Download Chapter 5

Journal #21

1.
2.
3.
4.
5.
If you stand on a scale while riding an elevator,
how would your weight differ from normal in
the following situations?
The elevator accelerates upward from rest.
The elevator accelerates downward from rest.
The elevator slows while going up.
The elevator slows while going down.
The elevator moves at a constant speed.
Chapter 5
Forces in Two Dimensions
Review from Chapter 2

Scalar Quantity – a quantity of
magnitude (size) only - without
direction; can be added, subtracted,
multiplied and divided just like ordinary
numbers


Ex: mass, volume, time
Vector Quantity - a quantity with both
magnitude (size) and direction

Ex: velocity, acceleration, force
Important Notes Regarding Vectors
Vector – an arrow whose length represents
the magnitude (how much) of a quantity, and
whose direction represents the direction
(which way) of that quantity
 You can move vectors around as long as you
do NOT change the magnitude (length) or
direction
 In calculations with vector quantities, vectors
of magnitude 1 and 1 do not always produce a
resultant vector of magnitude 2!

Vector Addition

Resultant – the result of adding 2 or
more vectors together
If 2 or more vectors run along the same
straight line in the same direction, ADD
them together to find resultant.
 If two vectors run along the same straight
line in opposite directions, SUBTRACT
them to find resultant.

Vectors at Right Angles (90º)
For 2 Vectors:
 Rearrange (move) the placement of one
vector so that they are drawn tip-to-tail

You can move either one and still get the same
answer!

Draw the resultant from the tail of the first
vector to the tip of the last vector.

If there are more than 2 vectors in a problem,
you should be able to simplify them down to
two directions before attempting this step!!!
Moving Vectors Around
Tip to Tail Method
Free Body Diagram
Tip to Tail Method
Mathematically Solving Right Angle Problems
Mathematically, when vector A is at a
right angle to vector B, use the
Pythagorean theorem to find resultant:
r2 = a2 + b2
 The sum of the squares of the
magnitudes of the two component
vectors is equal to the square of the
resultant.

Components of Vectors

You can break a single vector into its 2
components - vectors which make it
up. This process is called vector
resolution.

Any vector drawn on paper can be
broken into the horizontal and vertical
components which make it.
Components of a Vector
y component
x component
Example 1

During a cheer, Jessica moves 3m
forward and 5m to the right. What is her
displacement?
Example 2

On the first play of the game, Austin ran
the ball 10.0 m forward then 3.00 meters
to the right.
Example 3

A crate is being pushed forward by two
boys with a force of 20N each while a
third boy is pushing to the left with a
force of 25N. What is the resulting force
vector for the crate? (Start with a freebody diagram!)
Example answers
Ex. 1: c = 6 m forward to right
 Ex. 2: c = 10.4 m forward to right
 Ex. 3: b = 47 N forward and to left

Homework

P. 141 3 problems below
#79, 81, 82
 front side of vector wksheet – use a ruler

Journal #30

Solve for the missing side of the triangles
using the Pythagorean Theorem.
a2 + b2 = c2
25.0
5.0
x
x
11.2
4.0
x
20.3
7.8
Journal #30 answer
a2 + b2 = c2
11.2
5.0
x
25.0
x
x
20.3
11.2 2  20.32  c 2
4.0
4.0 2  5.0 2  c 2
16  25  c 2
41  c 2
6.4  c

125.44  412.09  c 2
537.53  c
23.2  c
2
7.8
a 2  7.8 2  25.0 2
a 2  25.0 2  7.8 2
a 2  564.16
a  24
Journal #31

Add all of the following vectors to find the
resultant:
5.0
4.0
3.0
3.0
Journal #31

Add all of the following vectors to find the
resultant:
5.0
4.0

c  (5.0 2  4.0 2 )
c  6.4
Answers to Homework
Answers to Homework
Answers to Homework
Journal #32
(TRY NOT TO USE YOUR NOTES)
 What are the 3 rules of adding vectors?
JOURNAL #32 ANSWER
If vectors have the SAME direction, ADD
the values and keep that direction.
 If vectors have OPPOSITE directions,
SUBTRACT the values and keep the
direction of the larger value.
 If vectors are at RIGHT ANGLES to
each other, use the Pythagorean
Theorem to solve for the resultant.

5.2: Friction

P. 126-130 in textbook
Friction
Friction is a force that opposes motion.
 There are two types:

Kinetic Friction
 Static Friction

Kinetic Friction

The friction exerted on one surface by
another when the two surfaces rub
against each other because one or both
of them are moving

Examples:
Hands rubbing together
 Pushing a box up a ramp
 Dragging a crate

Static Friction

The friction exerted on one surface by
another when there is no motion
between the two surfaces

Examples:
Leaning against a table but it doesn’t move
 Trying to push a couch but you can’t move it

Relating FN to Ff

Different surfaces cause different amounts of
friction between objects.
 If you were to plot a graph of Ff vs. FN for an
object, the slope of the line is called the
coefficient of friction (). This number is a
constant, regardless of the weight of the object
and can be found with the following formulas:
Ff
static 
FN
Ff
kinetic 
FN
Important notes about friction
When working with Ff, you will always
have to consider the rules of calculating
Fnet
 Usually, you will have to consider that
Fnet = F(forward motion) - Ff

Common Coefficients
Example 1 - P. 128, #18

You need to move a 105-kg sofa to a different location in the
room. It takes a force of 102N to start it moving. What is the
coefficient of static friction between the sofa and the carpet?
FN
F f Fthrust Fthrust
s 


FN
Fg
(mg)
Fthrust
Ff

Fg

102
s 
 0.0991
(105  9.80)
Example 2 - P. 128, #20

Suppose that a 52-N sled is resting on packed snow. The
coefficient of kinetic friction is only 0.12. If a person weighing
650N sits on the sled, what force is needed to pull the sled across
the snow at constant speed?
Ff
k 
, FN  Fg  (Fgsled  Fg person ) , Ff  FT
FN
FN
FN k  Ff (substitute)
Ff

Fg
FT
Fg k  FT
702  0.12  FT
84N in dir. of pull  FT
Example 3 - P. 130, #22

A 1.4-kg block slides across a rough surface such that it slows
down with an acceleration of 1.25 m/s2. What is the coefficient of
kinetic friction between the block and the surface?
FN
k 
Ff
FN
, Fnet  ma  Ff , FN  Fg  (mg )
Ff
Fnet ma a
k 



FN
Fg
mg g
Ff
Fg
a 1.25
k  
 0.128
g 9.80
Homework Assignment

P. 128, #17, 19, 21 (see p. 129 chart)