1 honors physics chapter 5 forces in two dimensions

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Honors Physics Chapter 5

Forces in Two Dimensions

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Honors Physics

Turn in H4 & W4 Lecture Q&A

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Vector and Scalar

Vector: – Magnitude: How large, how fast, … – Direction: In what direction (moving or pointing)– Representation depends on frame of reference– Position, displacement, velocity, acceleration, force,

momentum, …

Scalar: – Magnitude only– No direction– Representation does not depend on frame of reference– Mass, distance, length, speed, energy, temperature, charge, …

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Vector symbol

Vector: bold or an arrow on top and v V

– Typed: v and V or

Scalar: regular– v or V

v stands for the magnitude of vector v.

– Handwritten: and v V

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Graphical representation of vector: Arrow

An arrow is used to graphically represent a vector.

a b

– The direction of the arrow represents the direction of the vector.

head

tail

– The length of the arrow represents the magnitude of the vector. (Vector a is smaller than vector b because a is shorter than b.)

– When comparing the magnitudes of vectors, we ignore directions.

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Equivalent Vectors

Two vectors are identical and equivalent if they both have the same magnitude and are in the same direction.

A B

C– A, B and C are all equivalent vectors.

– They do not have to start from the same point. (Their tails don’t have to be at the same point.)

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Negative of Vector

Vector -A has the same magnitude as vector A but points in the opposite direction.

If vector A and B have the same magnitude but point in opposite directions, then A = -B, and B = -A

A -A

-A

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Adding Vectors

Graphical– Head-to-Tail (Triangular)

Algebraic (by components)

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Adding Vectors: Head-to-Tail

Head-to-Tail method:A

B

A

BA+B

Make sure arrows are parallel and of same length.

– Draw vector A– Draw vector B starting

from the head of A– The vector drawn from the

tail of A to the head of B is the sum of A + B.

Example: A + B

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A+B=B+A

A

A

A

B

B

BA+BB+A

A + B

How about B + A?

What can we conclude?

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A+B+C

A

B

A+B+C

C

A B C

Resultant vector:

from tail of first to head of last.

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A-B=A+(-B)

A B

-B

A

A-B

– Draw vector A– Draw vector -B from head

of A.– The vector drawn from the

tail of A to head of –B is then A – B.

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Think …

Two forces are acting on an object simultaneously, one is 4 N and the other is 5 N. a) What is the maximum possible resultant force on the object? How are these two forces oriented relative to each other when this happens?

b) What is the minimum possible resultant force on the object? How are these two forces oriented relative to each other when this happens?

9 N, same direction

1 N, opposite direction

4N 5N

9N

4N5N

1N

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What are the relations among the vectors?

A

CB

A

CB

a

bc

A + C = B

a + b + c = 0

a + b = -c

a

bc

A + C

a + b

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35o

a

ba+

b

-a

b-a

a+b: 4.3 unit, 50o North of East

b-a: 8.0 unit, 66o West of North

Using ruler and protractor, we find:

50o 66o

Example:

Vector a has a magnitude of 5.0 units and is directed east. Vector b is directed 35o west of north and has a magnitude of 4.0 units. Construct vector diagrams for calculating a + b and b – a. Estimate the magnitudes and directions of a + b and b – a from your diagram.

E

N

W

S

35o

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Law of cosine

Cabbac cos2222

b

a

c

C

2 2 2a b c Pythagorean’s Theorem:

C = 90o

a

b c

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Law of Sine

C

c

B

b

A

a

sinsinsin

a

c

b

B A

C

sin sin sinA B C

a b c or

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Example: 125-7You first walk 8.0 km north from home, then walk east until your displacement from home is 10.0 km. How far east did you walk?

E

N

0

b

ar

8.0 , 10.0 , ?a km r km b

2 2 2a b r

2 2b r a

2 210.0 8.0 6.0km km km

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Practice: 121-3A hiker walks 4.5 km in one direction, then makes an 45o turn to the right and walks another 6.4 km. What is the magnitude of her displacement?

a

bc45o

4.5 , 6.4 , ?a km b km c

180 45 135o o oC

2 2 2 2 cosc a b ab C

2 2

2 2

2 cos

4.5 6.4 2 4.5 6.4 cos135

10.

o

c a b ab C

km km km km

km

135o

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Vector Components

Drop perpendicular lines from the head of vector a to the coordinate axes, the components of vector a can be found:

cos

sinx

y

a a

a a

x

y

a

is the angle between the vector and the +x axis.

ax and ay are scalars.

ax

ay

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Useful Trigonometry

adjacent

opposite

hypotenuse

adjacent

hypotenuse

opposite

tan

cos

sin

adjacent side

hypotenuse

oppo

site

sid

e

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Finding components of a vector

Finding components of a vector

,x ya a a

– Resolving the vector– Decomposing the vector

ax is– the component of a in the x-direction– the x component of a.

Component form:

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Practice: A heavy box is pulled across a wooden floor with a rope. The rope makes an angle of 60o with the floor. A force of 75 N is exerted on the rope. What are the components of the force parallel and perpendicular to the floor?

T

Tx

||T

T

75 , 60 , ?, ?ox yT N T T Ty

75 cos60

38.

oN

N

75 sin60

65.

oN

N

xT cosT

yT sinT

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Vector magnitude and direction

The magnitude and direction of a vector can be found if the components (ax and ay) are given:

2 2magnitude: x ya a a

is the angle from the +x axis to the vector.

(for 3-D)2 2 2x y za a a

1 directi : non ta y

x

a

a

a

ax

ay

x

y

tan y

x

a

a

,a a

Magnitude and direction form:

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Example:A car is driven 125.0 km due west, then 65.0 km due south. What is the magnitude and direction of its displacement?

a

br

y

x

125.0 , 65.0 , ?, ?x yr a km r b km r

r

2 2

2 2

125.0 65.0 141

x yr r

km km km

1 1 65.0tan tan 27.5

125.0y o

x

r km

r km

Set up the frame of reference as to the right. Then

Magnitude is 141 km, at 27.5o South of West.

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Adding Vectors by Components

When adding vectors by components, we add components in a direction separately from other components.

r a b

3-D

x

y

ax

ay bx

by

a

b

r

y y yr a b xr x xa b

rx

ry

z z zr a b

2-D

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2.28 ,77.0oa N

xa

ba

y

x

W

2.28,103.0ob cos 2.28cos103 0.513

sin 2.28sin103 2.22

ox

y

b b N

b b N

a b

cos 2.28cos 77.0 0.513oa N

ya sin 2.28sin 77.0 2.22oa N

yyxx baba , 0.513 0.513, 2.22 2.22 0, 4.44N

Practice: 125-8A child’s swing is held up by two ropes tied to a

tree branch that hangs 13.0o from the vertical. If the tension in each rope is 2.28 N, what is the combined force (magnitude and direction) of the two ropes on the swing?

ab

a+b

b

2.28 , 2.28 , 90 13.0 77.0 , 90 13.0 103o o o o o oa ba N b N

4.44 ,N upward

Let x = horizontal, y = vertical, then

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Friction

Friction: force opposing the motion or tendency of motion between two rough surfaces that are in contact

s sf N

k kf N

Static friction is not constant and has a maximum:

depends on the properties of the two surfaces = 0 when one of the surfaces is smooth (frictionless.) s > k for same surfaces. has no unit.

o N: normal force between the two surfaces s (and k) is coefficient of static (and kinetic) friction.

Kinetic (or sliding) friction is constant:

,maxs sf N

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Fapp

f

0

Table

not m

oving

yet.

Static

fricti

on in

creas

es as

appli

ed

force

incre

ases

. fs =

F appsN Once table is moving, a smaller

constant kinetic friction. fk < fs, max

But what if the applied force increases just slightly?

•

•

sN

•

• •

Example:What happens to the frictional force as you increase the force pushing on a table on the floor?

Fapp f

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Example: 128-17A girl exerts a 36-N horizontal force as she pulls a 52-N sled across a cement sidewalk at constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance.

We only consider the motion in

the horizontal direction. N=

W=52N

Fp=36Nf

netF

36pf F N

f N 36

0.6952

f N

N N

pF f ma 0

52N

Fnet needs to include only forces in the horizontal direction. Define right to be the positive direction.

+

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Practice: 128-18You need to move a 105-kg sofa to a different location in the room. It takes a force of 102 N to start it moving. What is the coefficient of static friction between the sofa and the carpet?

N

W

Fp=102 Nf

netF

+

102pf F N

,maxsf ,max 1020.0991

1029s

s

f N

N N

pF f ma 0

Let right = + direction. Fnet includes only forces in the horizontal direction. Then

105 , 102 , ?pm kg F N

2105 9.8 1029

mN W mg kg N

s

N

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Equilibrium and Equilibrant Force

Concurrent forces: forces acting on the same object at the same time.

Equilibrium: Fnet = 0 Equilibrant Force: A force that produce equilibrium

when applied to an object.

A

BCC is the equilibrant

force of A + B because

(A + B) + C = 0.A + B + C = 0

C = - (A + B)

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Example

Find the equilibrant force c to a + b if a and b are given as followed. What is the relationship between c and a + b?

a

b b

a

c

c = - (a + b)

a + b

a + b + c = 0

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Example: Two forces act on an object. A 36-N force acts at 225o, a 48-N force acts at 315o. What would be the magnitude and direction of their equilibrant?

: 36 ,225oAF N

FA

FB

FA +F

B

?),315,48(),225,36( Co

Bo

A FNFNF

: 48 ,315oBF N

8.4Ax BxF F N

CF

36 cos225 25.5

36 sin 225 25.5

oAx

oAy

F N N

F N N

48 cos315 33.9

48 sin315 33.9

oBx

oBy

F N N

F N N

25.5 33.9 8.4N N N

Ay ByF F 25.5 33.9 59.4N N N

2 2 2 2( 8.4 ) (59.4 ) 60.Cx CyF F N N N

1 1 159.4tan tan tan 7.07 82 98

8.4Cy o o

Cx

F N

F N

FC

x

y

CxF

Ax BxF F

59.4Ay ByF F N CyF

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More Application on Force:Example: 135-40Stacie, who has a mass of 45 kg, starts down a slide that is inclined at an angle of 45o with the horizontal. If the coefficient of kinetic friction between Stacie’s shorts and the slide is 0.25, what is her acceleration?

W

N

x

y

Wx

Wy

f

netF

netF

N

f

sin cosmg mg ma

sin cosa g g

x:

y:

2 29.8 sin 45 0.25cos 45 5.20o om m

s s

xW f ma

yN W yma 0

yW cosmg N cosmg

W

sin cosg

xW sinmg

W

cosmg

sinW

yW cosW

What if frictionless incline?

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Practice:What is the acceleration of a box on a smooth (frictionless) incline that makes an angle of 30o with the horizontal, as in the diagram?

W

N

x

y

Wx

Wy

netF

netF

sinW ma

sina g

x:

y:

2 29.8 sin 30 4.9om m

s s

xW ma

yN W yma 0

W

xW sinmg

W

cosmg

sinW

yW cosW

sinmg ma

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Weight hangingWhat are the two tensions if the block has a mass of 5.0 kg and the upper string makes an angle of 30o with the horizontal?

T1

T2

Nx

y

5.0kg

•PT2x

T2y

T1

W

1T

y: .net yF 2 1yT T ma 0

1T

2T

2 yT 2 sinT

1 4998

sin sin 30oT N

N

Consider forces acting on point P:

+

Hanging mass:

netF 1T W ma 0

25.0 9.8 49

mkg N

s

W mg