Physics 231 Topic 3: Forces & Laws of Motion

Physics 231
Topic 3: Forces & Laws of Motion
+ Uniform Circular Motion
Wade Fisher
September 10-14 2012
MSU Physics 231 Fall 2012
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Quick Quiz
What aspects of your lectures do you find most useful?
A) Discussion of concepts
B) Demonstrations
C) Problem solving examples
D) Opportunity to ask questions
E) Opportunity to cover homework problems
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Recall from Last Lecture
Vectors and Scalars
Two Dimensional Motion
 Velocity in 2D
 Acceleration in 2D
Projectile motion
 Throwing a ball or cannon fire
Uniform Circular Motion
 Centripetal acceleration
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An example
A football player throws a ball with initial
velocity of 30 m/s at an angle of 30o
degrees w.r.t. the ground.
1) How far will the ball fly before hitting
the ground?
2) What about an angle of 60o?
3) At which angle is the distance thrown
maximum?
(A)
X(t) = 30 cos()t
Y(t) = 30 sin()t – ½gt2
(B)
Y(t) = 0 if…
t( 30 sin() - ½gt )=0
to=0 or tf = 30 sin()/(½g)
(C)
X(tf) = 30cos() tf
= 30x30 cos() sin() / ( ½g )
=900 sin(2) / g
(D)
if  = 30o X = 79.5 m
if  = 60o X = 79.5 m !!
(E)
MSU Physics 231 Fall 2012
Maximum if sin(2) is
maximum, so =45o
X(=45o) = 91.7 m
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Another example…
Calculate how far the ball goes.
Vertical direction
y(t) = y(0) + v0sin()t - ½gt2
11 = 37 + 10.1 sin(59)t - ½9.8t2
4.9t2 - 8.66t - 26=0
t = 3.35 (solve quadratic
equation)
Horizontal direction
x(t) = x(0) + vocos()t
Use time derived from
vertical direction
X(3.35)=0+10.1cos(59)3.35
X(3.35)=17.4 m
This time correspond to how
long it takes for the ball to
land on the second building.
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Clicker Quiz
You enter a circle intersection (roundabout) and travel at 20 mph around the
circle at constant velocity. Which direction
does your acceleration vector point?
A) Along the direction you’re moving
B) Towards the center of the circle
C) Away from the center of the circle
D) In the opposite direction you’re moving
E)
There is no acceleration
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Uniform Circular Motion
Consider a car at constant speed,
constrained to a circular track
• Velocity is always determined
by the direction the car is
facing at a given time
Velocity is constantly changing, so
this implies an acceleration
• But it’s just the direction of the
velocity that’s changing!
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Uniform Circular Motion
𝑣𝑓 − 𝑣𝑖 Δ𝑣
𝑎=
=
𝑡𝑓 − 𝑡𝑖
Δ𝑡
Sin( /2 ) = ( s/2 ) / r
Sin( /2 ) = ( v/2 ) / v
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Uniform Circular Motion
𝑣𝑓 − 𝑣𝑖 Δ𝑣
𝑎=
=
𝑡𝑓 − 𝑡𝑖
Δ𝑡
( s/2 ) / r = sin( /2 )
sin( /2 ) = ( v/2 ) / v
( s/2 ) / r = ( v/2 ) / v
( s/t ) v/r = ( v/t )
𝑎𝑐 =
𝑣
𝑣
𝑟
=
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𝑣2
𝑟
( s/t )  v
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Key Concepts: Forces &
Laws of Nature
Force and Mass
Newton’s Laws of Motion
 Inertia, F=ma, Equal/Opposite Reactions
Application of Newton’s Laws
 Force diagrams
 Component Forces
Friction and Drag
 Kinetic, static, rolling frictions
Uniform circular motion
Covers chapter 4 in Rex & Wolfson
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Isaac Newton (1642-1727)
“In the beginning of 1665 I found the…rule
for reducing any dignity of binomial to a series. The same
year in May I found the method of tangents and in November
the method of fluxions and in the next year in January had the
Theory of Colours and in May following I had the entrance
into the inverse method of fluxions and in the same year I
began to think of gravity extending to the orb of the
moon…and…compared the force requisite to keep the Moon
in her orb with the force of Gravity at the surface of the
Earth.”
“Nature and Nature’s Laws lay hid in night:
God said, Let Newton be! And all was light.”
Alexander Pope (1688-1744)
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Newton’s Laws
First Law: If the net force exerted on an object is zero the
object continues in its original state of motion; if it was at
rest, it remains at rest.
If it was moving with a certain velocity, it will keep on moving
with the same velocity.
Second Law: The acceleration of an object is proportional to
the net force acting on it, and inversely proportional to its
mass: F=ma
Third Law: If two objects interact, the force exerted by the
first object on the second is equal but opposite in direction
to the force exerted by the second object on the first:
F12=-F21
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Identifying the forces in a system
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Force, Mass and Weight
Forces are quantified in units of Newton (N).
1 N = 1 kgˑm/s2
F = ma
A force is a vector: it has direction.
But what’s this “mass” thing??
m = 1 kg
F = m a = 1 kgˑm/s2 = 1 Newton
a = -9.81 m/s2
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Two forces that luckily act upon
us nearly all the time.
Normal Force:
elastic force acting
perpendicular to the
surface the object is
resting on. Name: n
Gravitational Force
1) No net force: remains
at rest.
2) Fg = mg = n
3) Fmass-ground = -Fground-mass
Fg = mg (referred to as weight)
g = 9.81 m/s2
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question
5 kg
10 kg
5 kg
3 cases with different mass and size are standing on
a floor. which of the following is true:
a) the normal forces acting on the crates is the same
b) the normal force acting on crate b is largest
because its mass is largest
c) the normal force on the green crate is largest
because its size is largest
d) the gravitational force acting on each of the
crates is the same
e) all of the above
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Some handy things to remember.


90-

Choose your coordinate system in a clever way:
Define one axis along the direction where you expect
an object to start moving, the other axis perpendicular
to it (these are not necessarily the horizontal and
vertical direction.
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A first example
1) Draw all Forces acting on the red object
2) If =40o and the mass of the red object
equals 1 kg, what is the resulting
acceleration (assume no friction).
n = -FgL
FgL = mg cos

Fg//=mgsin
Fg=mg 
ma = F
F = mg sin = 6.3N
a = 6.3 m/s2
Balance forces in directions
where you expect no
acceleration; whatever is left
causes the object to
accelerate!
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Clicker
Question
question

A ball is rolling from a slope at angle . It is repeated but after
decreasing the angle . Compared to the first trial:
a)
b)
c)
d)
the normal force on the ball increases
the normal force gets closer to the gravitational force
the ball rolls slower
all of the above
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question
END LECTURE 1
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Forces seen previously
• Gravity: Force between massive objects
• Normal force: Elasticity force from supporting surface
n = -FgL
F = mg cos

F = mgsin


90-

Fg=mg 
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Decomposing Forces
An object has a mass of 30.0 kg and three
forces are acting on it as shown in the figure.
What is the magnitude and direction of
acceleration?
690 N
70o
24o
Net Force
300 N
mg
139 N
35.2o
Force
300 N
690 N
Weight
Resultant
x(N)
-122
236
0
114 N
y(N)
-274
648
-294
80 N
Total Force: F = (1142+80.02) = 139 N
Direction:
 = tan-1(Fy/Fx) = 35.2o
Acceleration: a = F/m = 139/30.0 = 4.65 m/s2
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Measuring mass and weight.
Given that gearth = 9.81 m/s2, gsun = 274 m/s2, gmoon = 1.67 m/s2,
what is the mass of a person on the sun and moon if his
mass on earth is 70 kg?
And what is his weight on each of the three surfaces?
The mass is the same on each of the surfaces
On Earth:
w = 686.7 N
On the Moon: w = 116.7 N
On the Sun: w = 19180 N
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Example
Buck Rogers travels from earth to a
planet which has a radius that is 2
times larger than that of earth and
has a mass 6 times that of earth.
By what factor does his weight
change relative to earth?
Fg~1/r2~1/4
Fg~M~6
Fg~M/r2=6/4=1.5
His weight is 1.5 times larger.
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QUIZ!
In the absence of friction, to keep an object that is
traveling with a certain velocity moving with exactly the
same velocity over a level floor, one:
a) does not have to apply any force on the object
b) has to apply a constant force on the object
c) has to apply an increasingly strong force on the object
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A 2000 kg sailboat is pushed by the tide of the
sea with a force of 3000 N to the East. Because
of the wind in its sail it feels a force of 6000 N
toward to North-West direction. What is the
magnitude and direction of the acceleration?
Horizontal
Due to tide: 3000 N
Due to wind: 6000cos(135)=-4243
Sum:
-1243 N
N
6000N
Problem
Vertical
0N
6000sin(135)=+4243
4243 N
Magnitude of resulting force:
Fsum=[(-1243)2+(4243)2]=4421 N
W
3000N
E
Direction: angle=tan-1(4243/-1243)
=1060 (calc: -730, add 1800)
F=ma so a = F/m = 4421/2000 = 2.21 m/s2
S
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Tension
T
The magnitude of the force T
acting on the crate, is the same
as the tension in the rope.
Spring-scale
You could measure the tension by inserting
a spring-scale...
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Tension
T1
T1=T2 ???
T2
M
As long as it’s one rope/string/cable, the
tension is the same at both ends!
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Newton’s second law and tension
m1
n
T
No friction.
T
Fg
What is the acceleration of
the objects?
m2
Fg
Object 1: F=m1a, so
Object 2: F=m2a, so
T = m1a
Fg-T = m2a & m2g-T = m2a
Combine 1&2 (Tension is the same): m2g - m1a = m2a
a = m2g/(m1+m2)
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Problem
What is the tension in the string and
what will be the acceleration of the
two masses?
Draw the forces: what is positive & negative?
For 3.00 kg mass: F = ma
T - 9.813.00 = 3.00a1
T = 9.813.00 + 3.00a1
T
T
For 5.00 kg mass: F = ma
T - 9.815.00 = 5.00a2
The masses are
T = 9.815.00 + 5.00a2
connected:
T = 9.815.00 - 5.00a1
a2 = -a1
9.813.00 + 3.00a1 = 9.815.00 - 5.00a1
8.00a1 = 9.812.00
Fg
Fg
T=36.8 N
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a=2.45 m/s2
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Example
A 1000-kg car is pulling a 300 kg trailer. Their acceleration
is 2.15 m/s2. Ignoring friction, find:
a) the net force the car
b) the net force on the trailer
c) the net force exerted by the trailer on the car
d) the resultant force exerted by the car on the road
1000
Fengine
300
Ftc=-Fct
Fengine = mtotal a = 1300*2.15 = 2795 N
Fct = mtrailer*2.15 = 645 N, so Ftc = -645 N
a) Fcar = 2795-645 = 2150 Fcar = mcar*2.15 = 2150
b) Ftrailer = Ftc = 645 N
2150
c) Ftc = -645 N
d) Ftotal = (21502+(-9.81*1000)2
mg=1000*9.8
= 1E+04 N
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Question
ceiling
A block of mass M is hanging
from a string. What is the
tension in the string?
M
a)
b)
c)
d)
e)
T=0
T=Mg with g=9.81 m/s2
T=0 near the ceiling and T=Mg near the block
T=M
one cannot tell
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question
END LECTURE 2
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EXTRA CREDIT QUIZ
M
What force must Sparty
apply in order to crush the
Fighting Irish?
a)
b)
c)
d)
e)
None, the Irish couldn’t fight their way out of a wet paper bag
None, without friction the only force is gravity!
None, MSU plays fair (especially when there’s no real fight)
Final Score H: 6 V:64
Sparty doesn’t appy force, Sparty only hands out humiliating defeat
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EXTRA CREDIT QUIZ
Force due to airblower.
n = -Fp
Fp = mg cos

F = mgsin
A small block is put on a
frictionless slope. Attached on the
block is an air-blower which keeps
the block from slipping down the
slope.
Which of the following is true:
Fg=mg 
a) The force by the airblower on the block equals the
total gravitational force on the block
b) The force by the airblower on the block is smaller than
the total gravitational force on the block
c) The force by the airblower on the block is larger than
the total gravitational force on the block
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Tension
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Quiz!
10 Kg
A homogeneous block of 10 kg is hanging with 2 ropes
from a ceiling. The tension in each of the ropes is:
a) 0 N
b) 49 N
c) 98 N
d) 196 N
e) don’t know
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T
900
T
1kg
A mass of 1 kg is hanging from a rope as shown in the figure.
If the angle between the 2 supporting wires is 90 degrees,
what is the tension in each rope?
TVerR
450
ThorR
TVerL
450
ThorL
Fg
Horizontal
left rope Tsin(45)
right rope -Tsin(45)
gravity
0
Sum:
0
Vertical
Tcos(45)
Tcos(45)
-19.81
2Tcos(45)-9.81
The object is stationary, so:
2Tcos(45)-9.81=0 so, T=6.9 N
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Friction
Friction are the forces acting on an object due to interaction
with the surroundings (air-friction, ground-friction etc).
Two variants:
• Static Friction: as long as an external force (F) trying to
make an object move is smaller than fs,max, the static
friction fs equals F but is pointing in the opposite direction:
no movement!
fs,max = sn
s = coefficient of static friction
• Kinetic Friction: After F has surpassed fs,max, the object
starts moving but there is still friction. However, the
friction will be less than fs,max!
Fk = kn
k = coefficient of kinetic friction
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Friction
N
20 kg
Fs
pull
A person wants to drag a crate of 20
kg over the floor. If he pulls the crate
with a force of 98 N, the crate starts
to move. What is s?
Fg
Answer: Fs = sN = sMg = 209.8s = 196s
Just start to move: Fpull-Fs = 0  Fs=Fpull  98=196s
so: s=0.5
After the crate starts moving, the person continues to pull with the
same force. Given k=0.4, what is the acceleration of the crate?
F=ma
F = 98-0.4Mg = 98 - 0.4209.8 = 98-78.4 = 19.6 N
19.6 = 20a  a = 0.98 m/s2
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General strategy
• If not given, make a drawing of the problem.
• Put all the relevant forces in the drawing, object by object.
• Think about the axis
• Think about the signs
• Decompose the forces in direction parallel to the motion and
perpendicular to it.
• Write down Newton’s law for forces in the parallel direction
and perpendicular direction.
• Solve for the unknowns.
• Use these solutions in further equations if necessary
• Check whether your answer makes sense.
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Example Problem
Force due to friction: Fstat or Fkin
n = -Fgp
Fgp = mg cos

FgL = mgsin
Fg=mg 
A) If s=1.0, what is the angle 
for which the block just starts to
slide?
B) The block starts moving. Given
that k=0.5, what is the
acceleration of the block?
A) Parallel direction:
mgsin - sn = 0 (F=ma)
Perpendicular direction: mgcos - n = 0 so n = mgcos
Combine:
mgsin - smgcos = 0
s = sin/cos = tan = 1 so =45o
B) Parallel direction: mgsin(45o) - kmgcos(45o) = ma (F=ma)
g (½2 – ½½2) = a so a = g¼2
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puck on ice
After being hit by a hockey-player, a puck is moving over
a sheet of ice (frictionless). The forces working on the
puck are:
a)
b)
c)
d)
e)
No force whatsoever
the normal force only
the normal force and the gravitational force
the force that keeps the puck moving
the normal force, the gravitational force and the force
that keeps the puck moving
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All the Forces Come Together
n
T
T
Fk
For the 7 kg block parallel to the
slope:
T - mgsin - kmgcos = ma
Fg
Fk = kmgcos
Solve for k
If a=3.30 m/s2 (the 12kg block
is moving downward), what is
the value of k?
k 
Fg
For the 12 kg block: T – Mg = -Ma
 M ( g  a )  mg sin   ma
 0.25
 mg cos 
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Ffriction
n
Example
1 kg
T
A
Fg
20o
T
Is there a value for the static
friction of surface A for which
these masses do not slide?
If so, what is it?
0.5 kg
Fg
0.5 kg mass: F=ma (only vertical)
T-mg=ma T-0.5g=0.5a
1 kg mass:
No sliding:
F=ma (parallel to the slope)
-Fg//-T+Ffriction=ma
-mgsin(q)-T+smgcos(q)=ma
-3.35-T+9.2s=a
a=0, so T=0.5g=4.9 (from 0.5kg mass equation)
-3.35-4.9+9.2s=0
s=0.9
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Airblower again
Force due to blower
n = -FgP
FgP = mg cos

FgL = mgsin
Fg=mg 
A cart is placed on a slope that
makes an angle  with the
horizontal. An airblower is
placed on the cart, creating a
force in the direction up the
slope.
If the cart and blower have a
combined weight of 10 kg and
=100, what is the force exerted
by the blower (ignore friction) is
the cart is stationary?
Fblower = FgL (direction along the slope)
Fblower = mgsin = 10 x 9.8 x sin(100) = 17 N
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Elevator
T
An object of 1 kg is hanging from a spring
scale in an elevator. What is the
weight read from the scale if the elevator:
a) moves with constant velocity
b) accelerates upward with 3 m/s2
c) accelerates downward with 3 m/s2
d) decelerates with 3 m/s2 while moving up
Before starting this:
1) Imagine standing in an elevator yourself
2) the scale reading is equal to the tension T
F=ma so… T-mg=ma and thus…T=m(a+g)
a)
b)
c)
d)
a=0
a=+3 m/s2
a=-3 m/s2
a=-3 m/s2
T = 1*9.8 = 9.8 N
T = 1(3+9.8) = 12.8 N (heavier)
T = 1(-3+9.8) = 6.8 N (lighter)
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T = 6.8
N231
PHY
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For Next Week
Rex & Wolfson Chapter 5:
Work and Energy
Homework Set 3: Covers Chapters 4
Due Wednesday 11PM
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