Transcript Mechanics 105 chapter 4

Mechanics 105
The laws of motion (chapter four)
Kinematics – answers the question “how?”
Statics and dynamics answer the question “why?”
Force
Newton’s 1st law (object at rest/motion stays that way)
Inertial mass
Newton’s 2nd law (F=ma)
Gravity
Newton’s 3rd law (Action-reaction)
Mechanics 105
Force
Intuitive concept of force
Contact and field forces
Most of the forces we experience are
due to gravitational or electromagnetic
Vector nature of forces – acceleration will
be in same direction as net force
Notation: F12 is the force exerted by object
1 on object 2
Mechanics 105
Newton’s 1st law
“In the absence of external forces, an object at rest remains at rest
and an object in motion continues in motion with a constant
velocity (that is, with a constant speed in a straight line).”
In other words, objects accelerate only if there is a net force on them
ConcepTest
Mechanics 105
Inertial frames & inertial mass

Inertial frame of reference = frame in which 1st law is valid
–



at rest or at constant velocity
Inertial mass is the resistance of an object to a change in motion
in response to an external force (basically it’s defined by F=ma)
SI unit: kg
not the same as weight (=force due to gravitational attraction of
earth)
Mechanics 105
Newton’s 2nd law
“The acceleration of an object is proportional to the net
force acting on it and inversely proportional to its

mass.”

 F  ma
i.e. the net (vector) force on an object is proportional to
the (vector) acceleration of the object
Units: “Newton” (N) = kgm/s2
Mechanics 105
Example
An object of mass m slides down a
frictionless inclined plane (angle  with
respect to the horizontal). What is its
acceleration?
m

Mechanics 105
Example
An object of mass m slides down a
frictionless inclined plane (angle  with
respect to the horizontal). What is its
acceleration?
Step one N
draw the force
vectors acting
on the object
Fg

Mechanics 105
Example
An object of mass m slides down a
frictionless inclined plane (angle  with
respect to the horizontal). What is its
acceleration?
Component along direction
of motion = Fgsin
N
Along this axis Fgsin=ma,
or a= Fgsin/m
Fg

what happens along the
direction perpendicular to the
direction of motion?
Mechanics 105
2nd law - ConcepTests
Mechanics 105
Gravitational force and weight


Fg  mg
What we call weight is the force of gravity
on an object at sea level, it is proportional to
the mass of the object
Note: this equation is valid even for no
acceleration – so the gravitational mass and
the inertial mass don’t have to be the same,
but they are.
Mechanics 105
A neutron walks into a bar; he asks the bartender, 'How much for a
beer?' The bartender looks at him, and says 'For you, no charge.‘
"We cannot learn without pain."
-Aristotle
"The only thing that interferes with my learning is my education."
-Albert Einstein
"Do not worry about your problems with mathematics, I assure you
mine are far greater."
-Albert Einstein
Mechanics 105
Newton’s 3rd law (action-reaction)
If two objects interact, the force F12 exerted by object 1 on object 2 is
equal in magnitude but opposite in direction to the force F21
exerted by object 2 on object 1, i.e.,


F12   F21
Conceptually, a little hard to grasp. Even for an accelerating object,
there are always equal and opposite forces. Even when it seems
like there is only a single object, i.e., an object in space, there
must be a corresponding object which “receives” the reaction force.
Mechanics 105
Applications of Newton’s laws
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Statics
Block on the incline from earlier, now being held up by a force F
If the block doesn’t move, the acceleration must be zero

T
y

N

W

N

T

 
x
W
Free body diagram
Mechanics 105
Applications of Newton’s laws
After making the free body diagram write out components of force
vectors, then apply Newton’s second law along each axis

T  Tx iˆ

N  N y ˆj

W  mg (sin iˆ  cos ˆj )
x:
F
x
mg sin   Tx  max  0
y:
F
y
 N y  mg cos   ma y  0
Mechanics 105
Applications of Newton’s laws
Atwood machine
assume m2>m1
m1 :
T  m1 g  m1a
m2 :
+y

T
m1
m1

m1 g
m2

T
m2

m2 g
T  m2 g  m2 a
Solving both for T
T  m1 ( g  a )  m2 ( g  a )
gives a 
(m2  m1 )
g
(m2  m1 )
substituti ng back gives
2m1m2
T
g
(m1  m2 )
Mechanics 105
Applications of Newton’s laws
One block pushing others
F
m1
m2
m3
Mechanics 105
Applications of Newton’s laws
Pulley’s
m1
m2
Mechanics 105
Applications of Newton’s laws
Combination with circular motion
m1
m2