Download Concept of Force and Newton`s Laws of Motion Concept of

Concept of Force
Force is a vector quantity
Concept of Force and
Newton’s Laws of Motion
8.01
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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
G
Apply two forces F1 and
Forces
G
F2 on a body,
the total force is the vector sum of the two forces:
G
G G
F total = F1 + F2
Notation: The force acting on body 1 due to the
G
interaction between body 1 and body 2 is denoted by F12
G
G
Ftotal ≡ ma
Gravitation
Electric and magnetic forces
Elastic forces (Hooke’s Law)
Frictional forces: static and kinetic friction,
fluid resistance
Example: The total force exerted on
m3 by m1 and m2 is:
G
G
G
F3total = F31 + F32
Contact forces: normal forces and static
friction
Tension and compression
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Free Body Diagram
Newton’s First Law
1. Represent each force that is acting on the object by an arrow
on a free body force diagram that indicates the direction of
GT G G
the force
F = F1 + F2 + ⋅⋅⋅
2. Choose set of independent unit vectors and draw them on
free body diagram.
3.
G
Decompose each force Fi
in terms of vector components.
G
Fi = Fi , x ˆi + Fi , y ˆj + Fi , z kˆ
4. Add vector components to find vector decomposition of the
total force
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.
G
G
∑F = 0
i
G
⇒ v = constant
FxT = F1, xT + F2, xT + ⋅⋅⋅
Fy T = F1, yT + F2, yT + ⋅⋅⋅
Fz T = F1, z T + F2, z T + ⋅⋅⋅
Reference Systems
Use coordinate system as a ‘reference frame’ to
describe the position, velocity, and acceleration
of objects.
Relatively Inertial Reference
Frames
Two reference frames.
Origins need not coincide.
One moving object has different
position vectors in different frames
G G G
r1 = R + r2
Relative velocity between the two reference frames
G
G
V = dR dt
is constant since the relative acceleration is zero
G
G
G
A = dV dt = 0
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Law of Addition of Velocities
Suppose the object is moving; then, observers in
different reference frames will measure different
velocities
Velocity of the object in Frame 1:
G
G
v1 = dr1 dt
Velocity of the object in Frame 2:
G
G
v 2 = dr2 dt
Newton’s First Law
Newton’s First Law in relatively inertial reference
frames: If there is no net force impressed on an
object at rest in Frame 2, then there is also no
net force impressed on the object in Frame 1.
An object that is at rest in Frame 2 is moving at a
constant velocity in reference Frame 1.
Velocity of an object in two different reference
G G
G
frames
dr1 R r2
=
+
dt dt dt
G G
G
v1 = V + v 2
Table Problem
Suppose two cars, Car A, and Car B, are
traveling along roads that are perpendicular to
each other. One observer is at rest with respect
to the ground. A second observer is in Car A.
According to the observer
ground, Car A
G on the
ˆ
is moving with velocity v A = vA j , and Car B is
G
ˆ
moving with velocity v B = vB i . What is the
velocity of Car B according to the observer in
Car A? Express the velocity of Car B according
to the observer in Car A as a vector (specify the
components) and give the direction and
magnitude.
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,
G
G
F = m a.
When multiple forces are acting,
G
N
G
∑ 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 .
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Force Law:
Newtonian Induction
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
G
G
parts.
F1,2 = −F2,1
• 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.
Action-reaction pair of forces cannot act on same
body; they act on different bodies.
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
G
Fgrav = mgrav g
where mgrav is the gravitational mass of the body and g
is a positive constant.
g = 9.80665 m ⋅ s −2
• 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:
G
| F | = k Δl
Direction: restoring spring to equilibrium
Hooke’s law holds within some reasonable range of extension or
compression
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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.
Lecture Demo:
Spring Constant
Group Problem: Hooke’s Law
and Free Body Diagram
The spring in configuration (a) is stretched a
distance Δx . The same spring in
configuration (b) will stretch a distance
1)
2)
3)
4)
5)
2 Δx
Δx
Δx/2
0
Not sure
1)
keq = k
2)
keq = k/2
3)
keq = 2k
4)
an unknown relation; there is not
enough information.
Force Laws: Contact Forces
Between Surfaces
The contact force between two
surfaces is denoted by the vector
G
G total
Fsurface
, hand ≡ C
Normal Force: Component of the
contact force perpendicular to
surface and is denoted by
G normal
G
Fsurface
, hand ≡ N
Friction Force: Component of the
contact force tangent to the surface
and is denoted by
G
G tangent
Fsurface
, hand ≡ f
G G G
C≡ N+f
Therefore the contact force can be
modeled as a vector sum
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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
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)
1) larger than
2) identical to
3) smaller than
the two forces form an interaction pair.
the net force on the car is zero.
neither of the above.
unsure
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 = μk N
Static Friction
Varies in direction and magnitude depending on
applied forces:
0 ≤ f s ≤ f s,max = μ s N
Static friction is equal to it’s maximum value
f s,max = μs N
where µk is the coefficients of friction.
Direction of fk: opposes motion
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Friction on Motorcycle Wheel
A motorcycle has mass
m and is accelerating
forwards with
acceleration a. The
coefficient of friction of
the rear wheel with the
road is μ. The weight W
is equally divided
between front and rear
wheels. The force of
friction on the rear wheel
is:
1.
2.
3.
4.
5.
6.
7.
8.
9.
W μ forwards
Wμ/2 forwards
Wμ backwards
Wμ/2 backwards
Ma forwards
Ma backwards
Ma/2 forwards
Ma/2 backwards
None of above
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 ,
G
G
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.
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