Force Components

FORCE VECTORS, VECTOR OPERATIONS & ADDITION OF FORCES 2D & 3D

FORCE VECTORS, VECTOR OPERATIONS & ADDITION OF FORCES 2D & 3D

Today’s Objective:

1. Resolve a 2-D vector into components.

2. Add 2-D vectors using Cartesian vector notations.

3. Represent a 3-D vector in a Cartesian coordinate system.

4. Find the magnitude and coordinate angles of a 3-D vector.

5. Add vectors (forces) in 3-D space.

2 D

3 D

APPLICATION OF VECTOR ADDITIONAPPLICATION OF VECTOR ADDITION

There are three concurrent forces acting on the hook due to the chains.

We need to decide if the hook will fail (bend or break)?

To do this, we need to know the resultant force acting on the hook.

FR

SCALARS AND VECTORS (Section 2.1)

SCALARS AND VECTORS (Section 2.1)

Scalars Vectors

Examples: Mass, Volume Force, Velocity

Characteristics: It has a magnitude It has a magnitude

(positive or negative) and direction

Addition rule: Simple arithmetic Parallelogram law

Special Notation: None Bold font, a line, an

arrow or a “carrot”

In these PowerPoint presentations, a vector quantity is represented like this (in bold, italics, and red).

I write them with an arrow on top

VECTOR OPERATIONS (Section 2.2)

VECTOR OPERATIONS (Section 2.2)

Scalar Multiplication

and Division

VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE

Parallelogram Law:

Triangle method (always ‘tip to tail’):

How do you subtract a vector?

How can you add more than two concurrent vectors graphically ?

Analyzing a Force Triangle Analyzing a Force Triangle

Force Triangle

If we form a force triangle that represents the forces in a given problem, we can analyze the triangle using trigonometry.

Right Triangles: If the triangle is a right triangle, we can use the following relationships:

22 BA R

A

B

adjacent

opposite )tan(

R

B

hypotenuse

opposite )sin(

R

A

hypotenuse

adjacent )cos(

A

BR

B A R

B A R

B A R

A B

Analyzing a Force Triangle Analyzing a Force Triangle

QPR

BPQQPR

cos2222

Law of cosines – Useful when:

• You know 2 side and the angle between

• You know all 3 sides and want to find the angles.

Law of sines – Useful when:

• You know 2 sides and an opposite angle

• You know 1 side and any two angles A

C

R

B

Q

A sinsinsin

Oblique or Obtuse Triangles: If the triangle is NOT a right triangle, we can use the following relationships:

b

A

B

R a

r

Example: Finding Resultant Forces using a force triangleExample: Finding Resultant Forces using a force triangle

To steady a sign as it is being lowered, two cables are

attached to the sign at A. Using trigonometry and

knowing that the magnitude of P is 300N, determine:

A) The required angle if the resultant R of the two

forces applied at A is to be vertical.

B) The corresponding value of R.

Example: Finding Resultant Forces using a force triangleExample: Finding Resultant Forces using a force triangle

Two structural members A and B are bolted to a bracket as shown.

Knowing that both members are in compression and that the force is

30 kN in member A and 20 kN in member B, determine, using

trigonometry, the magnitude and direction of the resultant of the

forces applied to the bracket by members A and B.

(Note: If a force is pushing, it is in compression.

If a force is pulling, it is in tension.)

ADDITION OF A SYSTEM OF COPLANAR FORCES (Section 2.4)

ADDITION OF A SYSTEM OF COPLANAR FORCES (Section 2.4)

• Each component of the vector is shown as a magnitude and a direction.

• The directions are based on the x and y axes. We use the “unit vectors” i and j to designate the x and y axes.

• We ‘resolve’ vectors into components using the x and y axis system.

For example,

F = Fx i + Fy j or F' = F'x i + ( F'y ) j

The x and y axis are always perpendicular to each other. Together, they can be directed at any inclination.

ADDITION OF SEVERAL VECTORSADDITION OF SEVERAL VECTORS

• Step 3 is to find the magnitude and angle of the resultant vector.

• Step 2 is to add all the x-components together, followed by adding all the y components together. These two totals are the x and y components of the resultant vector.

• Step 1 is to resolve each force into its components.

Break the three vectors into components, then add them.

FR = F1 + F2 + F3

= F1x i + F1y j F2x i + F2y j + F3x i F3yj

= (F1x F2x + F3x) i + (F1y + F2y F3y) j

= (FRx) i + (FRy) j

An example of the process:

You can also represent a 2-D vector with a magnitude and angle.

EXAMPLEEXAMPLE

Plan:

a) Resolve the forces into their x-y components.

b) Add the respective components to get the resultant vector.

c) Find magnitude and angle from the resultant components.

Given: Three concurrent forces acting on a tent post.

Find: The magnitude and angle of the resultant force.

EXAMPLE (continued)EXAMPLE (continued)

F1 = {0 i + 300 j } N

F2 = {– 450 cos (45°) i + 450 sin (45°) j } N

= {– 318.2 i + 318.2 j } NF3 = { (3/5) 600 i + (4/5) 600 j } N

= { 360 i + 480 j } N

EXAMPLE (continued)EXAMPLE

(continued)

Summing up all the i and j components respectively, we get,

FR = { (0 – 318.2 + 360) i + (300 + 318.2 + 480) j } N

= { 41.80 i + 1098 j } N

x

y

FR

Using magnitude and direction:

FR = ((41.80)2 + (1098)2)1/2 = 1099 N

= tan-1(1098/41.80) = 87.8°

Determine the resultant of the three forcesEXAMPLEEXAMPLE

Example – using calculatorsExample – using calculatorsUsing calculators to find resultants

Solutions to the problems shown above can be produced quickly using

calculators that can perform operations using complex numbers (or

numbers in polar and rectangular form).

Polar numbers - can be used to represent forces in terms of their

magnitude and angle

Rectangular numbers - can be used to represent forces in terms of unit

vectors

Example: Repeat the last example using the TI-85/86 or TI-89/92 calculator

(Determine the resultant of the 3 forces on the hook below.)

3D APPLICATIONS3D APPLICATIONS

In this case, the power pole has guy wires helping to keep it upright in high winds. How would you represent the forces in the cables using Cartesian vector form?

Many structures and machines involve 3-Dimensional Space.

APPLICATIONS (continued)APPLICATIONS (continued)

In the case of this radio tower, if you know the forces in the three cables, how would you determine the resultant force acting at D, the top of the tower?

CARTESIAN UNIT VECTORS CARTESIAN UNIT VECTORS

Characteristics of a unit vector :

a) Its magnitude is 1.

b) It is dimensionless (has no units).

c) It points in the same direction as the original vector (A).

For a vector A, with a magnitude of A, an unit vector is defined as

uA = A / A .

The unit vectors in the Cartesian axis system are i, j, and k. They are unit vectors along the positive x, y, and z axes respectively.

CARTESIAN VECTOR REPRESENTATIONCARTESIAN VECTOR REPRESENTATION

Consider a box with sides AX, AY, and AZ meters long.

The vector A can be defined as

A = (AX i + AY j + AZ k) m

The projection of vector A in the x-y plane is A´. The magnitude of A´ is found by using the

same approach as a 2-D vector: A’ = (AX2 + AY

2)1/2 .

The magnitude of the position vector A can now be obtained as

A = ((A´)2 + AZ2) ½ = (AX

2 + AY2 + AZ

2) ½

Using trigonometry, “direction cosines” are found using

These angles are not independent. They must satisfy the following equation.

cos ² + cos ² + cos ² = 1

This result can be derived from the definition of a coordinate direction angles and the unit vector. Recall, the formula for finding the unit vector of any position vector:

or written another way, u A = cos i + cos j + cos k .

DIRECTION OF A CARTESIAN VECTORDIRECTION OF A CARTESIAN VECTOR

These angles are measured between the vector and the positive X, Y and Z axes, respectively. Their range of values are from 0° to 180°

The direction or orientation of vector A is defined by the angles ά, β, and γ.

ADDITION OF CARTESIAN VECTORS

(Section 2.6)

ADDITION OF CARTESIAN VECTORS

(Section 2.6)

For example, if

A = AX i + AY j + AZ k and

B = BX i + BY j + BZ k , then

A + B = (AX + BX) i + (AY + BY) j + (AZ + BZ) k

orA – B = (AX - BX) i + (AY - BY) j + (AZ - BZ) k

Once individual vectors are written in Cartesian form, it is easy to add or subtract them. The process is essentially the same as when 2-D vectors are added.

IMPORTANT NOTESIMPORTANT NOTES

Sometimes 3-D vector information is given as:

a) Magnitude and the coordinate direction angles, or,

b) Magnitude and projection angles.

You should be able to use both these types of information to change the representation of the vector into the Cartesian form, i.e.,

F = {10 i – 20 j + 30 k} N .

EXAMPLEEXAMPLE

1. Using geometry and trigonometry, write F1 and F2 in Cartesian vector form.

2. Then add the two forces (by adding x, y and z components).

G

Given: Two forces F1 and F2 are applied to a hook.

Find: The resultant force in Cartesian vector form.

Plan:

Solution : First, resolve force F1.

Fz = 500 (3/5) = 300 lb

Fx = 0 = 0 lb

Fy = 500 (4/5) = 400 lb

Now, write F1 in Cartesian vector form (don’t forget the units!).

F1 = {0 i + 400 j + 300 k} lb

Now resolve force F2.

We are not given the direction angles.

So we need to find the horizontal component in the xy plane.

F2XY = 800 * cos(45°) = 565.69 lb

F2x = 565.69 * cos(45°) = 489.90 lb

F2y = 565.69 * sin(30°) = 282.84 lb

F2z = -800 * sin(45°) = -565.69 lb

F1 = {0 i + 400 j + 300 k} lb

F2 = {489.9 i + 282.8 j 565.7 k } lb

Now, R = F1 + F2 or

R = {489.9 i + 682.8 j 265.7 k} lb

R

ExampleExampleFind the magnitude and the coordinate direction angles of the resultant force.

POSITION VECTORS & FORCE VECTORS

DOT PRODUCT

POSITION VECTORS & FORCE VECTORS

DOT PRODUCTObjectives:

Students will be able to :

a) Represent a position vector in Cartesian coordinate form, from given

geometry.

b) Represent a force vector directed along a line.

c) determine an angle between

two vectors, and,

d) determine the projection of a vector

along a specified line.

APPLICATIONSAPPLICATIONS

This awning is held up by three chains. What are the forces in the chains and how do we

find their directions? Why would we want to know these things?

POSITION VECTOR POSITION VECTOR

Consider two points, A and B, in 3-D space. Let their coordinates be (XA, YA,

ZA) and (XB, YB, ZB ), respectively.

A position vector is defined as a fixed

vector that locates a point in space

relative to another point.

POSITION VECTOR POSITION VECTOR

The position vector directed from A to B, r AB , is defined as

r AB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m

Please note that B is the ending point and A is the starting point. ALWAYS subtract the starting

point from the ending point.

FORCE VECTOR DIRECTED ALONG A LINE

(Section 2.8)

FORCE VECTOR DIRECTED ALONG A LINE

(Section 2.8)

a) Find the position vector, rAB , along two points on that line.

b) Find the unit vector describing the line’s direction, uAB = (rAB/rAB).

c) Multiply the unit vector by the magnitude of the force, F = F uAB .

If a force is directed along a line, then we can

represent the force vector in Cartesian

coordinates by using a unit vector and the

force’s magnitude. So we need to:

EXAMPLEEXAMPLE

Plan:

1. Find the position vector rAC and the unit vector uAC.

2. Obtain the force vector as FAC = 420 N uAC .

Given: The 420 N force

along the cable AC.

Find: The force FAC in the Cartesian vector

form.

EXAMPLE (continued)EXAMPLE (continued)

(We can also find rAC by subtracting the coordinates of

A from the coordinates of C.)

rAC = (22

+ 32

+ 62

)1/2

= 7 m

Now uAC = rAC/rAC and FAC = 420 uAC N = 420 (rAC/rAC )

So FAC = 420{ (2 i + 3 j 6 k) / 7 } N

= {120 i + 180 j - 360 k } N

As per the figure, when relating A to C, we will have to go

2 m in the x-direction, 3 m in the y-direction, and -6 m in

the z-direction. Hence,

rAC = {2 i + 3 j 6 k} m.

ExampleExample Find the magnitude and the coordinate direction angles of the resultant

force. Plan:

1) Find the forces along CA and CB in the Cartesian vector form.

2) Add the two forces to get the resultant force, FR.

3) Determine the magnitude and the coordinate angles of FR.

DOT PRODUCT DOT PRODUCT

Students will be able to use the vector dot product to:

a) determine an angle between

two vectors, and,

b) determine the projection of a vector

along a specified line.

APPLICATIONSAPPLICATIONS

If the design for the cable placements

required specific angles between the

cables, how would you check this

installation to make sure the angles

were correct?

APPLICATIONSAPPLICATIONS

For the force F being applied to the wrench at Point A, what

component of it actually helps turn the bolt (i.e., the force

component acting perpendicular to the pipe)?

DEFINITIONDEFINITION

The dot product of vectors A and B is defined as A•B = A B cos .

The angle is the smallest angle between the two vectors and is always in a range of

0 to 180º.

Dot Product Characteristics:

1. The result of the dot product is a scalar (a positive or negative number).

2. The units of the dot product will be the product of the units of the A and B

vectors.

DOT PRODUCT DEFINITON

(continued)

DOT PRODUCT DEFINITON

(continued)

Finding a dot product using Cartesian coordinates:

By definition,

i • j = j • i = i • k = k • i = j • k = k • j = 0

i • i = j • j = k • k = 1

so

A • B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k)

= Ax Bx + AyBy + AzBz

USING THE DOT PRODUCT TO DETERMINE THE ANGLE BETWEEN TWO VECTORSUSING THE DOT PRODUCT TO DETERMINE THE ANGLE BETWEEN TWO VECTORS

For the given two vectors in the Cartesian form, one can find the angle by

a) Finding the dot product, A • B = (AxBx + AyBy + AzBz ),

b) Finding the magnitudes (A & B) of the vectors A & B, and

c) Using the definition of dot product and solving for , i.e.,

= cos-1

[(A • B)/(A B)], where 0º 180

º .

Example: Finding angles using dot productsExample: Finding angles using dot products Example: Consider the volleyball

net shown below. Determine the

angle formed by guy wires AB and

AC.

Using the Dot Product to determine the projection of a vector along a lineUsing the Dot Product to determine the projection of a vector along a line

Steps:

1. Find the unit vector, uaa´ along line aa´

2. Find the scalar projection of A along line aa´ by

A|| = A • uaa = AxUx + AyUy + Az Uz

You can determine the components of a vector parallel and perpendicular to a line using the dot

product.

3. If needed, the projection can be written as a vector, A|| , by using the unit vector uaa´ and

the magnitude found in step 2.

A|| = A|| uaa´

4. The scalar and vector forms of the perpendicular component can easily be obtained by

A = (A 2 - A||

2)

½ and

A = A – A||

(rearranging the vector sum of A = A + A|| )

Using the dot product to determine the projection of a vector

(continued)

Using the dot product to determine the projection of a vector

(continued)

EXAMPLE EXAMPLE

Plan:

1. Find rAO

2. Find the angle = cos-1

{(F • rAO)/(F rAO)}

3. Find the projection via FAO = F • uAO (or F cos )

Given: The force acting on the hook at point

A.

Find: The angle between the force vector and

the line AO, and the magnitude of

the projection of the force along the

line AO.

EXAMPLE (continued) EXAMPLE (continued)

rAO = {1 i + 2 j 2 k} m

rAO = (12 + 2

2 + 2

2)1/2

= 3 m

F = { 6 i + 9 j + 3 k} kN

F = (62 + 9

2 + 3

2)1/2

= 11.22 kN

= cos-1

{(F • rAO)/(F rAO)}

= cos-1

{18 / (11.22 * 3)} = 57.67°

F • rAO = ( 6)(1) + (9)(2) + (3)(2) = 18 kN m

Example:Example:

Plan:

1. Find rAO

2. Find the angle = cos-1

{(F • rAO)/(F rAO)}

3. The find the projection via FAO = F • uAO or F cos

Find: 1) The angle between the force vector and the pole

2) The magnitude of the projection of the force along the pole AO