CHAPTER 5

MOTION IN

TW

O DIM

ENSIONS

VECTORS

MATHEMAT

ICAL P

ROPERT

IES

SCALARS

A SCALAR is ANY quantity in physics that has MAGNITUDE, but NOT a direction associated with it.

Magnitude – A numerical value with units.

Scalar Example

Magnitude

Speed 20 m/s

Distance 10 m

Age 15 years

Heat 1000 calories

VECTOR

A VECTOR is ANY quantity in physics that has BOTH MAGNITUDE and DIRECTION.

Faxv

,,,

Vector Magnitude & Direction

Velocity 20 m/s, N

Acceleration 10 m/s/s, E

Force 5 N, West

Vectors are typically illustrated by drawing an ARROW above the symbol. The arrow is used to convey direction and magnitude.

POLAR NOTATION

Polar notation defines a vector by designating the vector’s magnitude |A| and angle θ relative to the +x axis. Using that notation the vector is written:

In this picture we have a force vector with magnitude 12 Newtons oriented at 210 degrees with the + x axis. It would be characterized as F = 12N < 210

SCALAR MULTIPLICATION

Multiplying a vector by a scalar will ONLY CHANGE its magnitude.

Thus if A = 12, Then 2A = 24

Multiplying a vector by “-1” does not change the magnitude, but it does reverse it's direction or in a sense, it's angle.

Thus if A = 12 , then -A = -12

If A = 12, then (-1/2)A = -6

-1/2A

UNIT VECTOR NOTATION

An effective and popular system used in engineering is called unit vector notation. It is used to denote vectors with an x-y Cartesian coordinate system.

UNIT VECTOR NOTATION

jiA ˆ3ˆ4

j= vector of magnitude “1” in the “y” direction

i = vector of magnitude “1” in the “x” direction= 4i

=3j

4i

3j

The hypotenuse in Physics is called the RESULTANT or VECTOR SUM.

The LEGS of the triangle are called the COMPONENTS

Horizontal Component

Vertical Component

NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TAIL.

UNIT VECTOR NOTATION

direction z in the 1 r unit vecto-ˆ

directiony in the 1r unit vecto-ˆ

directionx in the 1 r unit vecto-ˆ

k

j

i

jiJ ˆ4ˆ2

jiK ˆ5ˆ2

The proper terminology is to use the “hat” instead of the arrow. So we have i-hat, j-hat, and k-hat which are used to describe any type of motion in 3D space.

How would you write vectors J and K in unit vector notation?

APPLICATIONS OF VECTORS

VECTOR ADDITION – If 2 similar vectors point in the SAME direction, add them.

Example: A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started?

54.5 m, E 30 m, E+

84.5 m, E

Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION.

APPLICATIONS OF VECTORS

VECTOR SUBTRACTION - If 2 vectors are going in opposite directions, you SUBTRACT.

Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement relative to where he started?

54.5 m, E

30 m, W-

24.5 m, E

NON-COLLINEAR VECTORS

When 2 vectors are perpendicular, you must use the Pythagorean theorem.

kmc

c

bacbac

8.10912050

5595Resultant 22

22222

95 km,E

55 km, N

A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT.

BUT…..WHAT ABOUT THE VALUE OF THE ANGLE???

Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the VALUE of the angle.

30)5789.0(

5789.095

55

1

Tan

sideadjacent

sideoppositeTan

N of E

55 km, N

95 km,E

To find the value of the angle we use a Trig function called TANGENT.

109.8 km

mjmi

NofEkm

ˆ55ˆ95

30@8.109

EXAMPLE

A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

1.28)5333.0(

5333.015

8

/17158

1

22

Tan

Tan

smRv

smjsmi

WofNsm

/ˆ15/ˆ8

1.28@/17

15 m/s, N

8.0 m/s, W

Rv

The Final Answer :

VECTORS

If two vectors to be added are at an angle other than 90°, then you can use the law of Cosines or the law of Sines.

Vectors in Multiple Dimensions

Section

5.1

Where Ѳ is the cosine of the angle between length A and B.

Law of Cosines: R2 = A2 +B2 – 2ABcosѲ

VECTORS

Law of Sines: R / sin Ѳ = A / sin a = B / sin b

Vectors in Multiple Dimensions

Section

5.1

a = angle opposite length A

b = angle opposite length B

SOLVE FOR THE MAGNITUDE

1) 55 m north + 97 m northeast

2) 39 m/s south + 45 m/s southeast + 62 m/s north

Finding the Magnitude of the Sum of Two Vectors

The steps covered were:

Step 1: Analyze and sketch the Problem

Sketch the two displacement vectors, A and B, and the angle between them.

Step 2: Solve for the Unknown

When the angle is 90°, use the Pythagorean theorem to find the magnitude of the resultant vector.

When the angle does not equal 90°, use the law of cosines to find the magnitude of the resultant vector.

Section

5.1

Step 3: Evaluate the answer: check units, does it make sense

MAGNITUDE AND DIRECTION

If the direction is not one of the cardinals (north, south, east, west, up down, left right)

Then you must describe it using an angle from one of the cardinal directions

Example:

How far east does one travel if they move 1356 m/s 670 NE?

VECTORS

When the motion you are describing is confined to the surface of Earth, it is often convenient to have the x-axis point east and the y-axis point north.

When the motion involves an object moving through the air, the positive x-axis is often chosen to be horizontal and the positive y-axis vertical (upward).

If the motion is on a hill, it’s convenient to place the positive x-axis in the direction of the motion and the y-axis perpendicular to the x-axis.

Components of Vectors

Section

5.1

Two or more vectors (A, B, C, etc.) may be added by first resolving each vector into its x- and y-components.

The x-components are added to form the x-component of the resultant: Rx = Ax + Bx + Cx.Similarly, the y-components are added to form the y-component of the resultant:

Ry = Ay + By + Cy.

Vectors

Algebraic Addition of Vectors

Section

5.1

Repetitive, Tedious, but easy

VECTORS

Because Rx and Ry are at a right angle (90°), the magnitude of the resultant vector can be calculated using the Pythagorean theorem, R2 = Rx

2 + Ry2.

Algebraic Addition of Vectors

Section

5.1

QUESTION 1

Jeff moved 3 m due north, and then 4 m due west to his friends house. What is the magnitude of Jeff’s

displacement?

Section

5.1

Calculate the resultant of three vectors A, B, C as shown in the figure.

(Ax = Bx = Cx = Ay = Cy = 1 units and By = 2 units)

QUESTION 2

EXAMPLE

Johnny and Susie push a box containing a tied up physics teacher on ice towards the water’s edge. If the mass of teacher and box is 95 kg, and Susie pushes with 150 N [N] while Johnny pushes with 200 N 59˚ east of south, find:

A) net force

B) acceleration

C) how far the box was pushed in 2 minutes

2ND EXAMPLE

What is the acceleration of an object whose initial velocity is 20 m/s [S] and final velocity is 30 m/s [E]. It took 2 seconds to change from initial to final velocity.

SCALAR MULTIPLICATION

SCALAR MULTIPLICATION is multiplying a vector by a scalar.

Example:

A=8i+7j+9K

2A=16i +14j + 18k

THE SCALAR DOT PRODUCT

In unit vector notation, it looks a little different. Consider:

The "Dot" product between these is equal to:

THE SCALAR DOT PRODUCT

What is the SIGNIFICANCE of the dot product?

DOT PRODUCTS IN PHYSICS

cos

cos

xFxFW

BABA

Consider this situation: A force F is applied to a moving object as it transverses over a frictionless surface for a displacement, d.

It is |F|Cos θ ! Because it is parallel to the displacement d that causes the object to move

In fact if you apply the dot product, you get (|F|Cos θ)d, which happens to be defined as "WORK" (check your equation sheet!)

Work is a type of energy and energy DOES NOT have a direction, that is why WORK is a scalar or in this case a SCALAR PRODUCT (AKA DOT PRODUCT).

THE “CROSS” PRODUCT (VECTOR MULTIPLICATION)Multiplying 2 vectors sometimes gives you a VECTOR quantity which we call the VECTOR CROSS PRODUCT.

The cross product between A and B produces a VECTOR quantity. The magnitude of the vector product is defined as:

Where is the NET angle between the two vectors. As shown in the figure.

AB

kBA

BABA

ˆ30

150sin512sin

THE VECTOR CROSS PRODUCT

AB

What about the direction???? Positive k-hat??? We can use what is called the RIGHT HAND THUMB RULE.•Fingers are the first vector, A•Palm is the second vector, B•Thumb is the direction of the cross product.•Cross your fingers, A, towards, B so that they CURL. The direction it moves will be either clockwise (NEGATIVE) or counter clockwise (POSITIVE)

In our example, the thumb points OUTWARD which is the Z axis and thus our answer would be 30 k-hat since the curl moves counter clockwise.

CROSS PRODUCTS AND UNIT VECTORS

zyx

zyx

AAA

BBB

kji

AB

ˆˆˆ

The cross product between B and A produces a VECTOR of which a 3x3 matrix is need to evaluate the magnitude and direction.

You start by making a 3x3 matrix with 3 columns, one for i, j, & k-hat. The components then go under each appropriate column.

Since B is the first vector it comes first in the matrix

CROSS PRODUCTS AND UNIT VECTORS

zyx

zyx

AAA

BBB

kji

AB

ˆˆˆ

)()(ˆyzzy ABABi

)()(ˆzxxz ABABj

)()(ˆxyyx ABABk

You then make an X in the columns OTHER THAN the unit vectors you are working with. •For “i” , cross j x k•For “j” , cross i x k•For “k” , cross i x j

Let’s start with the i-hat vector: We cross j x k

Now the j-hat vector: We cross i x k

Now the k-hat vector: We cross i x j

EXAMPLE

44)4)(6()5)(4(ˆ i 8)5)(2()3)(6(ˆ j

20)3)(4()4)(2(ˆ k

kjiAB ˆ20ˆ8ˆ44

Let’s start with the i-hat vector: We cross j x k

Now the j-hat vector: We cross i x k

Now the k-hat vector: We cross i x j

The final answer would be:

THE SIGNIFICANCE OF THE CROSS PRODUCT

In this figure, vector A has been split into 2 components, one PARALLEL to vector B and one PERPENDICULAR to vector B. Notice that the component perpendicular to vector B has a magnitude of |A|sin θ

THEREFORE when you find the CROSS PRODUCT, the result is: i) The MAGNITUDE of one vector, in this case |B| and, ii) The MAGNITUDE of the 2nd vector's component that runs perpendicular to the first vector. ( that is where the sine comes from)

CROSS PRODUCTS IN PHYSICS

The cross product system will also be used in mechanics (rotation) as well as understanding the behavior of particles in magnetic fields.

sin

sin

rFrF

BABA

A force F is applied to a wrench a displacement r from a specific point of rotation (ie. a bolt). Common sense will tell us the larger r is the easier it will be to turn the bolt.

But which part of F actually causes the wrench to turn? |F| Sin θ

CROSS PRODUCTS IN PHYSICS

sin

sin

rFrF

BABA

What about the DIRECTION?

Which way will the wrench turn? Counter Clockwise

Is the turning direction positive or negative? Positive

Which way will the BOLT move? IN or OUT of the page? OUT

You have to remember that cross products give you a direction on the OTHER axis from the 2 you are crossing. So if “r” is on the x-axis and “F” is on the y-axis, the cross products direction is on the z-axis. In this case, a POSITIVE k-hat.

FORCES IN

TW

O

DIMENSIO

NS

FORCES IN TWO DIMENSIONS

An object is in equilibrium when the net force on it is zero. When in equilibrium the object is motionless or move with constant velocity.

Equilibrium also occurs when the resultant of three or more forces = a net force of zero.

The equilibrant force is equal in magnitude but opposite in direction to the resultant vector.

AB

A

BB

A Equilibrant

R

EXAMPLE PROBLEM

Creating Equilibrium

A 168-N sign is supported in a motionless position by two ropes that each make 22.5 angles with the horizontal. What is the tension in the ropes?

Sign

+y

22.5 22.5

Fg

FB

+x

FA

MOTION ALONG AN INCLINE PLANE

A tilted surface is called an inclined plane.

Objects accelerate down inclined planes because of an unbalanced force.

The two forces acting upon a crate which is positioned on an inclined plane (assumed to be friction-free) are the force of gravity and the normal force.

- The force of gravity (weight) acts in a downward direction; yet the normal force acts in a direction perpendicular to the surface.

Fgx = Fװ= Fgsin

Fgy = F= Fgcos

Fg = Weight= mg

EXAMPLE PROBLEMComponents of Weight for an Object on an incline.

A trunk weighing 562 N is resting on a plane inclined 30.0 above the horizontal. Find the components of the weight force parallel and perpendicular to the plane.

Skiing DownhillA 62-kg person on skis is going down a hill

sloped at 37 . The coefficient of kinetic friction between the skis and the snow is 0.15. How fast is the skier going 5.0 s after starting from rest?

• Force is a vector• Vectors are made of components

• The components are independent of each other

Section 1: Simple Breakdown of Forces

You can break down forces into several components easily. For example, the force F1 can be broken into two forces: Fx and Fy.

Section 2. Two Dimensional Forces into OneJack pushed a box with a force of 30 N at 0 degree and Michael

pushed it with a force of 40 N at 45 degrees. How can we find the net force acting on the box?

The first thing you have to do is to find all forces on x direction (x axis) only.

30 N + (cos 45 * 40) N = 58.3 N. East

Then, you will have to analyze all forces on y direction (y axis). 0 N + (sin 45 * 40) N = 28.3 N. North

Combined forces: use Pythagorean Theorem, we can calculate that

N NorthEast

is the magnitude (size) of the combined forces.

Section 3: One Dimensional Forces into TwoYou can also break down forces. For example, Fred pushed a box to the east and Jack pushed it to the north. If the net force is 100 N to north east by 45 degrees,

the force applied by Fred would beFFred = cos 45 * 100 = 70.7 N

and the force by Jack isFJack = sin 45 * 100 = 70.7 N

Section 4. Forces involving Gravity

This force, called the force of parallel (F//),causes the box to move forward. F// can be calculated by Fg * sin x.

To conclude, the mixture of the force of parallel and the force of friction determines how the box moves.

F// > fr box will slide.

F// = fr box will not slide.

F perpendicular is equal in magnitude to the normal forceF perpendicular = Fg cos (x)

QUESTION

Draw a free body diagram of a skier moving down an inclined slope

FORCE OF GRAVITY, NORMAL FORCEWhat force is making the skier

move down the slope?

What force is pushing the skier into the ground?

What direction is the normal force?

On an inclined plane, the magnitude of the normal force between the object and the plane will usually not be equal to the object’s weight.

PROBLEM 1

If the skier has a mass of 50 kg and the angle of the slope is 30 degrees from horizontal, find

1) What is the angle of inclination?

2) What is the amount of force pushing object into ground? (force perp?)

3) What is the amount of force pulling the skier down the hill? (force parallel)

PROBLEM #2

Skier from problem moves down 45 ˚hill.

Part 1) What is their acceleration, if hill was considered smooth?

Part 2) What is their acceleration if hill has a coefficient of friction of 0.21?

Part 3) How far down the hill will the skier travel in 33 seconds in part 2

PROBLEM 3

What minimum coefficient of friction value would be needed on the slope for the skier from problem #2 to be stuck (not accelerate down the slope?)

FRIC

TION R

EVISIT

ED

FRICTION

Frictional force depends on the materials that the surfaces are made of.

For example, there is more friction between skis and concrete than there is between skis and snow.

The normal force between the two objects also matters. The harder one object is pushed against the other, the greater the force of friction that results.

Values that determine the amount of friction

Section

5.2

The coefficients of friction are nearly independent of the area of contact

The forces themselves, Ff and FN, are at right angles to each other. The table here shows coefficients of friction between various surfaces.

Although all the listed coefficients are less than 1.0, this does not mean that they must always be less than 1.0.

Friction

Static and Kinetic Friction

Section

5.2

KINETIC FRICTION

The force of kinetic friction acts when the object is in motion

Frk = µk Fn

STATIC FRICTION, ƑS

Static friction acts to keep the object from moving

If F increases, so does ƒs

If F decreases, so does ƒs

Frs µs Fn

CALCULATION OF FRICTION

Ff = µ Fn

Ff or f = Force of frictionµ = Coefficient of friction

Fn = Normal force

FORCE OF FRICTION PROBLEMS Only the normal force determines the amount

of frictional force applied to a moving object

Generally, the normal force equals the weight of sliding object except at an angle then it is the perpendicular component of the weight

Other forces applied to an object may also effect friction if they effect the amount of force applied to ground by object

DETERMINING NORMAL FORCE

What force is dragging suitcase across floor?

What forces effect normal force on suitcase?

The applied force of woman must be divided up into 2 components

One horizontal, which tries to pull suitcase along ground, T//

One vertical, which tries to lift suitcase off ground, Tperp

Force applied by woman

T//

Tper

p

FN

W

DETERMINING NORMAL FORCE

X componentsFnet x = Tx – Fr = 0Tx = Fr

Y componentsFnet y = Ty + FN – W = 0FN = W - Ty