REVIEW what we did.

V

REVIEW

A vector can be broken down into two vectors at right angles

A MATHEMATICAL METHOD

90o

V

REVIEW

A MATHEMATICAL METHOD

The component that lies on the X-axis is called the horizontal or X - component

The X-component is found using Vx = Vcos

Vx = Vcos

V

REVIEW

A MATHEMATICAL METHOD

The component that lies on the Y-axis is called the vertical or Y- component

The Y-component is found using Vy = Vsin

Vy = Vsin

Vx = Vcos

Now that we know how to break vectors down into their X and Y components

we can solve for the resultant of many vectors using:

Let’s look at a sample problem!

travels 4.2 m at an angle of 37o north of east.

The first putt

10o east of north.

The second putt travels 3.9 m at

The third putt travels 2.6 m at an angle of 30o south of west.

The last putt heading 71o east of south travels 1.8 m before dropping into the

hole. What displacement was needed to sink the ball on the first putt?

A golfer, putting on a green, requires four shots to “hole the ball”.

To solve the problem, using the component method, we’ll carry

out the following sequence: 1. List the vectors

2. Solve for the X and Y components of each vector

3. Sum the X components and sum the Y components

5. Solve for the direction of the resultant using tangent

4. Solve for the magnitude of the resultant using the Pythagorean theorem

First, let’s list all of the vectors changing all of the directions to the number of degrees from east in a counterclockwise direction.

1. List

the ve

ctors

travels 4.2 m at an angle of 37o north of east.

The first putt

10o east of north.

The second putt travels 3.9 m at

The third putt travels 2.6 m at an angle of 30o south of west.

The last putt heading 71o east of south travels 1.8 m before dropping into the

hole. What displacement was needed to sink the ball on the first putt?

A golfer, putting on a green, requires four shots to “hole the ball”.

90o

0o

180o

270o

travels 4.2 m at an angle of 37o north of east.

The first putt

10o east of north.

The second putt travels 3.9 m at

The third putt travels 2.6 m at an angle of 30o south of west.

The last putt heading 71o east of south travels 1.8 m before dropping into the

hole. What displacement was needed to sink the ball on the first putt?

A golfer, putting on a green, requires four shots to “hole the ball”.

90o

0o

180o

270o

d1 = 4.2 m @ 37o

d2 = 3.9 m @ 80o

d3 = 2.6 m @ 210o

d4 = 1.8 m @ 341o

10o east of north 30

o south of west

71o east of south

37o north of east.4.2 m 3.9 m

2.6 m

1.8 m

Next, to make things easier to see, let’s set up a table that lists our vectors..

1. List

the ve

ctors

First label all of the columns

Vector(V)

X-component Solve Y-component Solve

Remember the X component is cosine

Vector(V)

X-component Solve Y-component SolveVector(V)

X-component Solve Y-component Solve(Vcos)

And the Y component is sine

Vector(V)

X-component Solve Y-component SolveVector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Next fill in the firstcolumn with the different vectors.

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Substitute into the equations for X and then Y

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

Remember to include units!

Now solve for the value of each X and Y component.Again, remember to include UNITS!

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Sum the X components and then the Y components.

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Sum () x = 3.48 m Y = 4.49 m

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Sum () x = 3.48 m Y = 4.49 m

The four vectors are now broken down into two.

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Sum () x = 3.48 m Y = 4.49 m

One vector 3.48 m long lies on the X-axis.

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Sum () x = 3.48 m Y = 4.49 m

The other vector, 4.49 m long, lies on the Y-axis.

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

Vector(V)

X-component Solve Y-component Solve(Vcos) (Vsin )

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Sum () x = 3.48 m Y = 4.49 m

We’ll plot these two vectorsand determine their

resultant.

Y

x = 3.48 m Y = 4.49 m

We’ll plot these two vectorsand determine their

resultant.

Xx = 3.48 m

Y =

4.4

9 m

Y

Xx = 3.48 m

The resultant of these two vectors is the same as the resultant of our original four vectors!.

Y =

4.4

9 m

Y

X

Y =

4.4

9 m

x = 3.48 m

The Pythagorean Theorem will be used to determine the

MAGNITUDE of the resultant.

c2 = a2 +b2

R2 = X2 + Y2

R = X2 + Y2

R = (3.48m)2 + (4.49m)2

R = 5.68 m

Y

X

Y =

4.4

9 m

x = 3.48 m

We have now found the magnitude of our resultant

to be 5.68 m long.

c2 = a2 +b2

R2 = X2 + Y2

R = X2 + Y2

R = (3.48m)2 + (4.49m)2

R = 5.68 m

Y

X

Y =

4.4

9 m

x = 3.48 m

The last step is to find the DIRECTION our resultant

is pointing.

Y

X

Y =

4.4

9 m

Y = 3.48 m

To find the angle we’ll use tangent.

Tan =adjacent

opposite

adj

opp = Tan-1

X

Y = Tan-1

3.48 m

4.49 m = Tan-1

= 52.2o

Y

X

Y =

4.4

9 m

Y = 3.48 m

The RESULTANT is5.68 m @ 52.2o.

Let’s go back to our original problem with

the results.

travels 4.2 m at an angle of 37o north of east.

The first putt

10o east of north.

The second putt travels 3.9 m at

The third putt travels 2.6 m at an angle of 30o south of west.

The last putt heading 71o east of south travels 1.8 m before dropping into the

hole. What displacement was needed to sink the ball on the first putt?

A golfer, putting on a green, requires four shots to “hole the ball”.

Let’s SUMMARIZE theCOMPONENT METHOD

List all the vectors.

Break each vector down into an X and Y component

using Cosine for X and Sine for Y.

Sum all of the X components and then sum all of the Y components

Using the Pythagorean Theorem solve for the MAGNITUDE of the resultant

Using Tangent solve for the DIRECTION of the resultant.

Let’s go over that once again.

Vector

(V)

X-componentSolve

Y-component Solve

x

Y

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

List all the vectors.

Break each vector down into an X and Y component

using Cosine for X and Sine for Y.

Vector

(V)

X-componentSolve

Y-component Solve

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

(Vcos) (Vsin )

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

x

Y

Sum all of the X components and then sum all of the Y components

Vector

(V)

X-componentSolve

Y-component Solve

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

(Vcos) (Vsin )

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

Sum () x = 3.48 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Y = 4.49 m

x

Y

Y = 3.48 m

Y =

4.4

9 m

Vector

(V)

X-componentSolve

Y-component Solve

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

(Vcos) (Vsin )

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

Sum () x = 3.48 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Y = 4.49 m

Using the Pythagorean Theorem solve for the MAGNITUDE of the resultant

c2 = a2 +b2

R2 = X2 + Y2

R = X2 + Y2

R = (3.48m)2 + (4.49m)2

R = 5.68 m

x

Y

Y = 3.48 m

Y =

4.4

9 m

Vector

(V)

X-componentSolve

Y-component Solve

4.2 m @ 37o

3.9 m @ 80o

2.6 m @ 210o

1.8 m @ 341o

(Vcos) (Vsin )

4.2 m cos 37o

3.9 m cos 80o

2.6 m cos 210o

1.8 m cos 341o

4.2 m sin 37o

3.9 m sin 80o

2.6 m sin 210o

1.8 m sin 341o

3.35 m

0.68 m

-2.25 m

1.70 m

Sum () x = 3.48 m

2.53 m

3.84 m

-1.30 m

-0.58 m

Y = 4.49 m

c2 = a2 +b2

R2 = X2 + Y2

R = X2 + Y2

R = (3.48m)2 + (4.49m)2

R = 5.68 m

x

Y

Y = 3.48 m

Y =

4.4

9 m

Using Tangent solve for the DIRECTION of the resultant.

Tan =adjacent

opposite

adj

opp = Tan-1

x

Y = Tan-1

3.48 m

4.49 m = Tan-1

= 52.2o