Transcript tire 130 80 17

NEWTON’S LAWS OF MOTION (EQUATION OF MOTION)
(Sections 13.1-13.3)
Today’s Objectives:
Students will be able to:
a) Write the equation of motion
In-Class Activities:
for an accelerating body.
• Check homework, if any
b) Draw the free-body and
• Reading quiz
kinetic diagrams for an
• Applications
accelerating body.
• Newton’s laws of motion
• Newton’s law of
gravitational attraction
• Equation of motion for a
particle or system of particles
• Concept quiz
• Group problem solving
• Attention quiz
APPLICATIONS
The motion of an object depends on the
forces acting on it.
A parachutist relies on the atmospheric
drag resistance force to limit his velocity.
Knowing the drag force, how can we
determine the acceleration or velocity of
the parachutist at any point in time?
APPLICATIONS (continued)
A freight elevator is lifted using a
motor attached to a cable and pulley
system as shown.
How can we determine the tension
force in the cable required to lift the
elevator at a given acceleration?
Is the tension force in the cable greater than the weight
of the elevator and its load?
NEWTON’S LAWS OF MOTION
The motion of a particle is governed by Newton’s three laws of
motion.
First Law: A particle originally at rest, or moving in a straight
line at constant velocity, will remain in this state if the resultant
force acting on the particle is zero.
Second Law: If the resultant force on the particle is not zero, the
particle experiences an acceleration in the same direction as the
resultant force. This acceleration has a magnitude proportional to
the resultant force.
Third Law: Mutual forces of action and reaction between two
particles are equal, opposite, and collinear.
NEWTON’S LAWS OF MOTION (continued)
The first and third laws were used in developing the
concepts of statics. Newton’s second law forms the
basis of the study of dynamics.
Mathematically, Newton’s second law of motion can be
written
F = ma
where F is the resultant unbalanced force acting on the
particle, and a is the acceleration of the particle. The
positive scalar m is called the mass of the particle.
Newton’s second law cannot be used when the particle’s
speed approaches the speed of light, or if the size of the
particle is extremely small (~ size of an atom).
NEWTON’S LAW OF GRAVITATIONAL ATTRACTION
Any two particles or bodies have a mutually attractive
gravitational force acting between them. Newton postulated
the law governing this gravitational force as
F = G(m1m2/r2)
where
F = force of attraction between the two bodies,
G = universal constant of gravitation ,
m1, m2 = mass of each body, and
r = distance between centers of the two bodies.
When near the surface of the earth, the only gravitational
force having any sizable magnitude is that between the earth
and the body. This force is called the weight of the body.
MASS AND WEIGHT
It is important to understand the difference between the
mass and weight of a body!
Mass is an absolute property of a body. It is independent of
the gravitational field in which it is measured. The mass
provides a measure of the resistance of a body to a change
in velocity, as defined by Newton’s second law of motion
(m = F/a).
The weight of a body is not absolute, since it depends on the
gravitational field in which it is measured. Weight is defined
as
W = mg
where g is the acceleration due to gravity.
UNITS: SI SYSTEM VS. FPS SYSTEM
SI system: In the SI system of units, mass is a base unit and
weight is a derived unit. Typically, mass is specified in
kilograms (kg), and weight is calculated from W = mg. If the
gravitational acceleration (g) is specified in units of m/s2, then
the weight is expressed in newtons (N). On the earth’s
surface, g can be taken as g = 9.81 m/s2.
W (N) = m (kg) g (m/s2) => N = kg·m/s2
FPS System: In the FPS system of units, weight is a base unit
and mass is a derived unit. Weight is typically specified in
pounds (lb), and mass is calculated from m = W/g. If g is
specified in units of ft/s2, then the mass is expressed in slugs.
On the earth’s surface, g is approximately 32.2 ft/s2.
m (slugs) = W (lb)/g (ft/s2) => slug = lb·s2/ft
EQUATION OF MOTION
The motion of a particle is governed by Newton’s second law, relating
the unbalanced forces on a particle to its acceleration. If more than one
force acts on the particle, the equation of motion can be written
F = FR = ma
where FR is the resultant force, which is a vector summation of all the
forces.
To illustrate the equation, consider a
particle acted on by two forces.
First, draw the particle’s freebody diagram, showing all
forces acting on the particle.
Next, draw the kinetic diagram,
showing the inertial force ma
acting in the same direction as
the resultant force FR.
INERTIAL FRAME OF REFERENCE
This equation of motion is only valid if the acceleration is
measured in a Newtonian or inertial frame of reference.
What does this mean?
For problems concerned with motions at or near the
earth’s surface, we typically assume our “inertial frame”
to be fixed to the earth. We neglect any acceleration
effects from the earth’s rotation.
For problems involving satellites or rockets, the
inertial frame of reference is often fixed to the stars.
SYSTEM OF PARTICLES
The equation of motion can be extended to include systems of
particles. This includes the motion of solids, liquids, or gas systems.
As in statics, there are internal forces and
external forces acting on the system.
What is the difference between them?
Using the definitions of m = mi as the
total mass of all particles and aG as the
acceleration of the center of mass G of
the particles, then maG = miai .
The text shows the details, but for a system of particles: F = maG
where F is the sum of the external forces acting on the entire
system.
KEY POINTS
1) Newton’s second law is a “Law of Nature”--experimentally
proven and not the result of an analytical proof.
2) Mass (property of an object) is a measure of the resistance
to a change in velocity of the object.
3) Weight (a force) depends on the local gravitational field.
Calculating the weight of an object is an application of
F = ma, i.e., W = m g.
4) Unbalanced forces cause the acceleration of objects. This
condition is fundamental to all dynamics problems!
PROCEDURE FOR THE APPLICATION OF THE
EQUATION OF MOTION
1) Select a convenient inertial coordinate system. Rectangular,
normal/tangential, or cylindrical coordinates may be used.
2) Draw a free-body diagram showing all external forces
applied to the particle. Resolve forces into their
appropriate components.
3) Draw the kinetic diagram, showing the particle’s inertial
force, ma. Resolve this vector into its appropriate
components.
4) Apply the equations of motion in their scalar component
form and solve these equations for the unknowns.
5) It may be necessary to apply the proper kinematic relations
to generate additional equations.
EXAMPLE
Given: A crate of mass m is pulled by a cable attached to a truck.
The coefficient of kinetic friction between the crate and
road is mk.
Find: Draw the free-body and kinetic diagrams of the crate.
Plan: 1) Define an inertial coordinate system.
2) Draw the crate’s free-body diagram, showing all
external forces applied to the crate in the proper
directions.
3) Draw the crate’s kinetic diagram, showing the inertial
force vector ma in the proper direction.
EXAMPLE (continued)
Solution:
1) An inertial x-y frame can be defined as fixed to the ground.
2) Draw the free-body diagram of the crate:
y
W = mg
The weight force (W) acts through the
crate’s center of mass. T is the tension
x
force in the cable. The normal force (N)
is perpendicular to the surface. The
friction force (F = uKN) acts in a direction
F = uKN
opposite to the motion of the crate.
N
3) Draw the kinetic diagram of the crate:
T
30°
ma
The crate will be pulled to the right. The
acceleration vector can be directed to the
right if the truck is speeding up or to the
left if it is slowing down.
GROUP PROBLEM SOLVING
Given: Each block has a mass m. The
coefficient of kinetic friction at all
surfaces of contact is m. A
horizontal force P is applied to the
bottom block.
Find: Draw the free-body and kinetic diagrams of each block.
Plan: 1)
2)
3)
GROUP PROBLEM SOLVING (continued)
Solution:
1) An inertial x-y frame can be defined as
2) Draw the _______________ of each block:
Block B:
y
x
.
Block A:
y
x
3) Draw the kinetic diagram of each block:
Block B:
Block A:
EQUATIONS OF MOTION: RECTANGULAR
COORDINATES (Section 13.4)
Today’s Objectives:
Students will be able to apply
Newton’s second law to
determine forces and
accelerations for particles in
rectilinear motion.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• Equations of motion using
rectangular (Cartesian)
Coordinates
• Concept quiz
• Group problem solving
• Attention Quiz
APPLICATIONS
If a man is pushing a 100 lb crate, how large a force F must
he exert to start moving the crate?
What would you have to know before you could calculate
the answer?
APPLICATIONS (continued)
Objects that move in any fluid have a drag force acting on
them. This drag force is a function of velocity.
If the ship has an initial velocity vo and the magnitude of the
opposing drag force at any instant is half the velocity, how
long it would take for the ship to come to a stop if its engines
stop?
EQUATION OF MOTION
The equation of motion, F = m a, is best used when the problem
requires finding forces (especially forces perpendicular to the
path), accelerations, velocities or mass. Remember, unbalanced
forces cause acceleration!
Three scalar equations can be written from this vector equation.
The equation of motion, being a vector equation, may be
expressed in terms of its three components in the Cartesian
(rectangular) coordinate system as
F = ma or Fx i + Fy j + Fz k = m(ax i + ay j + az k)
or, as scalar equations, Fx = max , Fy = may , and Fz = maz .
PROCEDURE FOR ANALYSIS
• Free Body Diagram
Establish your coordinate system and draw the particle’s
free body diagram showing only external forces. These
external forces usually include the weight, normal forces,
friction forces, and applied forces. Show the ‘ma’ vector
(sometimes called the inertial force) on a separate diagram.
Make sure any friction forces act opposite to the direction
of motion! If the particle is connected to an elastic spring,
a spring force equal to ks should be included on the
FBD.
PROCEDURE FOR ANALYSIS (continued)
• Equations of Motion
If the forces can be resolved directly from the free-body
diagram (often the case in 2-D problems), use the scalar
form of the equation of motion. In more complex cases
(usually 3-D), a Cartesian vector is written for every force
and a vector analysis is often best.
A Cartesian vector formulation of the second law is
F = ma or
Fx i + Fy j + Fz k = m(ax i + ay j + az k)
Three scalar equations can be written from this vector equation.
You may only need two equations if the motion is in 2-D.
PROCEDURE FOR ANALYSIS (continued)
• Kinematics
The second law only provides solutions for forces and
accelerations. If velocity or position have to be found,
kinematics equations are used once the acceleration is
found from the equation of motion.
Any of the tools learned in Chapter 12 may be needed to
solve a problem. Make sure you use consistent positive
coordinate directions as used in the equation of motion
part of the problem!
EXAMPLE
Given: WA = 10 lb
WB = 20 lb
voA = 2 ft/s
mk = 0.2
Find: vA when A has moved 4 feet.
Plan: Since both forces and velocity are involved, this
problem requires both the equation of motion and kinematics.
First, draw free body diagrams of A and B. Apply the
equation of motion . Using dependent motion equations,
derive a relationship between aA and aB and use with the
equation of motion formulas.
EXAMPLE (continued)
Solution:
2T
Free-body and kinetic
diagrams of B:
=
WB
mBaB
+   Fy = ma y
Apply the equation of
motion to B:
WB - 2 T = mB aB
20 a
20 - 2 T = 32
.2 B
(1)
EXAMPLE (continued)
Free-body and kinetic diagrams of A:
y
WA
x
T
N
=
mAaA
F = mkN
Apply the equations of motion to A:
+  F = ma

+  Fy = ma y = 0
x
x
N = WA = 10 lb
F = m N = 2 lb
k
F - T = mA aA
10
2 - T = 32 2 aA
.
(2)
EXAMPLE (continued)
Now consider the kinematics.
sA
Datums
A
sB
B
Constraint equation:
sA + 2 sB = constant
or
vA + 2 vB = 0
Therefore
aA + 2 aB = 0
aA = -2 aB
(3)
(Notice aA is considered
positive to the left and aB is
positive downward.)
EXAMPLE (continued)
Now combine equations (1), (2), and (3).
T = 22 = 7. 33 lb
3
ft 2 
ft
2
aA = -17.16 s = 17.16 s
Now use the kinematic equation:
vA
2
2
= voA + 2 aA ( sA - soA )
vA 2 = 2 2 + 2 (17.16 )(4 )
ft
vA = 11 . 9 s 
GROUP PROBLEM SOLVING
Given: The 400 kg mine car is
hoisted up the incline. The
force in the cable is
F = (3200t2) N. The car has
an initial velocity of
vi = 2 m/s at t = 0.
Find:
Plan:
The velocity when t = 2 s.
GROUP PROBLEM SOLVING (continued)
Solution:
1) Draw the free-body and kinetic diagrams of the mine car:
GROUP PROBLEM SOLVING (continued)
2) Apply the equation of motion in the x-direction:
3) Use kinematics to determine the velocity:
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL
COORDINATES (Section 13.5)
Today’s Objectives:
Students will be able to apply
the equation of motion using
normal and tangential
coordinates.
In-Class Activities:
• Check homework, if any
• Reading quiz
• Applications
• Equation of motion in n-t
coordinates
• Concept quiz
• Group problem solving
• Attention quiz
APPLICATIONS
Race tracks are often banked in the
turns to reduce the frictional forces
required to keep the cars from sliding
at high speeds.
If the car’s maximum velocity and a
minimum coefficient of friction
between the tires and track are
specified, how can we determine the
minimum banking angle (q) required
to prevent the car from sliding?
APPLICATIONS (continued)
Satellites are held in orbit around
the earth by using the earth’s
gravitational pull as the centripetal
force – the force acting to change
the direction of the satellite’s
velocity.
Knowing the radius of orbit of
the satellite, how can we
determine the required speed of
the satellite to maintain this orbit?
NORMAL & TANGENTIAL COORDINATES
When a particle moves along a
curved path, it may be more
convenient to write the equation of
motion in terms of normal and
tangential coordinates.
The normal direction (n) always points toward the path’s
center of curvature. In a circle, the center of curvature is the
center of the circle.
The tangential direction (t) is tangent to the path, usually set
as positive in the direction of motion of the particle.
EQUATIONS OF MOTION
Since the equation of motion is a
vector equation , F = ma,
it may be written in terms of the
n & t coordinates as
Ftut + Fnun = mat + man
Here Ft & Fn are the sums of the force components acting in
the t & n directions, respectively.
This vector equation will be satisfied provided the individual
components on each side of the equation are equal, resulting in
the two scalar equations: Ft = mat and
Fn = man .
Since there is no motion in the binormal (b) direction, we can also
write Fb = 0.
NORMAL AND TANGENTIAL ACCERLERATIONS
The tangential acceleration, at = dv/dt, represents the time rate of
change in the magnitude of the velocity. Depending on the direction
of Ft, the particle’s speed will either be increasing or decreasing.
The normal acceleration, an = v2/r, represents the time rate of change
in the direction of the velocity vector. Remember, an always acts
toward the path’s center of curvature. Thus, Fn will always be
directed toward the center of the path.
Recall, if the path of motion is defined
as y = f(x), the radius of curvature at
any point can be obtained from
r =
dy 2 3/2
[1 + ( ) ]
dx
d2y
dx2
SOLVING PROBLEMS WITH n-t COORDINATES
• Use n-t coordinates when a particle is moving along a
known, curved path.
• Establish the n-t coordinate system on the particle.
• Draw free-body and kinetic diagrams of the particle. The
normal acceleration (an) always acts “inward” (the positive ndirection). The tangential acceleration (at) may act in either the
positive or negative t direction.
• Apply the equations of motion in scalar form and solve.
• It may be necessary to employ the kinematic relations:
at = dv/dt = v dv/ds
an = v2/r
EXAMPLE
Given: At the instant q = 60°, the boy’s
center of mass G is momentarily
at rest. The boy has a weight of
60 lb. Neglect his size and the
mass of the seat and cords.
Find: The boy’s speed and the tension
in each of the two supporting
cords of the swing when q = 90°.
Plan:
1) Since the problem involves a curved path and finding the
force perpendicular to the path, use n-t coordinates. Draw
the boy’s free-body and kinetic diagrams.
2) Apply the equation of motion in the n-t directions.
3) Use kinematics to relate the boy’s acceleration to his speed.
EXAMPLE (continued)
Solution:
1) The n-t coordinate system can be established on the
boy at some arbitrary angle q. Approximating the boy
and seat together as a particle, the free-body and
kinetic diagrams can be drawn.
Free-body diagram
Kinetic diagram
n
n
2T
q
=
man
mat
W
t
t
T = tension in each cord
W = weight of the boy
EXAMPLE (continued)
2) Apply the equations of motion in the n-t directions.
(a) Fn = man => 2T – W sin q = man
Using an = v2/r = v2/10, w = 60 lb, and m = w/g = (60/32.2),
we get: 2T – 60 sin q = (60/32.2)(v2/10) (1)
(b) Ft = mat => W cos q = mat
=> 60 cos q = (60/32.2) at
Solving for at: at = 32.2 cos q (2)
Note that there are 2 equations and 3 unknowns (T, v, at).
One more equation is needed.
EXAMPLE (continued)
3) Apply kinematics to relate at and v.
v dv = at ds where ds = r dq = 10 dq
=> v dv = 32.2 cosq ds = 32.2 cosq (10 dq )
v
=>  v dv =
0
=>
90
 32.2 cosq dq
60
v2
2
90
= 32.2 sinq
=> v = 9.29 ft/s
60
This v is the speed of the boy at q = 90. This value can be
substituted into equation (1) to solve for T.
2T – 60 sin(90°) = (60/32.2)(9.29)2/10
T = 38.0 lb (the tension in each cord)
GROUP PROBLEM SOLVING
Given: A 200 kg snowmobile with
rider is traveling down the
hill. When it is at point A, it
is traveling at 4 m/s and
increasing its speed at 2 m/s2.
Find: The resultant normal force and resultant frictional force
exerted on the tracks at point A.
Plan: 1)
2)
3)
GROUP PROBLEM SOLVING (continued)
Solution:
1) The n-t coordinate system is established on the snowmobile
at point A.
GROUP PROBLEM SOLVING (continued)
2) Apply the equations of motion in the n-t directions:
GROUP PROBLEM SOLVING (continued)
3) Determine r
Determine q
EQUATIONS OF MOTION:
CYLINDRICAL COORDINATES (Section 13.6)
Today’s Objectives:
In-Class Activities:
Students will be able to
• Check homework, if any
analyze the kinetics of a
• Reading quiz
particle using cylindrical
coordinates.
• Applications
• Equations of motion using
cylindrical coordinates
• Angle between radial and
tangential directions
• Concept quiz
• Group problem solving
• Attention quiz
APPLICATIONS
The forces acting on the 100-lb boy can be analyzed using the
cylindrical coordinate system.
If the boy slides down at a constant speed of 2 m/s, can we
find the frictional force acting on him?
APPLICATIONS (continued)
When an airplane executes the vertical loop shown above, the
centrifugal force causes the normal force (apparent weight)
on the pilot to be smaller than her actual weight.
If the pilot experiences weightlessness at A, what is the
airplane’s velocity at A?
EQUATIONS OF MOTION:
CYLINDRICAL COORDINATES
This approach to solving problems has
some external similarity to the normal &
tangential method just studied. However,
the path may be more complex or the
problem may have other attributes that
make it desirable to use cylindrical
coordinates.
Equilibrium equations or “Equations of Motion” in cylindrical
coordinates (using r, q , and z coordinates) may be expressed in
scalar form as:
.. . 2
 Fr = mar = m(r – rq )
..
..
 Fq = maq = m(rq – 2rq)
..
 Fz = maz = mz
EQUATIONS OF MOTION (continued)
If the particle is constrained to move only in the r – q
plane (i.e., the z coordinate is constant), then only the first
two equations are used (as shown below). The coordinate
system in such a case becomes a polar coordinate system.
In this case, the path is only a function of q.
.. . 2
 Fr = mar = m(r – rq )
..
..
 Fq = maq = m(rq – 2rq)
Note that a fixed coordinate system is used, not a “bodycentered” system as used in the n – t approach.
TANGENTIAL AND NORMAL FORCES
If a force P causes the particle to move along a path defined
by r = f (q ), the normal force N exerted by the path on the
particle is always perpendicular to the path’s tangent. The
frictional force F always acts along the tangent in the opposite
direction of motion. The directions of N and F can be
specified relative to the radial coordinate by using angle y .
DETERMINATION OF ANGLE y
The angle y, defined as the angle
between the extended radial line
and the tangent to the curve, can be
required to solve some problems. It
can be determined from the
following relationship.
tan y =
r dq
dr
=
r
dr dq
If y is positive, it is measured counterclockwise from the radial
line to the tangent. If it is negative, it is measured clockwise.
EXAMPLE
Given: The ball (P) is guided along
the vertical circular
path.
.
..W = 0.5 lb, q2 = 0.4 rad/s,
q = 0.8 rad/s , rc = 0.4 ft
Find:
Force of the arm OA on the
ball when q = 30.
Plan: Draw a FBD. Then develop the kinematic equations
and finally solve the kinetics problem using
cylindrical coordinates.
Solution: Notice that r = rc cosq, therefore:
.
.
r = -2rc sinq q
.2
..
..
r = -2rc cosq q – 2rc sinq q
EXAMPLE (continued)
Free Body Diagram: Establish the r , q inertial coordinate
system and draw the particle’s free body diagram. Notice that
the radial acceleration is negative.
q
mg
q r
t
y
q
maq
q
=
Ns
q
NOA q
n
q = 30°
mar
EXAMPLE (continued)
Kinematics: at q = 30
r = 2(0.4) cos(30) = 0.693 ft
.
r = -2(0.4) sin(30)(0.4) = -0.16 ft/s
..
r = -2(0.4) cos(30)(0.4)2 – 2(0.4) sin(30)(0.8) = -0.431 ft/s2
Acceleration components are
.. . 2
ar = r – rq = -0.431 – (0.693)(0.4)2 = -0.542 ft/s2
..
..
aq = rq + 2rq = (0.693)(0.8) + 2(-0.16)(0.4) = 0.426 ft/s2
tan y = r/(dr/dq) where dr/dq = -2rc sinq
tan y = (2rc cosq)/(-2rc sinq) = -1/tanq
 y = 120
EXAMPLE (continued)
Kinetics:
 Fr = mar
0.5
Ns cos(30) – 0.5 sin(30) =
(-0.542)
32.2
Ns = 0.279 lb
 Fq = maq
NOA + 0.279 sin(30) – 0.5 cos(30) =
NOA = 0.3 lb
0.5
(0.426)
32.2
GROUP PROBLEM SOLVING
Given: A plane flies in a vertical loop
as shown.
vA = 80 ft/s (constant)
W = 130 lb
Find: Normal force on the pilot at A.
Plan:
Solution:
Kinematics:
GROUP PROBLEM SOLVING (continued)
Therefore v A =
Since r = 600 ft
· 80
= 0.133 rad
at A, q =
s
600
Since vA is constant, aq =
= - 42.67 ft
··
r=
·2
··
=
ar r rq =
= - 53.33 ft
s2
s2
GROUP PROBLEM SOLVING (continued)
Free Body Diagram & Kinetic Diagram
Kinetics:  Fr = mar