Transcript AP Physics Review - The Burns Home Page
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Vectors
Vectors
A vector is a quantity that involves both magnitude and direction
.
55
km/h [ N35E ]
A downward force of
3
Newtons
A scalar direction.
is a quantity that does not involve
55
km/h
18
cm long
Column Vectors
Vector a 2 up a 4 RIGHT
2 up b
Column Vectors
Vector b 3 LEFT
3 2
Describe these vectors
4
c a
1 2 3
b d
4 3
A
Alternative labelling
B D
EF
E
AB CD
G F C
GH
H
General Vectors
A vector has both Length and direction
k k k
All 4 vectors have same length and same direction
k
General Vectors
A C k F B D Line CD is Parallel to AB CD is TWICE length of AB 2k -k E Line EF is Parallel to AB EF is equal in length to AB EF is opposite direction to AB
Write these Vectors in terms of k
A k 2k 1½k C B E D ½k F -2k H G
Vector Notation
Vectors are often identified with arrows in graphics and labeled as follows:
We label a vector with a variable.
This variable is identified as a vector either by an arrow above itself :
A
Or By the variable being
BOLD
:
A
Displacement
Displacement Distance
by an object.
is an object’s change in position. is the total length of space traversed 1m 6.7m
3m 5m Displacement: 2 6.7
m
Distance: 5
m
3
m
1
m
9
m
Start Finish = 500 m Displacement = 0
m
Distance = 5 00
m
Vector Addition
R E D A B C D R E B A C B D A R E C A + B + C + D + E
= Distance
R
= Resultant = Displacement
R
Rectangular Components
y
Quadrant II
A
2
B
2
R
sin
-x
A
Quadrant I
R
R
cos
B
x
Quadrant III Quadrant IV sin
A
opp R hyp
cos
B R
adj hyp
tan
A
B opp adj -y
Resultant of Two Forces
• force: action of one body on another; characterized by its
point of application
,
magnitude
,
line of action
, and
sense
.
• Experimental evidence shows that the combined effect of two forces may be represented by a single
resultant
force.
• The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs.
• Force is a
vector
quantity.
P Q P -P
Vectors
• •
Vector
: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations .
Scalar
: parameters possessing magnitude but not direction. Examples: mass, volume, temperature • Vector classifications: -
Fixed
or
bound
vectors have well defined points of application that cannot be changed without affecting an analysis.
Free
vectors may be freely moved in space without changing their effect on an analysis.
Sliding
vectors may be applied anywhere along their line of action without affecting an analysis .
• •
Equal
vectors have the same magnitude and direction.
Negative
vector of a given vector has the same magnitude and the opposite direction.
P P Q Q P-Q Q
Addition of Vectors
• Trapezoid rule for vector addition • Triangle rule for vector addition P P • Law of cosines,
R
2
P
2
Q
2 2
PQ
cos
B R
• Law of sines, sin
A
Q
sin
B R
sin
C A
• Vector addition is commutative,
Q P
• Vector subtraction
P
P Q S
Addition of Vectors
• Addition of three or more vectors through repeated application of the triangle rule P Q S P 2P -1.5P
• The polygon rule for the addition of three or more vectors.
• Vector addition is associative ,
S
P
Q
S
• Multiplication of a vector by a scalar increases its length by that factor (if scalar is negative, the direction will also change.)
Resultant of Several Concurrent Forces
•
Concurrent forces
: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.
•
Vector force components
: two or more force vectors which, together, have the same effect as a single force vector.
Sample Problem
• Graphical solution - construct a parallelogram with sides in the same direction as
P
and
Q
and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal.
Two forces act on a bolt at
A
. Determine their resultant.
• Trigonometric solution use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.
Sample Problem Solution
Q P
R
98 N R • Graphical solution - construct a parallelogram with sides in the same direction as
P
and
Q
and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal.
Sample Problem Solution
• Trigonometric solution From the Law of Cosines,
R
2
P
2
Q
2 2
PQ
cos
B
40N 60N 2
R
97.73N
From the Law of Sines, sin
A
Q
sin
B R
sin
A
sin
B Q R A
sin155 60N 97.73N
15.04
20
A
35.04
Sample Problem
• Find a graphical solution by applying the Parallelogram Rule for vector addition. The parallelogram has sides in the directions of the two ropes and a diagonal in the direction of the barge axis and length proportional to 5000 N .
A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 5000 N directed along the axis of the barge, determine a) the tension in each of the ropes for α = 45 o , • Find a trigonometric solution by applying the Triangle Rule for vector addition. With the magnitude and direction of the resultant known and the directions of the other two sides parallel to the ropes given, apply the Law of Sines to find the rope tensions.
Sample Problem
• Graphical solution Parallelogram Rule with known resultant direction and magnitude, known directions for sides.
T 1 T 2 30 45 5000N 45 30
T
1 3700
N T
2 2600
N
• Trigonometric solution Triangle Rule with Law of Sines
T
1 sin 45
T
2 sin 30 5000
N
sin105 45 T 2 5000N 105 T 1 30
T
1 3700
N T
2 2600
N
Rectangular Components of a Force: Unit Vectors
• May resolve a force vector into perpendicular components so that the resulting parallelogram is
F x
and
F y rectangular vector components
• Define perpendicular
unit vectors
are parallel to the
x
and
y
axes.
i
ˆ an d ˆ
j
which • Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.
F
F x i
ˆ
F y
ˆ
j F x
and
F y
are referred to as the
components
of
F x
and
F y scalar
F
S j y
Addition of Forces by Summing Components P S Q
Q i x Q j y P j y
R
R i x
• Wish to find the resultant of 3 or more concurrent forces,
R S
• Resolve each force into rectangular components
R i x
R j y
P i x
P x
P j y
Q x
Q i x x
Q j y P y
S i x Q y
S y S j y
j
• The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.
R x
P x
F x Q x
S x R y
P y
F y Q y
S y
• To find the resultant magnitude and direction,
R
R
2
x
R y
2 tan 1
R R x y
Sample Problem
Four forces act on bolt
A
as shown. Determine the resultant of the force on the bolt.
Plan: • Resolve each force into rectangular components.
• Determine the components of the resultant by adding the corresponding force components.
• Calculate the magnitude and direction of the resultant.
Sample Problem Solution
• Resolve each force into rectangular components .
force mag F
1
F
2
F
3
F
4 150 80 110 100 129.9
27.4
0 96.6
R x
199.1
75.0
75.2
R y
110.0
25.9
14.3
• Determine the components of the resultant by adding the corresponding force components.
• Calculate the magnitude and direction.
R
199.1
2 14.3
2 tan 14.3 N 199.1N
R
199.6N
4.1
Equilibrium of a Particle
• • When the resultant of all forces acting on a particle is zero, the particle is in
equilibrium
.
Newton’s First Law
: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.
• Particle acted upon by two forces: - equal magnitude - same line of action - opposite sense • Particle acted upon by three or more forces: - graphical solution yields a closed polygon - algebraic solution
R
F x
F
0 0
F y
0
Free-Body Diagrams
T AB 50 A 30 T AC
Space Diagram
: A sketch showing the physical conditions of the problem.
736N
Free-Body Diagram
: A sketch showing only the forces on the selected particle.
Sample Problem
In a ship-unloading operation, a 3500-lb automobile is supported by a cable. A rope is tied to the cable and pulled to center the automobile over its intended position. What is the tension in the rope?
Plan of Attack: • Construct a free-body diagram for the particle at the junction of the rope and cable.
• Apply the conditions for equilibrium by creating a closed polygon from the forces applied to the particle. • Apply trigonometric relations to determine the unknown force magnitudes.
T AB 2 A 30 T AC
Sample Problem
SOLUTION: • Construct a free-body diagram for the particle at
A
.
• Apply the conditions for equilibrium in the horizontal and vertical directions.
horizontal
T AB
T AB x
T AC T AC x
0 0 Vertical
T AB y T AB
3500
lb
T car
T AC y T AC
0 0 3500lb
T AB 2 A 3500lb 30
Sample Problem
T AC
T AB
T AC
0
T AB
3500
lb
T AC
0 cos 30
T AB
T AB
T AC
3500
lb T AC
T AC
cos 2 3500
lb
0.979839
T AC T AC
0.02016015
T AC
141.12105
lb
141.12105
lb
144
lb
Sample Problem
• Solve for the unknown force magnitudes using Sine Law.
T AB T AC 58 120 3500lb 2
T AB
sin120
T AC
sin 2
3500lb sin 58
T AB
3570lb
T AC
144lb Sometimes the Sine Law / Cosine Law is faster than component vectors. Intuition should tell you which is best.
Sample Problem
It is desired to determine the drag force at a given speed on a prototype sailboat hull. A model is placed in a test channel and three cables are used to align its bow on the channel centerline. For a given speed, the tension is 40 lb in cable
AB
and 60 lb in cable
AE
. Determine the drag force exerted on the hull and the tension in cable
AC
.
PLAN OF ATTACK: • Choosing the hull as the free body, draw a free-body diagram. • Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.
• Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.
T AC T AB =40 60.26
69.44
A F D T AE =60
Sample Problem
SOLUTION: • Choosing the hull as the free body, draw a free-body diagram. tan 7 ft 4 ft 60.25
1.75
tan 1.5 ft 4 ft 20.56
0.375
• Express the condition for equilibrium for the hull by writing that the sum of all forces must be zero.
R
T AB
T AC
T AE
F D
0
T AB
T AC
T AC A T AE =60
Sample Problem
T AC
F D • Resolve the vector equilibrium equation into two component equations. Solve for the two unknown cable tensions.
T AE
T AC
T
60 0.936305
T AC
42.889
AC
19.842597
F D
T AC
0.351188
T AC
0.35
1188 34.73142
42.
889 34.7
3142 19.66
lb
Sample Problem
This equation is satisfied only if all vectors when combined, complete a closed loop.
Rectangular Components in Space
is contained in the plane
OBAC
.
horizontal and vertical components.
F h
F
sin
y F y
F
cos
y F
rectangular components
F x
F h
cos
F
sin
y
cos
F y
F h
sin
F
sin
y
sin
Rectangular Components in Space
• • With the angles between and the axes,
F x
F
cos
x F y
F
cos
F
y F z
F
F F i x
F j y
cos
x i
F k z
cos
y j
cos
z k
F
cos
z
F
cos
x i
cos
y j
cos
z k
is a unit vector along the line of
F
action of
F
and are the direction cosines for
F
Rectangular Components in Space
Direction of the force is defined by the location of two points, 1 , 1 , 1 and 2 ,
y z
2 , 2 d is the length of the vector F
d
vector joining
M
and
N d x
d i x
2
d j y x
1
d
d k z
y y
2
y
1
F
F
d
1
d i x
d j y
d k z
F x
Fd x d F y
Fd y d d z z
2
z
1
F z
Fd z d
Sample Problem
The tension in the guy wire is 2500 N. Determine: a) components
F x , F y , F
z of the force acting on the bolt at
A
, b) the angles
x ,
y ,
z
defining the direction of the force PLAN of ATTACK: • Based on the relative locations of the points
A
and
B
, determine the unit vector pointing from
A
towards
B
.
• Apply the unit vector to determine the components of the force acting on
A
.
• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.
Sample Problem
SOLUTION: • Determine the unit vector pointing from
A
towards
B
.
AB
40 m 80 m 30 m
k AB
40 m 94.3 m 80 m 40
i
80
j
30 94.3
94.3
94.3
0.424
i
0.848
j
0.318
k
30 m 2
k
• Determine the components of the force.
F
F
2500 N 0.424
i
1060 N 0.848
j
2120 N 0.318
k
795 N
k
Sample Problem
• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.
cos
x i
cos
y j
cos
z k
0.424
i
0.848
j
0.318
k
x
y
z
115.1
32.0
71.5