richard j. terwilliger by let’s look at some examples
TRANSCRIPT
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