Topic 4: Forces & Laws of Motion
Alex Brown
September 23-28 2015
MSU Physics 231 Fall 2015 1

MSU Physics 231 Fall 2015 2

What’s up? (Wednesday Sept 23)
1) Homework set 03 due Tuesday Sept 29rd 10 pm
2) Learning Resource Center (LRC) in PBS 1248 note expanded hours
Monday 9 am to 8 pm
Tuesday 9 am to 9 pm
Friday 9 am to 4 pm
3) First midterm in here during class time on Wednesday Sept 30
Covers Chapters 1-4 and homeworks 01-03
MSU Physics 231 Fall 2015 3

• The 1 st 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
4
4

Vectors and Scalars
Two Dimensional Motion
 Velocity in 2D
 Acceleration in 2D
Projectile motion
 Throwing a ball or cannon fire
MSU Physics 231 Fall 2015 5

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
MSU Physics 231 Fall 2015 6

“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)
7
7

(C)
(D)
(A)
(B)
(E) v f
 v
0
 at
 x
 v o t

1
2 at
2
 x
 v f t

1
2 at
2
 x

1
2
( v
0
 v f
) t
 v t
 x

1
2 a
( v f
2  v
0
2
)
MSU Physics 231 Fall 2015
8

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:
F
12
=-F
21
9
9


F
AB

F
BA

F
 BA
F
AB
 a
A
 m
 a
B B
 m
A

 a

F
AB
 m
B
A
 a

Force of

Force of
A on B (causes B to move)
B on A (causes A to move)
B

Newtons
 

F
AB

3rd law
 m
A
 a
A

m m
B
A
 a
B a a
A
B
 m m
B
A

1 magnitudes
MSU Physics 231 Fall 2015 10

11
11

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
12
12

1 N = 1 kgˑm/s 2

F net
 m
 a
vector
13

F g
= m g where g = 9.8 m/s 2
For an object with m = 1 kg F g
= m g = 9.8 kgˑm/s 2 = 9.8 Newton
F = ma = mg means that a = g the direction is down
14


F net
 m
 a

F net


F

F net


F
1

F
2


F
3
.....
MSU Physics 231 Fall 2015 15

 n Normal Force: elastic force acting perpendicular to the surface the object is resting on. Name: n
1)

F net


F g

 n

0
2) F g
= mg = n (magnitudes)
3)

F g
 
 n

F g
Gravitational Force (F g
)
F g
= mg (referred to as weight) g = 9.81 m/s 2
16
16

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
17
17

24 o
300 N
690 N
70 o
Net Force
139 N
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?
Force
300 N
690 N x(N)
-122
236
Weight
Resultant 114 N
0 y(N)
-274
648
-294
80 N mg
35.2
o
Net Force: F =

(114 2 +80.0
2 ) = 139 N
Direction:

= tan -1 (F y
/F x
) = 35.2
o
Acceleration: a = F/m = 139/30.0 = 4.65 m/s 2
MSU Physics 231 Fall 2015 18



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).
MSU Physics 231 Fall 2015 19

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).
MSU Physics 231 Fall 2015 20

n
1) Draw all Forces acting on the red object
2) If

=40 o and the mass of the red object equals 1 kg, what is the resulting acceleration in the direction of motion.?
F gx
= mg sin

(assume no friction).
F gy
= mg cos


F g
=mg

Balance forces in directions where you expect no acceleration; whatever is left causes the object to accelerate!
n = F gy ma = F
(magnitudes) net a = g sin

= F gx
= mg sin
 a = 6.3 m/s 2
MSU Physics 231 Fall 2015 21

question

A ball is rolling from a slope at angle

. It is repeated but after decreasing the angle

. Compared to the first trial: a) the normal force on the ball increases b) the normal force gets closer to the gravitational force c) the ball rolls slower d) all of the above
MSU Physics 231 Fall 2015 22

question

A ball is rolling from a ramp at angle

= 10 o . The length of the ramp is
L=1.5 m. What is the time to reach the bottom (in s) a) 2.71
b) 1.33
c) 0.55
d) 0.20
MSU Physics 231 Fall 2015 23

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...
MSU Physics 231 Fall 2015 24


F g

 w
MSU Physics 231 Fall 2015 25

T
1
T
2
T
2
= - T
1
(vectors)
T
2
= T
1
(magnitudes)
M
As long as it’s one rope, the tension is the same at both ends!
MSU Physics 231 Fall 2015 26

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/s 2 (start going up) c) accelerates downward with 3 m/s 2 (start going down) d) decelerates with 3 m/s 2 while moving up
(end going up)
T
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) a=0 T = 1*9.8 = 9.8 N b) a=+3 m/s 2 T = 1(3+9.8) = 12.8 N (heavier) c) a=-3 m/s 2 T = 1(-3+9.8) = 6.8 N (lighter) d) a=-3 m/s 2 T = 6.8 N 27
27

n
T
1
No friction.
m
1
F
1g
T
2 m
2
What is the acceleration of the objects?
T
2
= T
1
(magnitudes)
F
2g
F
2g
= m
2 g
Object 1 (red):

F
1
Object 2 (green):

F
2
= m
1 a = T
1
= m
2 a = F
2g
- T
2
= m
2 g – T
1
= m
2 g - m
1 a
Solving for a a = m
2 g /(m
1
+m
2
)
MSU Physics 231 Fall 2015 28

T
1
T m
1
F g
F g m
2
T
T
2
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 = T
1
= T
2
Object 1 (left): T – m
1 g = m
1 a
Object 2 (right): – T + m
2 g = m
2 a two equations and two unknowns a = g (m
2
- m
1
)/(m
1
+ m
2
) = 2.45 m/s 2
T = m
1
(a+g) = 36.8 N
MSU Physics 231 Fall 2015 29

x x
Clicker question:
Walking on the moon. The acceleration on the moon is 1/6 that of the earth. If m
2
=140 kg what does m
1 acceleration of m
2 have to be to make the the same as that on the moon?
x x
A) 80
B) 100
C) 120
D) 140
E) 160 m
2 m
2 g – m
1 g = (m
1
+ m
2
) a a = g/6 solve for m
1 m
1
= m
2
(g - a)/(a + g) = = m
= 100
2
MSU Physics 231 Fall 2015
(5/7)
30

A 1000-kg car is pulling a 300 kg trailer. Their acceleration is 2.15 m/s 2 . 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
F tc
F ct
300
F F
F tc
= F ct
MSU Physics 231 Fall 2015 31

A 1000-kg car is pulling a 300 kg trailer. Their acceleration is 2.15 m/s 2 . 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
F tc
F ct t
300 F tc
= F ct
F F
Force provided by engine F e
F ct
= m t a = 645 N, so F tc
= (m c
+m
= 645 N t
) a = 2795 N a)  F c b)  F t c) F tc
= F
= F e ct
- F tc
= (2795 – 645) N = 2150 N
= 645 N
= 645 N
MSU Physics 231 Fall 2015 32

ceiling
A block of mass M is hanging from a string. What is the tension in the string?
M a) T=0 b) T=Mg with g=9.81 m/s 2 c) T=0 near the ceiling and T=Mg near the block d) T=M e) one cannot tell
MSU Physics 231 Fall 2015 33

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
MSU Physics 231 Fall 2015 34

T
90 0
T
1kg
T x
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?
T
45 y
0
T
45 y
0 x y
T left rope T sin(45) T cos(45)
T right rope -T sin(45) T cos(45) gravity 0 -1

9.81
sum: 0 2 T cos(45) - 9.81
T x
F g
The object is stationary, so:
2 T cos(45) - 9.81 = 0 so, T=6.9 N
MSU Physics 231 Fall 2015 35

After being hit by a hockey-player, a puck is moving over a sheet of ice (frictionless). The forces on the puck are: a) No force whatsoever b) the normal force only c) the normal force and the gravitational force d) the force that keeps the puck moving e) the normal force, the gravitational force and the force that keeps the puck moving
36
36

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 f s,max
, the static friction f s equals F but is pointing in the opposite direction :
- so they cancel and there is no movement!
f s,max
=  s n  s
= coefficient of static friction n = magnitude of the normal force
MSU Physics 231 Fall 2015 37

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 f s,max
, the object starts moving but there is still friction . However, the friction will be less than f s,max f k
=  k n  k
= coefficient of kinetic friction n = magnitude of normal force f k points in the opposite direction to the motion
MSU Physics 231 Fall 2015 38

39
39

MSU Physics 231 Fall 2015 40

f s
20 kg n 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
?
F g
Answer: f smax
=
 s n =
 s
Mg = 20

9.8
 s
= 196
 s
Just start to move: F pull
-f smax
= 0
 f smax
=F pull so:
 s
=0.5

98=196
 s
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/s 2
MSU Physics 231 Fall 2015 41

• 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.
MSU Physics 231 Fall 2015 42

Force due to friction: f s or f k n = -F gy
A) If
 s
=1.0, what is the angle
 for which the block just starts to slide?
F gy
= mg cos


F gx
= mg sin

B) The block starts moving. Given that
 k
=0.5, what is the acceleration of the block?
F g
=mg

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:
 s m g sin

-
 s m g cos

= 0
= sin

/cos

= tan

= 1 so

=45 o
B) Parallel direction: m g sin(45 o ) -
 k a = 3.47
m g cos(45 o ) = ma (

F=ma)
MSU Physics 231 Fall 2015 43

x x
Clicker question:
Walking on the moon. The acceleration on the moon is 1/6 that of the earth. If m
2
=140 kg what does m
1 acceleration of m
2 have to be to make the the same as that on the moon?
x x
A) 80
B) 100
C) 120
D) 140
E) 160 m
2 m
2 g – m
1 g = (m
1
+ m
2
) a a = g/6 solve for m
1 m
1
= m
2
(g - a)/(a + g) = = m
= 100
2
MSU Physics 231 Fall 2015
(5/7)
44

m
2
=
 
x x The 12 kg block is moving downward, what is the value of a?
m
1
=
 -m
2 g sin  + m
1 g
= (m
1
+ m
2
) a solve for a a
 m
1 g

( m
1 m
2
 g m
2 sin
)


4 .
0
MSU Physics 231 Fall 2015 45

n m
2
=
T
2
T
1
F
2g m
1
=
F
1g
MSU Physics 231 Fall 2015 46

f k n
F
2g
T
2
T
1
If a=3.30 m/s 2 what is the value of
 k
?
(the 12 kg block is moving downward),
-f k
= (m
1
-m
2 g sin  + m
1
+ m
2
) a g with f k
=
 k m
2 g cos

F
1g solve for
 k
 k

( m
1
 m
2
) a

 m
2 m
2 g g sin cos

  m
1 g

0 .
25
MSU Physics 231 Fall 2015 47

f smax
A
F
2g n
1 kg
T
20 o
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
F
1g
-f s
= (m
1
+ m
2
+ m g sin  + m
1
2
) a = 0 g with f smax
=
 s m
2 g cos
 solve for
 s
 s

 m
2
 g m
2 sin g
 cos

 m
1 g

0 .
90
MSU Physics 231 Fall 2015 48

Rotational motion and forces centripetal acceleration must caused by a net force acting
F c
= m a c
MSU Physics 231 Fall 2015 49

Example from Homework set 02
“weight” is due to the normal force of the wall on the person n= F c
=mv 2 /R = m w 2
R = m g w  ( g/R) 1/2 = 0.61 radians/s
MSU Physics 231 Fall 2015 50

Ball on string – F c provided by tension in the rope
T = F c
= mv 2 /R
T= Mg
MSU Physics 231 Fall 2015 51

Car on curve without friction – F c provided by normal force n
F c
= mv 2 /R
MSU Physics 231 Fall 2015 mg
52

Car on curve without friction – F c provided by normal force n mg
F c
= mv 2 /R
MSU Physics 231 Fall 2015 mg tan
  v
2
Rg
53

Car on curve with friction –
F c provided by friction
R
R n mg
F c
= mv 2 /R = f s
 s v
2

Rg
MSU Physics 231 Fall 2015 54

Moon in orbit around the earth –
F net provided by gravitational force between the moon and the earth
MSU Physics 231 Fall 2015 55

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
Vertical
0 N
Due to wind: - 6000 cos(45)=- 4243 6000 sin(45)= 4243
Sum: -1243 N 4243 N
6000N
N
W
S
3000N
E
Magnitude of resulting force:
F sum
=

[(-1243) 2 +(4243) 2 ] = 4421 N
Direction: angle with respect to north
= tan -1 (1243/4243) =16.3
0
F=ma so a = F/m = 4421/2000 = 2.21 m/s 2
MSU Physics 231 Fall 2015 56