Physics 231
Topic 4: Forces & Laws of Motion
Alex Brown
September 23-28 2015
MSU Physics 231 Fall 2015
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What’s up? (Wednesday Sept 23)
1) Homework set 03 due Tuesday Sept 29rd 10 pm
2) Learning Resource Center (LRC) in 1248 BPS
Monday 9 am to 1 pm and 3 pm to 5 pm
Tuesday 9 am to 9 pm
Friday 9 am to 1 pm
3) First midterm in here during class time on Wednesday Sept 30
Covers Chapters 1-4 and homeworks 01-03
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• The 1st exam will be Wednesday September 30.
• The exam will take place right here in BPS 1410.
• About 15 multiple choice questions that you should be able to answer in
1 to 5 min each.
• We will have assigned seating, so you need to show up 10min early to
guarantee your seat.
• If you are sick, you MUST have a note from a doctor on the doctor’s
letterhead or prescription pad.
• You CANNOT use cell phones during the exam.
• You can (should!) bring a hand written equation sheet - 8.5x11 sheet of
paper (both sides).
• Bring your calculator and several no. 2 pencils.
• Bring your MSU picture ID.MSU Physics 231 Fall 2015
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Recall from Last Chapter
Vectors and Scalars
Two Dimensional Motion
 Velocity in 2D
 Acceleration in 2D
Projectile motion
 Throwing a ball or cannon fire
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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
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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(A)
v f  v0  at
What causes the
Acceleration?
(B)
x  v o t  at
(C)
x  v f t  at
(D)
x  12 (v0  v f )t  v t
(E)
x 
1
2
1
2
1
2a
2
2
(v f  v 0 )
2
2
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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 is moving with a given 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: a = F/m or 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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

FAB  m B a B  Force of A on B (causes B to move)


FBA  m A a A  Force of B on A (causes A to move)


FBA  - FAB  Newtons 3rd law




FAB  m B a B   FAB  m A a A
aA
mB
mB 


 1 magnitudes
aA  aB
aB
mA
mA
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Identifying the forces in a system
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question
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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Force, Mass and Weight
Forces are quantified in units of Newton (N).
1 N = 1 kgˑm/s2


Fnet  m a
A force is a vector: it has direction.
The net force causes acceleration in the same
direction as the net force.
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Force due gravity
Fg = m g
where g = 9.8 m/s2
For an object with m = 1 kg
F = ma = mg
Fg = m g = 9.8 kgˑm/s2 = 9.8 Newton
means that a = g
the direction is down
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

Fnet  m a
If there is one force acting then


Fnet  F
If there are many forces acting then



Fnet  F1  F2  F3 .....
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Two forces that act upon us
nearly all the time.

n
Gravitational Force (Fg)
Fg = mg (referred to as weight)
g = 9.81 m/s2



1) Fnet  Fg  n  0
2) Fg = mg = n (magnitudes)


3) Fg  n
Normal Force:
elastic force acting
perpendicular to the
surface the object is
resting on. Name: n

Fg
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Clicker 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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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
Net 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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inclinded plane - angles 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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inclinded plane - angles to remember
y
x
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
n
Fgy = mg cos

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 in the direction of motion.?
(assume no friction).
Fgx = mg sin
Balance forces in directions
where you expect no
acceleration; whatever is left
causes the object to accelerate!
Fg=mg 
n = Fgy (magnitudes)
ma = Fnet = Fgx = mg sin
a = g sin
a = 6.3 m/s2
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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
question

A ball is rolling from a ramp at angle  = 10o. The length of the ramp is
L=1.5 m. What is the time to reach the bottom (in s)
a)
b)
c)
d)
2.71
1.33
0.55
0.20
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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


Fg  w
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Tension
T1
T2 = - T1
T2 = T1
T2
(vectors)
(magnitudes)
M
As long as it’s one rope, the tension is the
same at both ends!
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Elevator
T
a)
b)
c)
d)
a=0
a=+3 m/s2
a=-3 m/s2
a=-3 m/s2
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 (start going up)
c) accelerates downward with 3 m/s2 (start going down)
d) decelerates with 3 m/s2 while moving up
(end going 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)
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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Newton’s second law and tension
n
m1
T1
No friction.
T2
What is the acceleration of
the objects?
F1g
m2
F2g
T2 = T1 (magnitudes)
F2g = m2g
Object 1 (red):
 F1 = m1 a = T1
Object 2 (green):  F2 = m2 a = F2g - T2 = m2 g – T1 = m2 g - m1a
Solving for a
a = m2 g /(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?
T
T1
T
T2
m1
T = T1 = T2
Object 1 (left):
T – m1g = m1a
Object 2 (right): – T + m2g = m2a
two equations and two unknowns
a = g (m2- m1)/(m1+ m2) = 2.45 m/s2
m2
T = m1(a+g) = 36.8 N
Fg
Fg
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Walking on the moon. The acceleration on the moon is 1/6 that
of the earth. If m2=140 kg what does m1 have to be to make the
acceleration of m2 the same as that on the moon?
A) 80
B) 100
C) 120
D) 140
E) 160
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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 on car (c)
b) the net force on the trailer (t)
c) the net force exerted by the trailer on the car
1000
Ftc Fct
Fengine
Fe
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Ftc= Fct
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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 on car (c)
b) the net force on the trailer (t)
c) the net force exerted by the trailer on the car
c
1000
Ftc Fct
Fengine
Fe
t
300
Ftc= Fct
Force provided by engine Fe = (mc+mt ) a = 2795 N
Fct = mt a = 645 N, so Ftc = 645 N
a) Fc = Fe - Ftc = (2795 – 645) N = 2150 N
b) Ft = Fct = 645 N
c) Ftc = 645 N
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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
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
T
900
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?
Ty
450
Ty
450
Tx
Tx
Fg
T left rope
T right rope
gravity
sum:
x
y
T sin(45)
T cos(45)
-T sin(45)
T cos(45)
0
-19.81
0
2 T cos(45) - 9.81
The object is stationary, so:
2 T cos(45) - 9.81 = 0 so, T=6.9 N
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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 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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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:
- so they cancel and there is no movement!
fs,max = sn
s = coefficient of static friction
n = magnitude of the normal force
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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:
• 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
n = magnitude of normal force
fk points in the opposite direction to the motion
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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: fsmax = sn = sMg = 209.8s = 196s
Just start to move: Fpull-fsmax = 0  fsmax=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 (x)
and perpendicular (y) 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: fs or fk
n = -Fgy
Fgy = mg cos

Fgx = 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 (x):
m g sin - s n = 0 (F=ma)
Perpendicular direction (y): m g cos - n = 0 so n = m g cos
Combine:
m g sin - s m g cos = 0
s = sin/cos = tan = 1 so =45o
B) Parallel direction: m g sin(45o) - km g cos(45o) = ma (F=ma)
a = 3.47
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All the Forces Come Together
n
T
T2
If a=3.30 m/s2 (the 12 kg block
is moving downward), what is
the value of k?
fk
Fg
fk = kmgcos
f k   k mg cos 
F2g
Solve for k
 f k  mg sin   Mg  ( M  m)a
( M  m)a  mg sin   Mg
k 
 mg cos 
 0.25
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fsmax
n
Example
1 kg
T
A
Fg
20o
Is there a value for the static
friction of surface A for which
these masses do not slide?
If so, what is it?
T
0.5 kg
Fg
No sliding:
same as previous problem with a=0 and angle=-20
-3.35 - 4.9 + 9.2s=0
s = 0.9
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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: - 6000 cos(45)=- 4243
Sum:
-1243 N
N
6000N
Problem
Vertical
0N
6000 sin(45)= 4243
4243 N
Magnitude of resulting force:
Fsum = [(-1243)2+(4243)2] = 4421 N
W
3000N
E
Direction: angle with respect to north
= tan-1(1243/4243) =16.30
F=ma so a = F/m = 4421/2000 = 2.21 m/s2
S
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Rotational motion and forces
centripetal acceleration must caused by a net force acting
Fnet = m ac
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Example from Homework set 02
“weight” is due to the normal force of the wall on the person
n= Fnet=mv2/R
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Ball on string – Fnet provided by tension in the rope
T= Fnet=mv2/R
T= Mg
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Car on curve without friction – Fnet provided by normal force
n
mg
Fnet=mv2/R
mg
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Car on curve with friction –
Fnet provided by friction
n
Fnet=mv2/r = fs
mg
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Moon in orbit around the earth –
Fnet provided by gravitational force between the moon and the earth
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