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This Week
Textbook -- Read Chapter 4, 5Competent Problem Solver - Chapter 4
Pre-lab Computer Quiz
What’s on the next Quiz?
Kinematics - EverythingForces - to lecture before quiz
Check Your Understanding atend of each Laboratory is agood guide
Work problems as if taking the quiz:
Assigned homework - minimumif you can do them easilyProblems at end of chapter 3 and 4of CPS
Check out sample quiz on web by Thurs.
Other textbook problems
What you missed on first quiz
The combined effect of ALL forces on anobject that determines its acceleration.The effects of forces in perpendiculardirections are independent
Forces are vector quantitiesCombining Forces means Vector Addition
Theory of Forces
The Mathematics:
Define a coordinate systemwith perpendicular x and y axes
Add x components of forces
and independently
Add y components of forces
+y
+x F
1
F
2
Example: 2 forces act on an object, howdoes it move?
+y
+x
F
1
F
2
Adding Forces
F = F
1
+ F
2
means
F
x
= F
1x
+ F
2x
and
F
y
= F
1y
+ F
2y
F
1x
F
2x
+y
+x
ΣΣΣΣ
F
x
=F
1x
2x
F
1y
F
2y
Σ Σ
Σ Σ
F
y
=F
1y
2y
ΣΣΣΣ
F
θθθθ
F
F
x
2
F
y
2
tan
θ θ
θ θ =
F
y
F
x
First: Take the vectors apart
Second: Put the resulting vector together
The sum of the x components of Forces
F
x
the x component of acceleration
The sum of the y components of Forces
F
y
the y component of acceleration
The sum of the z components of Forces
F
z
the z component of acceleration
F
x
causes a
x
F
y
causes a
y
F
z
causes a
z
Review
affects ONLY affects ONLY
affects ONLY
The relationship between
Forces on an object andAcceleration of an object
was found to be amazingly simple The x component of acceleration of an object
is proportional to
the sum of the x components of all of theforces on that object from the interactions ofall other objects.
The constant of proportionality is
The mass of the object
Forces and Motion
The same is true for the y and z components
F
x
= m a
x
F
x
= m a
x
F
y
= m a
y
F
z
= m a
z
Where x, y, z are threeperpendicular directions.
all other objects
"your" object with
m is the mass of “your” object
F = m a
a is the acceleration of "your" object
F = m a
means:
Vector Notation
Known as Newton’s 2nd Law
F is the sum of the interactions of
If an object is accelerating
Is there always a force in thedirection of the acceleration?
(a) Yes(b) No
If no, give an example
m
F
1
F
2
a
There is no “real” force in the direction of a
Analyze the Forces
Finding all of the forces and their values
Use diagrams to clarify
Simple case: A book on a table
Draw all relevant forces.Did you get them all?How do they combine?How are they related?
Isolate the object you are interested in
Free-body Diagram of Book
Draw only the forces on that object
Only those forces determine its acceleration
First Object: the book
Step 1: Isolation
There appear to be 2 important objects here:
The book
The table
Only the forces on the book
W
b
N
tb
W
b
: gravitational pull of the
Earth on the book (weight ofthe book)N
tb
: contact push of the table
on the book (perpendicular tothe table/book contact)
Free-body Diagram of Table Second Object: the table One more relevant Object: the Earth
W
t
N
bt
N
Et
W
t
: gravitational pull of
the Earth on the table(weight of the table)
N
bt
: contact push of
the book on the table(perpendicular to thebook/ table contact) N
Et
: contact push of the
ground on the table(perpendicular to theground/table contact)
Contact pushes by a surface are usuallycalled normal forces
Normal is an old fashioned wordfor perpendicular
Third Object: the EarthFree-body Diagram of Earth
W
bE
: gravitational pull of
the book on the Earth
W
tE
: gravitational pull of
the table on the Earth
N
tE
: contact push of the
table on the ground(perpendicular to theground/table contact)
W
bE
W
tE
N
tE
How are these forces related to each other?
Relationship of Forces on
a Single Object
For the book:
F
y
= m a
y
W
b
N
tb
N
tb
- W
b
N
tb
= W
b
For the table:
F
y
= m a
y
W
t
N
bt
N
Et
N
Et
= W
t
+ N
bt
W
bE
W
tE
N
tE
For the Earth:
F
y
= m a
y
N
tE
= W
bE
+ W
tE
+y
A 20 kg block rests on a frictionless table. A
cord attached to the block extendshorizontally to a pulley at the edge of thetable. When another block of unknown massis hung at the end of the cord after it passesover the pulley, the hanging block acceleratesdownward at 2.7 m/s
2
and pulls the other
block with it. Calculate the mass of thehanging block and the tension in the cord.
Example
As might be stated in your text
You have been hired as a consultant for anew movie about an expedition to the SouthPole. In one scene, the explorers are on aglacier that comes to an end with a steepcliff. They need to get their supplies downthe cliff. For safety, each box of supplies isroped to another box with a 30 m rope. Onebox breaks away and falls off the cliff. The20 kg box it is roped to is pulled over thehorizontal ice of the glacier towards the edgeof the cliff. It is important that the hero of thestory save this box. You calculate that thiscan be done if this it has an acceleration of2.7 m/s
2
. Now you need to know the mass of
the other box and the necessary strength ofthe rope. As it might appear on a quiz
A
B
Find mass of hanging block andtension in cord
Use forces on B to relate acceleration to m
B
2.7 m/s
2
20 kg
2.7 m/s
2
Question:
Approach: Use forces on A to relate acceleration to T
A
For massless rope T
A
= T
B
m
B
Assumptions: massless cord,
frictionless ice
Since B accelerates, force of rope on B isless than its weight
W
A
N
A
T
A
W
B
T
B
A and B are tied together so they havethe same magnitude of acceleration
You recognizethis from the lab
FOCUS THE PROBLEM
+y
+x
The Massless Rope
A very useful approximation
You pull a block with a rope by exertinga force P and the block accelerates
P
The rope pulls on the block with a force T
T
How are P and T related?
The block pulls on the rope with a force T.
3rd Law Pairs
T
Free body diagram of the rope
P
T
F = m
rope
a
F = 0
Approximate m
rope
Thus T = P
The force exerted on the rope equals the forcethe rope exerts on the block
a
a
Free-body Diagram of A
W
A
N
A
T
Force Diagram of A
+y
+x
W
A
N
A
T
Relevant Equations:
F
x
= m a
x m
A
= 20 kg
a
B
= 2.7 m/s
2
Target quantities: m
B
and T
Block A
T = m
A
a
N
A
- W
A
Block B
T - W
B
= m
B
(-a)
Free-body Diagram of B
T
W
B
Force Diagram of B
+y
+x
T W
B
a
a
W = mg
Physics Description
F
y
= m a
y
Find T
Find m
B
unknowns
consider the motion of object B.
m
B
W
B
= m
B
g
T, W
B
Find W
B
-m
B
a = T - m
B
g
Must consider object A to getmore information
T - W
B
= m
B
(-a)
For object A
T = m
A
a
Note: 3 unknowns, 3 equations
ok
Plan the Solution
(20kg) (2.7 m/s )
2
2
2
(9.8 m/s - 2.7 m/s )
m
B
7.6 kg
m
B
Now for the other target, T
Looking at the mathematical plan
equation
T = m
A
a
T = (20 kg) (2.7 m/s )
2
T = 54 kg m/s
2
= 54 N
m
A
a - m
B
g = -m
B
a
m
A
a = m
B
g - m
B
a
m
A
a = m
B
(g - a)
Check units
accel. units cancelgiving mass units, ok
m
A
a
(g - a)
= m
B
Execute the Plan
Putting [3] and [2] into [1]
Is it properly stated?
Quantities in the mathematics aredescribed in the picture
Quantities solved for have appropriate units
Is it unreasonable?Since B accelerates downward,
T < W
B
W
B
= m
B
g
W
B
= 7.6 kg (10 m/s )
2
W
B
= 76 N
54 N < 76 N
Is it complete?
Yes, found both T and m
B
mass in Kgforce in N
The mass of B (7.6 kg) is reasonable for asmall box
Evaluate the Solution
This is reasonable
As the engineering advisor to anarcheological team, you are trying to figureout how the ancient Egyptians could lift largeblocks of stone from a quarry. The team hasfound evidence of wooden disks which couldhave been used as pulleys. The team leaderhas suggested that a three pulley system withone fixed pulley and two movable pulleysmight have been used. You to have beenassigned determine if the ropes used by theEgyptians would have been strong enoughfor such a system to lift a 1000 kg block ofstone using the pulley system sketchedbelow. You also want to know if one personcould lift a block using that system.
Example
T
2
T
1
T
4
M
What is the force on each rope?
Assume massless ropes
A
B
C
+y
+x
Assume frictionless, massless pulleys
1000 kg
Question:Approach
F
y
on each pulley is 0 (no acceleration)
Assume that block is pulled up ata constant velocity.
There are 4 ropes in the problem
Use Newton’s 3rd Law to relate forceexerted on a rope to force exerted by therope by other objects.
What is force needed to lift block?
Objects are: block and 3 pulleys
Focus
P T
3
Free - body diagrams
T
1
W
T
2
T
2
T
1
A
T
2
T
3
T
3
B
T
4
T
3
T
3
C
Block
Pulley A
Pulley B
Pulley C
M = 1000 kga = 0
Target variables: T
1
, T
2
, T
3
, T
4
Relevant Equations:
F
y
= m a
y
Block: T
1
- W = 0
Pulley A: 2T
2
- T
1
Pulley B: 2T
3
- T
2
Pulley C: T
4
- 2T
3
W = Mg
Describe the Physics
Find T
1
T
1
Find T
3
consider the motion of the block
unknowns
T
1
Find T
2
T
2
consider the motion of pulley A
consider the motion of pulley B
T
3
T
1
= M g
2 T
2
-T
1
2T
3
-T
2
T
2
= Mg/
T
3
= Mg/
Find T
4
consider the motion of pulley C
T
4
T
4
- 2T
3
T
4
= Mg/
Plan
Riding in a friend’s sports car you feel the
seat pushing on your back when it pulls awayfrom a stop light. You decide you want toknow the 1000 kg car’s acceleration. Howmany g's are you "pulling"? You notice thereis a die hanging from the rear view mirror.Very retro. When the car leaves theintersection the die makes an angle of 15 °with the vertical. Later you measure themass of that die to be 100 grams. What wasthe acceleration of the car?
Example
15°
T
a
Are we sure thatthe angle of thedie is as drawn?
Picture
Mass of die : 100 gramsMass of car: 1000 kg
What is the acceleration of car?
Initial velocity of car : 0
Question:Approach:
Acceleration of car is acceleration of die
Relate acceleration to force on the die Consider forces on die
By Earth (Weight)By string (tension)
The acceleration in one direction isindependent of the forces in perpendiculardirections
Use components
Focus
W
Free body diagramof die
T
W
W
Force diagram of die
T
θθθθ
+y
m = 100 gr
a
y
F
x
= m a
x
F
y
= m a
y
W = m g
target variable: a
x
Relevant equations:
T
x
= ma
x
T
y
y
sin
θθθθ
T
x
T
cos
θ θ
θ θ
T
y
T
tan
θθθθ
T
x
T
y
Physics Description
T
x
T
y
Find a
x
Horizontal motion of the die
unknowns
a
x T
x
Find T
x
T
T
x
= ma
x
tan
θθθθ
T
x
T
y
Find T
y
T
y
T
y
3 unknowns, 3 equations
done
T
y
= mg
tan
θθθθ
T
x
mg
mg tan
θθθθ
= T
x
mg tan
θθθθ
= ma
x
g tan
θθθθ
= a
x
Execute:
a
x
tan (15°) g
a
x
= 0.27 g
Plan
Vertical motion of the die
a
x
is in same units as g which is an
acceleration so units are correctThe question is answered since thecar is accelerating in the x directionand the die is moving with the car.A car which goes from 0 to 50 mph in10 seconds has a high acceleration
v
a
av
t
a
av
mihr
sec
This car’s acceleration is about that so theanswer is reasonable
Is the answer unreasonable?
mi
hr
ft
mi
hr
min
min
sec
10 sec
a
av
since g = 32 ft
sec
2
a
av
7 ft
sec
2
< 1/4 g
Evaluate
