Honors Physics Chapter 5
Forces in Two
Dimensions
1
Honors Physics



2
Turn in H4 & W4
Lecture
Q&A
Vector and Scalar

Vector:
–
–
–
–

Scalar:
–
–
–
–
3
Magnitude: How large, how fast, …
Direction: In what direction (moving or pointing)
Representation depends on frame of reference
Position, displacement, velocity, acceleration, force,
momentum, …
Magnitude only
No direction
Representation does not depend on frame of reference
Mass, distance, length, speed, energy, temperature, charge, …
Vector symbol

Vector: bold or an arrow on top
– Typed: v and V or v and V
– Handwritten: v and V

Scalar: regular
–

4
v or V
v stands for the magnitude of vector v.
Graphical representation of vector:
Arrow
An arrow is used to graphically represent a vector.

–
The direction of the arrow represents the direction of the vector.
–
The length of the arrow represents the magnitude of the vector.
(Vector a is smaller than vector b because a is shorter than b.)
head
a
–
5
tail
b
When comparing the magnitudes of vectors, we ignore
directions.
Equivalent Vectors

Two vectors are identical and equivalent if they both
have the same magnitude and are in the same
direction.
A
B
C
6
–
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.)
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
7
Adding Vectors

Graphical
–

8
Head-to-Tail (Triangular)
Algebraic (by components)
Adding Vectors: Head-to-Tail

Head-to-Tail method:
Example: A + B
–
–
–
A
B
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.
B
A
9
Make sure arrows are
parallel and of same length.
A+B=B+A
B
A
A+B
A
How about
B + A?
B
B
A
10
What can we
conclude?
A+B+C
B
A
C
C
B
A
Resultant vector:
11
from tail of first to head of last.
A-B=A+(-B)
A
B
A
–
-B
–
–
12
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.
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?
9 N, same direction
4N
5N
9N
b) What is the minimum possible resultant force on the
object? How are these two forces oriented relative to
each other when this happens?
4N
13
1 N, opposite direction
1N
5N
What are the relations among the
vectors?
B
B
C
A
A+C=B
A
b
c
C
c
b
a + b = -c
a+b+c=0
14
a
a
Example:
N
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.
-a
b
35o
66o
50o
a
Using ruler and protractor, we find:
a+b: 4.3 unit, 50o North of East
b-a: 8.0 unit, 66o West of North

15
35o
W
E
S
Law of cosine
c 2  a 2  b 2  2ab cos C
b
c
C = 90o
C
a
a
Pythagorean’s Theorem:
16
c
b
a 2  b2  c 2
Law of Sine
a
b
c


sin A sin B sin C
a
C
or
b
B
A
c
17
sin A sin B sin C


a
b
c
Example: 125-7
You 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?
N
a  8.0km, r  10.0km, b  ?
a b  r
2
2
a
2
r
E
0
b  r 2  a2

18
b
10.0km  8.0km  6.0km
2
2
Practice: 121-3
A 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  4.5km, b  6.4km, c  ?
a
C  180  45  135
o
o
o
o
135o 45
c
c  a  b  2ab cos C
2
2
2
c  a 2  b 2  2ab cos C

 4.5km    6.4km 
 10.km
19
2
2
 2  4.5km  6.4km  cos135o
b
Vector Components

Drop perpendicular lines from
the head of vector a to the
coordinate axes, the
components of vector a can
be found:

 ax  a cos 

a y  a sin 


20

 is the angle between the
vector and the +x axis.

ax and ay are scalars.
y
a
ay

ax
x
Useful Trigonometry

opposite
sin   hypotenuse

adjacent

cos



hypotenuse


opposite
 tan  
adjacent

21

adjacent side
Finding components of a vector

Finding components of a vector
–
–

ax is
–
–

22
Resolving the vector
Decomposing the vector
the component of a in the x-direction
the x component of a.
Component form:
a   ax , a y 
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  75N ,  60o , Tx  ?, Ty  ?
T||  Tx  T cos
 75N cos60o
 38. N
T  Ty  T sin 
 75N sin 60o
 65. N
23
Ty
T

Tx
y
ay
Vector magnitude
and direction


ax
x
The magnitude and direction of a vector can be
found if the components (ax and ay) are given:
 magnitude: a  a 2  a 2
2
2
2

a

a

a
x
y
x
y
z



a
 tan   a y
1 y

ax  direction:   tan
ax

 is the angle from the +x axis to the vector.
24
a
Magnitude and direction form:
a   a,  
(for 3-D)
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?
Set up the frame of reference as to the
right. Then
rx  a  125.0km, ry  b  65.0km, r  ?,   ?
r  rx 2  ry 2

ry
rx
2
2
 tan 1
65.0km
 27.5o
125.0km
Magnitude is 141 km, at 27.5o South of West.
25

b
r
y
125.0km    65.0km   141km
  tan 1
a
x
y
ry
Adding Vectors
by Components

a
r
bx
ax
When adding vectors by components, we add
components in a direction separately from other
components.
r  a  b
x
x
x
r  ab
26
ay
b
by


  ry

rz


 a y  by
 az  bz
2-D
3-D
rx
x
y
Practice: 125-8
A 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,Wwhat is
the combined force (magnitude and direction)
of the two ropes on the swing?
b
b
a
Let x = horizontal, y = vertical, then
a  2.28 N , b  2.28 N ,  a  90o  13.0o  77.0o , b  90o  13.0o  103o
o
a
cos


2.28cos
77.0
 0.513 N
a x 
o
a   2.28 N , 77.0   
a y  a sin   2.28sin 77.0o  2.22 N

b   2.28,103.0o 
bx  b cos   2.28cos103o  0.513 N

by  b sin   2.28sin103  2.22 N
  a  b , a  b  
a  b  x x y y  0.513  0.513, 2.22  2.22    0, 4.44N 
27
a  b  4.44 N , upward
x
Friction
Friction: force opposing the motion or tendency of motion
between two rough surfaces that are in contact

Static friction is not constant and has a maximum:
f s ,max  s N  f s  s N

Kinetic (or sliding) friction is constant: f k 
o
o
N: normal force between the two surfaces
s (and k) is coefficient of static (and kinetic) friction.



28
k N

 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.
Example:
What happens to the frictional
force as you increase the force
pushing on a table on the floor?
 sN
•
• Once table is moving, a smaller
constant kinetic friction. fk < fs, max
•
•
•
29
f
But what if the applied force increases
just slightly?
f
0
Fapp
 sN
Fapp
Example: 128-17
A 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.


Fnet needs to include only forces
in the horizontal direction. Define
right to be the positive direction.
Fnet  Fp  f  ma  0
 f  Fp  36 N
f  N  
30
+
We only consider the motion in the
horizontal direction.
f 36 N

 0.69
N 52 N
N= 52N
f
Fp=36N
W=52N
Practice: 128-18
You 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?
+
Let right = + direction. Fnet includes only
forces in the horizontal direction. Then
m  105kg , Fp  102 N ,   ?
N  W  mg  105kg  9.8
N
f
m
 1029 N
2
s
Fnet  Fp  f  ma  0
 f  Fp  102 N
f s ,max   N  s 
31
f s ,max
N
102 N

 0.0991
1029 N
Fp=102 N
W
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.
C
C is the equilibrant
force of A + B because
(A + B) + C = 0.
32
C = - (A + B)
B
A
A+B+C= 0
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?
b
c
a
a
33
c = - (a + b)
a+b+c=0
b
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?
FA  (36 N ,225o ), FB  (48 N ,315o ), FC  ?

FAx  36 N cos 225o  25.5N

FA : 36 N ,225o  
o
F

36
N
sin
225
 25.5N

Ay

o

F

48
N
cos315
 33.9 N

Bx
FB : 48 N ,315o  
o
F

48
N
sin
315
 33.9 N

By

 FAx  FBx  25.5N  33.9N  8.4N

 FAy  FBy  25.5N  33.9N  59.4N
 FCx    FAx  FBx   8.4 N

 FCy   FAy  FBy  59.4 N

34

 FC  FCx 2  FCy 2  ( 8.4 N )2  (59.4 N )2  60. N


F
1 Cy
1 59.4 N
tan

tan
 tan 1  7.07   82o  98o



FCx
8.4 N

y
FC
x
FA
FB
More Application on Force:
Example: 135-40
Stacie, 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?
y
x: Fnet  Wx  f  ma
y: Fnet  N  Wy  ma y  0
 N  Wy  mg cos 
N
f
Wx
f   N   mg cos 
mg sin    mg cos   ma
W  Wx  W sin 

 mg sin 


 W  Wy  W cos

 mg cos 
x
 a  g sin    g cos 

Wy
W
 g  sin    cos  
 9.8
35
m
m
o
o
sin
45

0.25cos
45

5.20


s2
s2
What if frictionless
incline?
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?
y
x: Fnet  Wx  ma
y: Fnet  N  Wy  ma y  0
N
 W sin   ma
Wx
 mg sin   ma
 a  g sin 
 9.8
36
m
m
o

sin
30

4.9
s2
s2
W  Wx  W sin 

 mg sin 


 W  Wy  W cos

 mg cos 
x

Wy
W
Weight hanging
What 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?
y
Hanging mass:
Fnet  T1  W  ma  0
 T1  W  mg
m
 5.0kg  9.8 2  49 N
s
T2

T2x
T2y
P
x
•
N
T1
Consider forces acting on point P:
y: Fnet . y  T2 y  T1  ma  0
 T1  T2 y  T2 sin 
37
T1
49 N
 T2 

 98 N
o
sin  sin 30
T1
5.0kg
W
+