Transcript Presentation

Unit 6
Analytical Vector Addition
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
1
Multiplying a Vector by a Scalar
A
B
A
C
½A
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A
B = 2A
C = -1/2 A
2
Adding “-” Vectors
 Add “negative” vectors
by keeping the same
magnitude but adding
180 degrees to the
direction of the original
vector.
B
C
-B
A
D
C=A+B
D=A-B
D = A + (- B)
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3
Components of Vectors
A = Ax + Ay
A

Ay
Ax
Recall: Vectors are always added “head to tail.”
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4
Components of Vectors
Finding the components when you know A.
A = Ax + Ay
Ax
 cos   Ax  A cos 
A
Ay
A
 sin   Ay  A sin 
Recall:  is measured
from the positive x axis.
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5
From the + x Axis
If A = 3.0 m and  = 45, find Ax + Ay.
Ay  A sin 

Ax  A cos
Ax  A cos  3.0m cos( angle ?)
A
 x
Ay
A
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Ax  A cos  3.0m cos( 45)
Ax  2.1m
Ay  A sin   3.0m sin( 45)
Ax  2.1m
6
Components of Vectors
Finding the vector magnitude and direction
when you know the components.
A  Ax2  Ay2
tan  
Ay
Ax
   arctan
Ay
Ax
Recall:  is measured
from the positive x axis.
Caution: Beware of the tangent function.
Always consider in which quadrant the vector lies when
dealing with the tangent function.
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7
If Ax = 2.0 m and Ay. = 2.0 m, then Find A and .
A A  A  2 2  8
2
x

2
y
2
2
A
 x
Ay
A
2
  arctan
 arctan
 45
Ax
2
Ay
  360   315
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
8
Adding Vectors With Components.


A  72.4 @ 58


B  57.3 @ 216

C  17.8 @ 270
   
D  A B C

C

D

B

A
Ay  A sin 
Ax  A cos
A  Ax2  Ay2
  arctan
Ay
Ax
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9
Component Template
Follow this type of methodology when doing these problems.
Magnitude
A=72.4m
B = 57.3 m
C = 17.8 m
Angle
58
216 
270 
R = 12.7 m
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Rx
38.37 m
-43.36 m
0.0 m
Rx = -7.99 m
Ry
61.4 m
-33.68 m
-17.80 m
Ry = 9.92m
Angle = 129 degrees
10
Hand Glider Trip
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11
Analytical Vector Addition – Hand Glider
 Find the Final displacement of the
hand glider using analytical vector
addition.
 This problem is similar to the
following problems: WS 21, 1; WS
22, 1; and WS 23, 4.
Vector
Magnitude
Angle
x component
y component
Quad
--------------
--------------
--------------
--------------
--------------
--------------
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12
Analytical Vector Addition
 Three Physics track robots pull on a book as
shown.
 They pull with the following forces: 7.0 N
@ 45, 8.0 N @ 180, and 5.0 N @ 270 .
 Find the net force applied to this most
valuable book.
 This problem is similar to WS 21, 2.
Vector
Magnitude
Angle
x component
y component
Quad
--------------
--------------
--------------
--------------
--------------
--------------
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13
Statics – Hanging Sign
 Draw the Static FBD for the sign below
 What do we need to do with the tension (T)?
 Resolve it into its components (Tx & Ty).
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14
Statics – Hanging Sign
 Draw the Static FBD for the sign below
 What do we need to do with the tension (T)?
 Resolve it into its components (Tx & Ty).
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15
Inclined Plane Problems
 Draw the FBD for the piano on the inclined plane.
 What will we have to do with the Normal Force (N) and the force of
friction (Ff)?
 Resolve them into their x and y components.
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16
Inclined Plane Problems
 Would you like to do less work?
 How could we do this problem by resolving only one force?
 Try rotating the FBD so that the N is in the y plane and the Ff is in the x
plane.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
17
Analytical Vector Addition
 Use the table below when performing analytical vector addition.
 Do WS 23 numbers 1 & 2.
Vector
Magnitude
Angle
x component
y component
Quad
--------------
--------------
--------------
--------------
--------------
F1
F2
F3
F4
-------------F
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18
Relative Velocity
Vr
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19
Relative Velocity
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20
Relative Velocity
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21
Relative Velocity
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22
Relative Velocity
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23
Relative Velocity
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24
Relative Velocity
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25
Relative Velocity
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26
Relative Velocity
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27
Relative Velocity
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28
Relative Velocity
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29
Relative Velocity
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30
Relative Velocity
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31
Analytical Vector Addition
 Do WS 23 number 3.
 This problem is similar to WS
24, 3 and WS 22 , 2.
Vector
Magnitude
Angle
x component
y component
Quad
--------------
--------------
--------------
--------------
--------------
F1
F2
F3
F4
-------------F
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32
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33
Analytical Vector Addition
 A
Vector
Magnitude
Angle
x component
y component
Quad
--------------
--------------
--------------
--------------
--------------
F1
F2
F3
F4
-------------F
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34
Setting the Standard
 When we do problems involving
kinematics, it is important that
we stick to a standard when
imputing data into the knowwant table.
 This standard enables us to take
into account the vector nature of
acceleration, velocity,
displacement, etc.
 Here is a diagram we will use in
order to help us correctly input
data into the table.
 This standard is based upon the
Cartesian Coordinate system.
 If a body travels West, then what
sign would you give its velocity?
 If a body travels at an angle of
90 degrees, then what sign
would you give its velocity?
3-3
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35
Dot Products
 Write each of the three vectors
given in their unit vector notation.
 A = 15.0 m @ 30
 B = 22.0 m @ 225
 C = 9.0 m @ 267
 Calculate the Dot Products below.
A B 
AC 
BB 
BC 
CA
3-3
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36
Dot Products – Finding the angle





Given the vectors below, find the angles between the following vectors.
A and C.
B and A.
C and E.
D and E.
A  3iˆ  2 ˆj  4kˆ
B  3.5iˆ  4 ˆj  2kˆ
C  1.5iˆ  4 ˆj  2kˆ
D  5 ˆj
E  1.8iˆ  2.2kˆ
3-3
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37
3-D Cartisian Coordinate System
+y
A  Ax  Ay  Az

+x
A A  A  A
2
x
2
y
2
z
+z
1-17
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38
Unit Vectors
y
A = Axi + Ayj + Azk
j
-k
-i
i
k
z
-j
x Note: Remember to put the
“^” over the hand written
vector when writing unit
vectors.
1-18
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39
Scalar or “Dot” Product
B
B

A
AB
A
BA
The Dot product gives the projection of one vector onto another.
You can also use the dot product to find the angle between the vectors.
BA = Projection of B onto A.
AB = Projection of A onto B.
1-19
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40
Scalar or “Dot” Product
B

A
The Dot product results in a Scalar quantity.
A  B  AB cos   BA cos   B  A
1-20
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41
Scalar or “Dot” Product & Unit Vectors
You “multiply” the dot product in a similar way as below.
4 x  2 y x  3 y   4 x 2  12 xy  2 yx  6 y 2  4 x 2  14 xy  6 y 2
A = Axi + Ayj
B = Bxi + Byj
A•B = (Axi + Ayj) • (Bxi + Byj)
A•B = Ax i • Bxi + Ax i • Byj+ Ayj • Bxi + Ayj • Byj
However,
i•i=j•j=k•k=1
i•j=i•k=j•k=0
A•B = Ax Bx + Ay By
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1-21
42
Scalar or “Dot” Product
B

A
One use for the dot product is to
determine the angle between two
vectors.
A  B Ax Bx  Ay By 
  arccos

AB
AB
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1-22
43
Vector or “Cross” Product
R1

B

A
R1  A  B
B
A
R2
R2  B  A
The Cross product results in a VECTOR quantity.
Right hand rule: Place the fingers of your right hand in the
direction of the first vector in the cross product. Rotate your
fingers towards the second vector. Your thumb tells you the
direction of the resultant vector.
1-23
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44
Vector or “Cross” Product
R1

B
A
R1  A  B
The Cross product results in a VECTOR quantity.
The magnitude of the vector is given by
R1  AB sin 
WARNING: AB sin  DOES NOT EQUAL BA sin 
A x B DOES NOT EQUAL B x A
However, A x B = - B x A
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
1-24
45
Vector or “Cross” Product
AxB = (Axi + Ayj) x (Bxi + Byj)
AxB = Ax i x Bxi + Ax i x Byj+ Ayj x Bxi + Ayj x Byj
However,
ixi=jxj=kxk=0
i x j = -j x i = k
j x k = -k x j = i
K x i = -i x k = j
AxB = Ax i x Byj+ Ayj x Bxi
AxB = (AxBy )i x j + (AyBx)j x i
AxB = (AxBy )k - (AyBx)k
1-25
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46
Vector or “Cross” Product
Determinant method of solving for the cross product.
A = Axi + Ayj + Azk
AxB=
i
Ax
Bx
j
Ay
By
B = Bxi + Byj + Bzk
k
Az
Bz
i
Ax
Bx
j
Ay
By
A x B = (Axi + Ayj + Azk) x (Bxi + Byj + Bzk)
A x B = AyBzi - AyBxk + AzBxj - AzByi + AxByk - AxBzj
A x B = (AyBz –AzBy)i + (AzBx-AxBz)j + (AxBy-AyBx)k
A x B = Rx i + Ry j + Rz k
1-26
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47
Spherical Coordinates
r xy
z
y  r sin 
r
x  r cos
r   sin 
z   cos 

y    sin   sin 
x    sin   cos

y
r
z

x
y
x
  xiˆ  yjˆ  zkˆ
y
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48
Advanced Physics Unit 5
Applications of
Newton’s Laws
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49
Newton’s Laws – A Review
 Newton’s First Law - An object remains at rest, or in uniform motion in a





straight line, unless it is compelled to change by an externally imposed
force.
Newton’s first law describes an Equilibrium Situation.
An Equilibrium Situation is one in which the acceleration of a body is
equal to zero.
Newton’s Second Law – If there is a non-zero net force on a body, then it
will accelerate.
Newton’s Second Law describes a Non-equilibrium Situation.
A Non-equilibrium Situation is one in which the acceleration of a body is
not equal to zero.
 Newton’s Third Law - for every action force there is an equal, but opposite,
reaction force.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
50
Free Body Diagrams – A Review
 When solving problems involving forces, we must draw FBDs of all bodies involved





in the force interactions.
Since torque is related to force, we must modify the FBD concept to apply to bodies
upon which a torque acts.
Before we carry out this modification, lets review problems involving force using
FBDs.
If the crate started from rest, then which way did it accelerate?
Draw the FBD for the crate.
What type of a situation is depicted below?
Dig
Dug
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51
Free Body Diagrams – Inclined Plane (WS 14 # 8)
 A block slides down an inclined plane as
shown.
 Draw the FBD for the block as it slides down
the ramp at a constant speed.
 Write the Newton’s laws in vector form for
the block in both the horizontal and vertical
directions.
 Now convert from vector form to math form.
Fx  FFx  N x  max
Fy  FFy  N y  W  ma y
Fx   FFx  N x  max
Fy  FFy  N y  W  ma y
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
52
WS 14 Problems 1-4





A mass rest on an inclined plane as shown.
What type of friction is acting on the mass?
Now suppose the mass begins to slide down the plane.
What type of friction is acting on the mass as it slides?
Draw the FBD for the mass while at rest and while
sliding down the plane.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
53
Static Friction v. Kinetic Friction
 Static friction exists when an object wants to move but is held in
place by the force of friction.
 This force of friction is greater than the component of the weight
acting down the plane.
 If we continue to rotate the plane, the component of the weight
acting down the plane will eventually become larger than the
normal force.
 When this happens, the object will begin to slide changing from
static friction to kinetic friction.
Force of Friction v. Time
1
0
Friction (N)
-1 0
2
4
6
8
10
-2
-3
-4
-5
-6
-7
Time (s)
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
54
Like WS 14 Problem 5
 MeanyBot and PhysicsBot are moving a crate as shown.
 MeanyBot is pulling with a force F2 = 10,000 N and the PhysicsBot
is pushing with a force of F1 = 6,000 N.
 Additionally, the coefficient of kinetic friction, k, is 0.459.
 The mass of the crate is 1000 kg.
 Determine the net force and the acceleration of the crate.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
55
WS 14 Problem 9
 Derive the equations needed to
determine the tension and the
acceleration of the weights (m1<m2)
on the Atwood’s machine shown to
the right..
 What type of a situation do we have
when the masses first begin to move?
 Define this situation with its two
predominate characteristics.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
56
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
57
WS 14 Problem 10 – Elevator Problems
 An elevator (m = 675.0 kg) ascending at a rate of 8.5 m/s
comes to a stop in a distance of 22.0 m.
 Find the Tension in the three cables supporting the weight of
the elevator and the acceleration experienced by the elevator.
 What type of a situation is the elevator in while coming to a
stop?
 Now suppose Dr. Physics (m = 62.5 kg) is standing on a
scale inside the elevator.
 After three seconds of descending, the elevator begins
traveling at a constant speed of 9.0 m/s.
 What does the scale say that Dr. Physics weighs while he
descends?
 What situation is the elevator in once it begins traveling at a
constant velocity?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
58
WS 15 Problem 1
 Block A below weighs 90.0 N.
 The coefficient of static friction
between the block and the table is s
= 0.30.
 Block B weighs 15.0 N.
 The system is in equilibrium.
 Draw and label the FBDs for Body A
& Body B.
 What is meant by the term
“equilibrium” above?
 Find the friction force acting on Block
A.
TB
N
TA
WA
WB
45°
A
B
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59
WS 15 Problem 2
 A wooden block (m1) rests on
a plane inclined at an angle of
.
 This block is attached to
mass m2 held at a height of y
above the ground.
 The coefficient of friction
between the block and the
incline is K.
 Derive the equations (in
terms of m1, m2, K, y, , and
g) needed to calculate the
tension in the string, the
acceleration of the system,
and the time needed for to
hit the ground.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
T2
N
T1
W1
W2
60
WS 15 Problem 3
 Derive the equations needed to determine the
tension in each chain given the angle , that the
angle between chain 2 and the post is 90, and the
fact that the weight of the bug zapper is W.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
T1

T2
61
WS 15 Problem 4
 Derive the equations needed to
determine m2 and the tensions in
the strings given angles  and  that
the mass of weight one is m1.
 Suppose m1 = 10.0 kg. What are
the tensions and what is the value
of m2?
 Suppose m1 = 6.0 kg. What are the
tensions and what is the value of
m2?
 Suppose m1 = 2.9 kg. What are the
tensions and what is the value of
m2?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.

T1
T3
m2
T2

m1
62
222222
 A wrapped box (m1) rests on
a table and is attached to a
hanging weight (m2) as
shown.
 The coefficient of friction
between the box and the table
is K.
 The weight is released
pulling the box to the right as
shown.
 Derive the equations (in
terms of m1, m2, K, and g)
needed to calculate the
tension in the string and the
acceleration of the system.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
T2
N
T1
W1
W2
63
Relative Velocity
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64
Relative Velocity
T1
T2
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

65
Torque
 a
  F l
1  W1  l1
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66
Torque
 a
  F l
1  W1  l1
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67
Torque
 a
  F l
1  W1  l1
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
68
Static Friction v. Kinetic Friction
 Static friction exist when an object wants to move but is held in
place by the force of friction.
 This force of friction is greater than the component of the weight
acting down the plane.
 If we continue to rotate the plane, the component of the weight
acting down the plane will eventually become larger than the
normal force.
 When this happens, the object will begin to slide changing from
static friction to kinetic friction.
Force of Friction v. Time
1
0
Friction (N)
-1 0
2
4
6
8
10
-2
-3
-4
-5
-6
-7
Time (s)
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
69
Torque
 a
  F l
1  W1  l1
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70
Torque
 a
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71
Torque
 a
  F l
1  W1  l1
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72
Advanced Physics Unit 5 Exam
QUESTION 1 - 3
QUESTION 4
QUESTION 5
QUESTION 6
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73
Unit 5 Exam Problems 1-3
 The graph below shows the force of friction verses the pull time.
 What type of friction is represented in the portion of the graph that is blue in
color?
 What type of friction is represented in the portion of the graph that is red in
color?
 What is physically happening at the point where the graph changes from blue
to red?
Force of Friction v. Time
1
0
Friction (N)
-1 0
2
6
4
8
10
-2
-3
-4
-5
-6
-7
Time (s)
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
RETURN
74
Unit 5 Exam Problem 4
 Under what conditions would a Elevator passenger appear to
weigh more than his or her actual weight: while accelerating
upwards, while accelerating downwards, or while riding at a
constant speed?
 Justify your answer using verbal explanations or equations
as needed.
RIDE
RETURN
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
75
Unit 5 Exam Problem 5
 A wrapped box (m1) rests on
a table and is attached to a
hanging weight (m2) as
shown.
 The coefficient of friction
between the box and the table
is K.
 The weight is released
pulling the box to the right as
shown.
 Derive the equations (in
terms of m1, m2, K, and g)
needed to calculate the
tension in the string and the
acceleration of the system.
RETURN
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
76
Unit 5 Exam Problem 6
 Derive the equations needed to
determine the tension in each
chain given angles  and  and
the fact that the weight of the
bug zapper is W.

T1

T2
RETURN
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
77