Transcript Force Law

Concept of Force and
Newton’s Laws of Motion
8.01
W02D2
Today’s Reading Assignment:
Young and Freedman Sections 4.1-4.6,
5.1-5.3
Review: Relatively Inertial
Reference Frames
Two reference frames with the
zero relative acceleration
r
r
V  dR dt A  dV dt  0
One moving object has different
position vectors in different frames
r1  R  r2
Law of addition of velocities
r r
r
v1  V  v 2
Acceleration in either reference frame is the same
r r
r
r
a1  A  a 2  a 2
Concept Question: Relatively
Inertial Reference Frames
Suppose Frames 1 and 2 are relatively inertial reference frames.
1) An object that is at rest in Frame 2 is moving at a constant
velocity in reference Frame 1.
2) An object that is accelerating in Frame 2 has the same
acceleration in reference Frame 1.
3) An object that is moving at constant velocity in Frame 2 is
accelerating in reference Frame 1.
4) An object that is accelerating in Frame 2 is moving at constant
velocity in reference Frame 1.
5) Two of the above
6) None of the above
C.Q. Answer: Newton’s First
Law
Answer: (5). Both (1) and (2) are correct.
An object that is at rest in Frame 2 is moving at a
constant velocity in reference Frame 1. The
accelerations are the same in relatively inertial
reference frames because the relative velocity of
the two reference frames is constant hence the
relative acceleration of the two reference frames
is zero.
Inertial Mass

Inertial mass as a ‘quantity of matter’

Standard body with mass ms and SI units [kg]

Mass of all other bodies will be determined relative to the mass of our
standard body.

Apply the same action to the standard body and an unknown body

Define the unknown mass in terms of ratio

mu as

ms au
Mass ratio is independent of the method used to produce the uniform
accelerations.
Definition of Force
Force is a vector quantity
F
total
 ma
The magnitude of the total force is defined to be
r total
r
F
ma
The direction of the vector sum of all the forces on a body is the
same as the direction of the acceleration.
The SI units for force are newtons (N):
1 N  1 kg  m  s2
Superposition Principle
Apply two forces
F1 and F2 on a body,
the total force is the vector sum of the two forces:
F total  F1  F2
Notation: The force acting on body 1 due to the
interaction between body 1 and body 2 is denoted by F
12
Example: The total force exerted on
body 3 due to the interactions with
bodies 1 and 2 is:
total
3
F
 F31  F32
Examples of Forces
• Gravitation
• Electric and magnetic forces
• Elastic forces (Hooke’s Law)
• Frictional forces: static and kinetic friction, fluid
resistance
• Contact forces: normal forces and static friction
• Tension and compression
Newton’s First Law
Every body continues in its state of rest, or of
uniform motion in a right line, unless it is
compelled to change that state by forces
impressed upon it.
r r
r
 Fi  0  v  constant
Newton’s First Law in relatively inertial reference
frames:
If the vector sum of forces acting on an object at
rest in Frame 2 is zero then the vector sum of
forces acting on the object moving at constant
speed in Frame 1 is also zero
Newton’s Second Law
The change of motion is proportional to the motive
force impresses, and is made in the direction of the right line
in which that force is impressed,
F  m a.
When multiple forces are acting,
N
 F  m a.
i
i 1
In Cartesian coordinates:
N
F
i 1
x,i
 m ax ,
N
F
i 1
y,i
 m ay ,
N
F
i 1
z,i
 m az .
Newton’s Third Law
To every action there is always opposed an equal
reaction: or, the mutual action of two bodies upon
each other are always equal, and directed to contrary
parts.
F1,2  F2,1
Action-reaction pair of forces cannot act on same
body; they act on different bodies.
Force Law:
Newtonian Induction
• Definition of force has no predictive content.
• Need to measure the acceleration and the mass in order to define the
force.
• Force Law: Discover experimental relation between force exerted on
object and change in properties of object.
• Induction: Extend force law from finite measurements to all cases
within some range creating a model.
• Second Law can now be used to predict motion!
• If prediction disagrees with measurement adjust model.
Empirical Force Law: Hooke’s
Law
Consider a mass m attached to a spring
Stretch or compress spring by
different amounts produces
different accelerations
Hooke’s law:
| F |  k l
Direction: restoring spring to equilibrium
Hooke’s law holds within some reasonable range of extension or
compression
Table Problem: Hooke’s Law
Consider a spring with negligible mass
that has an unstretched length 8.8 cm. A
body with mass 150 g is suspended from
one end of the spring. The other end (the
upper end) of the spring is fixed. After a
series of oscillations has died down, the
new stretched length of the spring is 9.8
cm. Assume that the spring satisfies
Hooke’s Law when stretched. What is the
spring constant?
Force Law: Gravitational Force
near the Surface of the Earth
Near the surface of the earth, the gravitational
interaction between a body and the earth is mutually
attractive and has a magnitude of
Fgrav  mgrav g
where mgrav is the gravitational mass of the body and g
is a positive constant.
g ; 9.81 m  s 2
Force Laws: Contact Forces
Between Surfaces
The contact force between two
surfaces is denoted by the vector
r
r total
Fsurface,hand  C
Normal Force: Component of the contact force
perpendicular to surface and is denoted by
normal
Fsurface
, hand  N
Friction Force: Component of the contact force tangent to
the surface and is denoted by
tangent
Fsurface
, hand  f
Therefore the contact force can be modeled as a vector
r r r
sum
C Nf
Concept Question: Car-Earth
Interaction
Consider a car at rest. We can conclude that
the downward gravitational pull of Earth on the
car and the upward contact force of Earth on it
are equal and opposite because
1.
2.
3.
4.
the two forces form a third law interaction pair.
the net force on the car is zero.
neither of the above.
unsure
Concept Question: Normal Force
Consider a person standing in an elevator that
is accelerating upward. The upward normal
force N exerted by the elevator floor on the
person is
1.
2.
3.
larger than
identical to
smaller than
the downward force of gravity on the person.
Kinetic Friction
The kinetic frictional force fk is proportional to the normal
force, but independent of surface area of contact and the
velocity.
The magnitude of fk is
fk  k N
where µk is the coefficients of friction.
Direction of fk: opposes motion
Static Friction
Varies in direction and magnitude depending on
applied forces:
0  fs  fs,max  s N
Static friction is equal to it’s maximum value
fs,max  s N
Tension in a Rope
The tension in a rope at a distance x from one end
of the rope is the magnitude of the action-reaction
pair of forces acting at that point ,
T ( x)  Fleft,right ( x)  Fright,left ( x)
Table Problem: Tension in a
Rope Between Trees
Suppose a rope is tied rather tightly between two trees
that are separated by 30 m. You grab the middle of the
rope and pull on it perpendicular to the line between the
trees with as much force as you can. Assume this force
is 1000 N, and the point where you are pulling on the
rope is 1 m from the line joining the trees. What is the
magnitude of the tension in the rope?
Concept of System: Reduction
• Modeling complicated interaction of objects by isolated a
subset (possible one object) of the objects as the system
• Treat each object in the system as a point-like object
• Identify all forces that act on that object
Free Body Diagram
•
Represent each force that is acting on the object by an arrow
on a free body force diagram that indicates the direction of
the force
T
F  F1  F2 
•
Choose set of independent unit vectors and draw them on
free body diagram.
•
Decompose each force Fi
in terms of vector components.
Fi  Fi , x ˆi  Fi , y ˆj  Fi , z kˆ
•
Add vector components to find vector decomposition of the
total force
T
T
T
Fx  F1, x  F2, x  
FyT  F1, yT  F2, yT  
FzT  F1, zT  F2, zT  
Newton’s Second Law: Strategy
• Treat each object in the system as a point-like object
• Identify all forces that act on that object, draw a free body
diagram
• Apply Newton’s Second Law to each body
• Find relevant constraint equations
• Solve system of equations for quantities of interest
Worked Example: Pulley and
Inclined Plane
A block of mass m1, constrained to move along a plane
inclined at angle ϕ to the horizontal, is connected via a
massless inextensible rope that passes over a massless
pulley to a bucket to which sand is slowly added. The
coefficient of static friction is μs. Assume the gravitational
constant is g. What is the mass of the bucket and sand
(m2) just before the block slips upward?
Solution: Pulley and Inclined
Plane
Sketch and coordinate
system
Free body force diagrams
Newton’s Second Law
r
r
a1  0
r
r
a2  0
Constraint Conditions
f s  f s,max  s N
T  T1,r  T2,r
Newton’s Second Law
Object 1:
φ
i1 : T  m1g sin   s N  0
φ
j : N  m g cos   0
1
Object 2:
1
φ
j2 : T  m2 g  0
Simplification: m2 g  m1g sin   s m1 g cos   0
Solution:
m2  m1 (sin   s cos  )
Concept Question: Varying
Tension in String
A block of mass m1, constrained to move along a
plane inclined at angle ϕ to the horizontal, is
connected via a massless inextensible rope that
passes over a massless pulley to a bucket to
which sand is slowly added. The coefficient of
static friction is μs. Assume the gravitational
constant is g. What happens to the tension in the
string just after the block begins to slip upward?
1.
2.
3.
4.
5.
6.
Increases
Decreases
Stays the same
oscillates
I don’t know
None of the above
Table Problem: Two Blocks
and Two Pulleys
Two blocks 1 and 2 of mass m1 and m2 respectively are attached by a string
wrapped around two pulleys as shown in the figure. Block 1 is accelerating to the
right on a fricitonless surface. You may assume that the string is massless and
inextensible and that the pulleys are massless. Find the accelerations of the
blocks and the tension in the string connecting the blocks.
Lecture Demo:
Block with Pulley and
Weight on Incline B24
http://scripts.mit.edu/~t
sg/www/index.php?pag
e=demo.php?letnum=B
%24&show=0
Table Problem: Painter on
Platform
A painter pulls on each rope
with a constant force F. The
painter has mass m1 and the
platform has mass m2.
a. Draw free body force
diagrams for the platform
and painter. Is there
something wrong with this
picture?
b. Find the acceleration of the
platform.
Source: why you need to know
physics
Next Reading Assignment:
W02D3
Young and Freedman (Review) 5.1-5.3
Appendix:
Newton’s Second Law
Detailed Problem Solving
Strategy
Methodology for Newton’s 2nd Law
I.
Understand – get a conceptual grasp of the
problem
Sketch the system at some time when the system is in motion.
Draw free body diagrams for each body or composite
bodies:
Each force is represented by an arrow indicating the direction
of the force
Choose an appropriate symbol for the force
II. Devise a Plan
Choose a coordinate system:
•
Identify the position function of all objects and unit vectors.
•
Include the set of unit vectors on free body force diagram.
Apply vector decomposition to each force in the free body diagram:
Fi  ( Fx )i ˆi  ( Fy )i ˆj  ( Fz )i kˆ
Apply superposition principle to find total force in each direction:
ˆi : F total   F    F  
x
x 1
x 2
ˆj : F total   F    F  
y
y
y
1
2
kˆ : Fz total   Fz 1   Fz 2 
II. Devise a Plan:
Equations of Motion
• Application of Newton’s Second Law
total
F  F1  F2    m a.
• This is a vector equality; the two sides are equal in
magnitude and direction.
ˆi :  F    F  
x 1
x 2
 m ax
ˆj :  F    F  
y
y
 m ay
kˆ :  Fz 1   Fz 2 
 m az
1
2
II. Devise a Plan (cont’d)
Analyze whether you can solve the system of
equations
•
Common problems and missing conditions.
•
Constraint conditions between the components of the
acceleration.
•
Action-reaction pairs.
•
Different bodies are not distinguished.
Design a strategy for solving the system of
equations.
III. Carry Out your Plan
Hints:
Use all your equations. Avoid thinking that
one equation alone will contain your
answer!
Solve your equations for the components of
the individual forces.
IV. Look Back
• Check your algebra
• Substitute in numbers
• Check your result
• Think about the result: Solved problems
become models for thinking about new problems.