Transcript PPT_W02D1_mac

Vectors & Concept of Force
8.01
W02D1 Fall 2006
Coordinate System
Coordinate system: used to describe the position
of a point in space and consists of
1. An origin as the
reference point
2. A set of coordinate
axes with scales and
labels
3. Choice of positive
direction for each axis
4. Choice of unit vectors
at each point in space
Cartesian Coordinate System
Vector
A vector is a quantity
that has both
direction and
magnitude. Let a
vector be denoted
by
r
the symbol A
r
The magnitude of A
r
is denoted by | A | A
Vector Addition
r
r
Let A and B rbe two
r rvectors. Define a
 A  B ,the “vector
new vector C
r
r
addition” of A and B by the geometric
construction shown in either figure
Summary: Vector Properties
Addition of Vectors
r r r r
ABBA
r r
r r
r r
( A  B)  C  A  (B  C)
1.
Commutativity
2.
Associativity
3.
Identity Element for Vector Addition
4.
r
r
r
r
Inverse Element for Vector Addition  A such that A   A  0
r
0
such that
r r r r r
A0 0 A  A
 
Scalar Multiplication of Vectors
1.
Associative Law for Scalar Multiplication
2.
Distributive Law for Vector Addition
3.
4.
r
r
r
r
b(cA)  (bc)A  (cb A)  c(bA)
r r
r
r
c( A  B)  c A  c B
r
r
r
Distributive Law for Scalar Addition
(b  c) A  b A  c A
r r
Identity Element for Scalar Multiplication: number 1 such that
1A A
Application of Vectors
(1) Vectors can exist at any point P in space.
(2) Vectors have direction and magnitude.
(3) Vector Equality: Any two vectors that have the
same direction and magnitude are equal no matter
where in space they are located.
Vector Decomposition
Choose a coordinate system
with an origin and axes. We
can decompose a vector
into component vectors
along each coordinate axis,
for example along the x,y,
and z-axes of a Cartesian
coordinate system. A vector
at P can be decomposed
into the vector sum,
r r
r
r
A  Ax  A y  Az
Unit Vectors and Components
The idea of
multiplication by real
numbers allows us to
define a set of unit
vectors at each point
ö
in space
(öi, öj, k)
with |öi | 1,|öj| 1, | kö | 1
Components:
r
A  ( Ax ,Ay ,Az )
r
r
A x  Ax öi, A y  Ay öj,
r
ö
A z  Az k
r
ö
A  Ax öi  Ay öj  Az k
Vector Decomposition in Two
Dimensions
Consider
a vector
r
A  ( Ax ,Ay ,0)
x- and y components:
Ax  Acos( ), Ay  Asin( )
A
Magnitude:
Direction:
Ax2  Ay2
Asin( )

 tan( )
Ax Acos( )
Ay
  tan 1 ( Ay / Ax )
Vector Addition
A  Acos(A ) ˆi  Asin(A ) ˆj
B  Bcos(B ) ˆi  Bsin(B ) ˆj
r r r
Vector Sum: C  A  B
Components
Cx  Ax  Bx ,
Cy  Ay  By
Cx  C cos(C )  Acos( A ) Bcos( B )
Cy  C sin(C )  Asin( A )  Bsin( B )
r
C  ( Ax  Bx ) öi  ( Ay  By ) öj  C cos( C ) öi  C sin( C ) öj
Example: Displacement Vector
At 2 am one morning a person runs 250 m along
the infinite corridor at MIT from Mass Ave to the
end of Building 8, turns right at the end of the
corridor and runs 178 m to the end of Building 2,
and then turns right and runs 30 m down the hall.
1. Construct a vector diagram that represents this
2.
motion. Indicate your choice of unit vectors.
What is the direction and magnitude of the straight
line between start and finish?
Example: Displacement Vector
Solution
Total displacement:
r
r
r
r
r  r1  r2  r3
 250 m öj  178 m öi  30 m öj
 178 m öi  220 m öj
Magnitude:
r
r  ((178 m)2  (220 m)2 )1/ 2  283 m
Direction:
r
r
  tan (( r) y / ( r) x )
1
 tan 1 (220 m / 178 m)  51.0o
Concept of Force
• Force is a vector quantity
r total
r
F  ma
• The magnitude of the total force is defined to be
F = (mass) x (magnitude of the acceleration)
• The direction of the total force on a body is the
same as the direction of the acceleration.
• The SI units for force are newtons (N):
1 N = 1kg . m/s2
Superposition
Principle
r
r
Apply two forces
F1
and
F2 on a body,
the total force is the vector sum of the two forces:
Ftotal  F1  F2
Notation: The force acting on body 1 due to the
r
interaction between body 1 and body 2 is denoted by F12
Example: The total force exerted on
m3 by m1 and m2 is:
r total r
r
F3  F31  F32
Force Law
• 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
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
r
Fgrav  mgrav g
where mgrav is the gravitational mass of the body and g
is a positive constant.
g  9.80665 m  s
2
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
Concept Question: Hooke’s Law
Two identical springs with spring
constant k are attached to each other.
A block of mass m is suspended from
the lower spring. What is the
equivalent spring constant keq of the
system of two springs? Note the block
displaces the same amount in the two
figures.
1)
keq = k
2)
keq = k/2
3)
keq = 2k
4)
an unknown relation; there is not
enough information.
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.
r
F1,2  force on 1 due to interaction between 1 and 2
r
r
F1,2  F2,1
•Action-reaction pair of forces cannot act on same
body; they act on different bodies.
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 r
r
normal
Fsurface,hand  N
•Friction Force: Component of the contact
force tangent to the surface and is denoted
by
r
r
tangent
Fsurface,hand
f
•Therefore the contact force can be
modeled as a vector sum
r r r
C Nf
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
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. the two forces form an interaction pair.
2. the net force on the car is zero.
3. neither of the above.
4. unsure
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)
Concept Question: Tension in
Rope
You are trying to pull a rock resting on the ground with a
heavy rope (the rope has non-zero mass). Just before the
rock slips and starts to move, the magnitude of the tension
in the rope is
1)
greater than the magnitude of the pulling
force?
2)
equal to the magnitude of the pulling force?
3)
less than the magnitude of the pulling force?
4)
Not enough information is given to answer.
Static Equilibrium for Forces
(1) The sum of the forces acting on a body at
rest is zero
Ftotal  F1  F2  ...  0
Measuring Forces: Statics


The science of statics
investigates how the forces
can act is such a way, and is
not concerned with the
motions produced by the
individual forces. With this in
mind, we can measure forces
using a statics procedure.
We shall choose a spring
that satisfies Hooke’s Law for
spring forces between 0.2 N
and 10 N.
Linear Calibration of Scale
 apply a standard force by suspending
a body of mass 0.102 kg
 gravitation force acting on body is mg
= 1.0 N
 spring stretches and exerts a force
pulling the body up
r
r
Fspring  Fgrav
 mark off 1 N on the scale at the
position of the reference point on
spring
 A second body of twice the mass is
now suspended from the spring and
we label the position of the reference
point by 2.0 N . Continue scale
linearly.
Force Measurements
 Attach an unknown body to the spring and measure the
spring force.
 Use Newton’s Third Law, deduce the force that the spring
exerts on the body.
 We are measuring the force with which the spring pulls
the body, which is equal by Newton’ s Third Law to the
force with which the body pulls the spring.
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 pointlike object
• Identify all forces that act on that object
Free Body Diagram
For each object in static equilibrium the sum of individual forces acting on object is zero
rT r r
r
F  F1  F2   0
1.
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
2.
Choose set of independent unit vectors and draw them on free body diagram.
3.
Decompose each force in terms of vector components.
4.
Use free body diagram to add vector components to find vector decomposition of the
total force which is equal to zero (static equilibrium)
r
T ˆ
T ˆ
ˆ
F  Fx i  Fy j  Fz k  0
T
5.
T
Set each component of the total force equal to zero, for example the x-component of the
total force is zero
Fx T  F1, x T  F2, x T    
Methodology for Newton’s 2nd Law:
Statics
I.
Understand – get a conceptual grasp of the problem

Think about the problem.

Sketch the system at some time when the system is in
motion.

Choose a coordinate system and identify the position
function of all objects.

Quantify the constraint conditions.

Draw free body diagrams for each body in the problem.
II. Devise a Plan
Draw free body diagrams for each body:
•
Include the set of unit vectors
•
Each force represented by an arrow indicating
the direction of the force
•
Choose an appropriate symbol for the force
Choosing directions for the forces
•
If you solve for a force and find that the force is
negative then the force points in the opposite
direction as your choice of direction for the
force.
II. Devise a Plan:
Static Equilibrium of Forces
• Application of Static Equilibrium of Forces
r total r r
r
F  F1  F2    0
• This is a vector equality so the two sides are equal
in magnitude and direction
     ...  0
öi : F  F
x 1
x
2
öj : F  F
y
y
2
kö : Fz 1  Fz
2
     ...  0
1
     ...  0
II. Devise a Plan - (con’t)
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
ˆk : F total   F    F   ...
z
z 1
z 2
II. Devise a Plan (con’t)
Analyze whether you can solve the system of
equations
•
Common problems and missing conditions.
•
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.
Group Problem: Hooke’s Law
and Free Body Diagram
The spring in configuration (a) is stretched a
distance x . The same spring in
configuration (b) will stretch a distance
1.2 x
2.x
3.x/2
4.0
5.Not sure