Dynamics II Presentation 12.14.12x

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Nov 14, 2013 (3 years and 8 months ago)

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Dynamics II

Contact Forces

and

Acceleration

Chapter Objectives


State Newton’s the three laws of motion and give examples
that illustrate each law.


Explain what is meant by the term “net force”.


Use the methods of vector arithmetic to determine the net
force acting on an object.


Define each of the following terms: mass, inertia, and weight.
Distinguish between mass and weight.


Identify the SI units for force, mass, and acceleration.


Draw an accurate free body diagram locating each of the
forces acting on an object or a system of objects.


Use free body and force diagrams help you to solve word
problems.

After studying the material of this chapter, the student should
be able to:

2

Mr. Gary

Key Terms and Phrases

dynamics

is the study of the causes of motion

force

is any kind of push or pull on an object

net force
refers to the vector sum of all the forces acting on an object.

Newton’s first law of motion
is also known as Galileo’s law of inertia, where
inertia refers to the tendency of an object to resist any change in its state of
motion. Inertia is measured by measuring an object’s mass.

Newton’s second law of motion
refers to an object’s motion when a net force
does not equal zero. The net force ( ) will cause an object to accelerate or
decelerate. The rate of acceleration

is directly proportional to the
magnitude of the net force and inversely proportional to the object’s mass
(m),
(i.e., )
.

Newton’s third law of motion
states that whenever one object exerts a force
on a second object, the second object exerts an equal but opposite force on
the first.

Weight

is a measure of the force of gravity on an object.


Mr. Gary

3

a
  
a F m or F=ma
F

Summary: Newton’s Laws of Motion

Mr. Gary

4

Newton’s first law of motion
is also known as Galileo’s law of inertial, where inertia refers
to the tendency of an object to resist any change in its state of motion. Inertia is measured
by measuring an object’s mass.


Newton’s second law of motion
refers to an object’s motion when a net force does not
equal zero. The net force will cause an object to accelerate or decelerate. The rate of
acceleration is directly proportional to the magnitude of the net force and inversely
proportional to the object’s mass, m
:



The SI unit of force is the newton, N, and for mass it is the kilogram, kg. A net force of
1
N
ewton
will
cause
a 1 kg object to accelerate at 1 m/s
2
, thus

1 N = 1 kg m/s
2
.


Newton’s third law
is the
“action
-
reaction
” law with which most students are familiar.
However, it is necessary to be very careful in interpreting the meaning of the this law. The
action force and the reaction force are equal and opposite but do not act on the same
object. A good way to remember the law is the statement given in your textbook:
“Whenever one object exerts a force on a second object, the second object exerts an equal
but opposite force on the first.”

a
   
a F ma or F ma
Mr. Gary

5

A
B
B
C
Vector
A
……………. Pulling
Force.......


Vector
B
……………. Normal Force; Contact
Force……….


Vector
C
………………. Force of Gravity;
Weight……….

P
F
N
F
W
F
Friction Box


P
F
N
F
N
F
W
F
f
F
x net x
x
x
x component
F F =ma
F

 

0
P f
x
F cos F cos ma
   
0
P f
f
f
f
f
F cos F cos 0
Plug and Chug
(300N)cos(30 ) F cos(180 ) 0
267.30N ( 1)F 0
267.30N F 0
F 267.30N
  
   
  
 

y net y
y
y
y component
F F ma
F

  

0
N W
y
P
F sin F sin F sin ma
    
0
P N W
N W
F sin F sin F sin 0
Now, Plug and Chug
(300N)sin(30 ) F sin(90 ) F
   
   
mg
N
N
N
N
2
N
2 2
N
2
N
sin(270 ) 0
150 N+(1)F ( 1)mg 0
150N+F mg 0
Rearrange for F
F mg 150N
F (200kg)(9.8m s ) 150N
F (1960 kgm s ) (150
1810 kgm s 1810
kgm s

)
N
F
 
  
 
 
 
 
 
30

P
F 300N

200kg
f
F
N
F
N
F
W
F
P
F 300

N
F
f
F cos

P
F cos

P
F sin

x
y
30

Mr. Gary

8

x
x net
x
x
x-component
F F ma
F
  

0
P f
x
F cos F cos ma
   
0
P f
P f
f
f
f
f
F cos F cos 0
Now, Plug and Chug
F cos F cos 0
Now, Plug and Chug
(200 N) cos(315 ) F cos(180 ) 0
46.69 N + (-1)F 0
46.69 N - F
46.69
0
N
F

  
  
   



y net y
y
y P N W
y
P N W
N W
y component
ΣF = F = ma
ΣF F sin F sin F sin ma
F sin F sin F sin 0
Now, Plug and Chug
(200 N)sin(315 ) F sin(90 ) F

    
   
   
mg
2
N
2
N
2
N
N
sin(270 ) 0
( 194.47 N) + (+1)F ( 1)(150 kg)(9.8m s ) 0
( 194.47 N) + F (1470 kgm s ) 0
F (194.47 N) (1470 kgm s
1664.
)
47 N
F
 
   
  
 

P
F 200N

P
F 200N

N
F
N
F
f
F
f
F
W
F
c
v
c
v
W
F
45

315

m 150kg

f N
Determine F andF
P
F sin

P
F cos

There’s More!!

Coefficient of Friction

Coefficient of Friction

Mr. Gary

10

The coefficient of friction:



the coefficient of friction is a scalar quantity.



it is an electrostatic bond between and object and surface over which the object


passes.



this electrostatic bond is the result of the attraction of the valence electrons


that interact between a surface and a body.



mathematically it is the ratio of the normal force and friction force.



is dependent on only the “normal force.”



the answer provides us with information as to how stickiness of a surface.



the general mathematical expression is written as:



μ
s

represents the coefficient static friction.



μ
k

represents the coefficient of kinetic friction.

f
f N
N
F
F =
μF where μ = .
F
General Mathematical Expression

Mr. Gary

11

f
f N
N
f P
n g P
P g P
Representing the x-component, we have
and representing the y
F
F =
μF = μ =
F
F = F cos
θ
F = F -F si
-component, we have
Thus, solving for
μ, we get
n
θ
F
F
cos
θ = μ(F -F sinθ
μ
)
=
P
g P
cos
θ
F -F sin
θ
Key Terms and Phrases

friction

is a contact force that opposes the relative motion of two
surfaces as they slide past each other. The frictional force depends on
the coefficient of friction,
μ
, and normal force

.

coefficient of friction
,
μ
,

is a pure number without physical units and
varies with the types of surfaces that are in contact. When the object is
at rest, the coefficient of static friction
μ
s

is used to determine the
magnitude of the frictional force just before the object starts to move.
When the object is moving, the coefficient of kinetic friction
μ
k

is used.

Mr. Gary

12

N
F
Gary’s Check List for Solving Problems


Read each problem carefully and READ IT AGAIN.


Write down conceptual meanings from the wording of each problem.


(i.e., constant velocity, constant acceleration, …, etc.). Oftentimes,
certain word combinations may look the same but have radically
different meanings.


Record all numerical data from the problem and any data inferred
(i.e., gravity,


Use both the reading and any diagram to size up a problem.


Sketch a problem from the reading and provide a free body diagram
to show all forces acting on a body.


spring constant, centripetal acceleration,…, etc.).


In what units should your supporting and final answers be written.


Determine which numbers, if any, need to be converted and perform


that/those operations first.




13

Mr. Gary

Mr. Gary

14

A force of 200 N is pulled by a rope which is at an angle of 30
°

to the horizontal. The box has a mass of 150 kg. (a) Make a
sketch of the problem. (b) Construct a free body diagram from the sketch. (c) Calculate the x, y


components for all the
forces on the block. (d) Calculate the coefficient kinetic friction between the block and surface.


x
x net
x
x
x-component
F F ma
F
  

0
P f
x
F cos F cos ma
   
0
P f
P f
f
f
f
f
F cos F cos 0
Now, Plug and Chug
F cos F cos 0
Now, Plug and Chug
(200 N) cos(315 ) F cos(180 ) 0
46.69 N + (-1)F 0
46.69 N - F
46.69
0
N
F

  
  
   



y net y
y
y
y component
F F ma
F

  

0
N W
y
P
F sin F sin F sin ma
    
0
P N W
N W
F sin F sin F sin 0
Now, Plug and Chug
(300N)sin(30 ) F sin(90 ) F
   
   
mg
N
N
N
N
2
N
2 2
N
2
N
sin(270 ) 0
150 N+(1)F ( 1)mg 0
150N+F mg 0
Rearrange for F
F mg 150N
F (200kg)(9.8m s ) 150N
F (1960 kgm s ) (150
1810 kgm s 1810
kgm s

)
N
F
 
  
 
 
 
 
 
P
F
P
F
f
F
f
F
N
F
N
F
N
F
W
F
W
F
P
F cos

P
F sin

f N
k
k
f
k
N
k
k
Determine the coefficient of
kinetic Friction
F F
Rearrange for
F
F
Now, Plug and Chug
(46.69 N)
(18
0.2
N
0
10
6
)


 
 
 
Mr. Gary

15

Newton’s first

law

If

the net force on an object
equals zero, then the object’s
浯瑩潮o睩ll 湯琠c桡hg攬ei⹥⸬†.†† † ††
†† †† †
.

Newton’s
獥c潮o

law

An object’s acceleration is
r敬eW敤 W漠瑨o

湥琠W潲c攠慮a 瑨攠
object’s mass.

Newton’s third

law

If object 1 exerts a force on

object 2, then object 2 exerts
an equal but opposite force on
object 1.

Weight

An object’s weight is directly
proportional to its mass and
gravitational acceleration.



force of friction

Frictional force depends on the
coefficient of friction

and the
normal force.

0
net
F
F = 0
 
0
a

a F m
F ma
 
 
12 21
F F
 
W
g
F mg
F mg


fr N
F F
 