Upload
others
View
9
Download
0
Embed Size (px)
Citation preview
INTRODUCTION
According to Newton’s law, a particle will acceleratewhen it is subjected to unbalanced force. Kinetics is thestudy of the relations between unbalanced forces andresulting changes in motion.
The three general approaches to the solution of kineticsproblems are:problems are:
a) Direct application of Newton’s law (called the force-
mass-acceleration method)
b) Work and energy principles
c) Impulse and momentum methods
The basic relation between force and acceleration is found inNewton’s second law, the verification of which is entirely experimental.
Newton’s second law can be stated as follows:
If the resultant force acting on a particle is not zero, the particle
will have an acceleration proportional to the magnitude of the
resultant and in the direction of this resultant force.
We subject a particle to the action of a single force F1 and we
measure the acceleration a1 of the particle. The ratio F1/a1 of the
magnitudes of the force and the acceleration will be some number C1.
We then repeat the experiment by subjecting the same particle to a
different force F2 and measuring the corresponding acceleration a2.
The ratio F2/a2 of the magnitudes will again produce a number C2. The
experiment is repeated as many times as desired.experiment is repeated as many times as desired.
We draw two important conclusions from the results of these
experiments. First, the ratios of applied force to corresponding
acceleration all equal the same number. Thus,
Ca
F
a
F
a
F
n
n ==== ...2
2
1
1 , a constant
We conclude that the constant C is a measure of some invariable
property of the particle. This property is the inertia of the particle,
which is its resistance to rate of change of velocity. For a particle
of high inertia (large C), the acceleration will be small for a given
force F. On the other hand, if the inertia is small, the acceleration
will be large. The mass m is used as a quantitative measure of
inertia, and therefore, we may write the expression
kma
FC ==
inertia, and therefore, we may write the expression
where k is a constant introduced to account for the units used.
Thus, we may express the relation obtained from the experiments
as
kmaF =
where F is the magnitude of the resultant force acting on the
particle of mass m, and a is the magnitude of the resulting
acceleration of the particle.
The second conclusion is that the acceleration is always in the
direction of the applied force.
(Equation of Motion)
In SI unit system, k=1.
akmFrr
=
Primary Inertial System
(Birincil (Temel) Eylemsizlik Sistemi)
Although the results of ideal experiment are obtained
for measurements made relative to the “fixed” primary
inertial system, they are equally valid for measurements
made with respect to any nonrotating reference systemmade with respect to any nonrotating reference system
which translates with a constant velocity with respect to
the primary system. Newton’s second law holds equally
well in a nonaccelerating system, so that we may define
an inertial system as any system in which equation of
motion is valid.
If the ideal experiment described were performed on
the surface of the earth and all measurements were
made relative to a reference system attached to the
earth, the measured results would show a slight
discrepancy from those predicted by the equation of
motion, because the measured acceleration would not bemotion, because the measured acceleration would not be
the correct absolute acceleration. These discrepancy
would dissappear when we introduced the corrections
due to the acceleration components of the earth.
These corrections are negligible for most engineering
problems which involve the motions of structures and
machines on the surface of the earth.
An increasing number of problem occur, particularly inAn increasing number of problem occur, particularly in
the fields of rocket and spacecraft design, where the
acceleration components of the earth are of primary
concern.
The concept of time, considered an absolute quantity
in Newtonian theory, received a basically different
interpretation in the theory of relativity announced
by Einstein. Although the difference between the
mechanics of Newton and Einstein is basic, there is amechanics of Newton and Einstein is basic, there is a
practical difference in the results given by the two
theories only when velocities of the order of the
speed of light (300x106 m/s) are encountered.
Solution of Problems
1) The acceleration is either specified or can be determined directly
from known kinematic conditions. We then determine the
corresponding forces which act on the particle by direct
substitution into the equation of motion.
amFrr
=∑
We encounter two types of problems.
2) The forces acting on the particle are specified and we must
determine the resulting motion. If the forces are constant, the
acceleration is also constant and is easily found from the equation
of motion. When the forces are functions of time, position or
velocity, the equation of motion becomes a differential equation
which must be integrated to determine the velocity and
displacement.
∑
Constrained and Unconstrained Motion(Serbest ve Kısıtlanmış Hareket) (Degree of Freedom-Serbestlik Derecesi)
There are two physically distict types of motion.
The first type is unconstrained motion
where the particle is free of mechanical
guides and follows a path determined by
initial motion and by the forces which
are applied to it from external sources.
An airplane or rocket in flight and an
electron moving in a charged field are
examples of unconstrained motion.
The second type is constrained motion where the path of
the particle is partially or totally determined by
restraining guides. A marble is partially constrained to
move in the horizontal plane. A train moving along its
track and a collar sliding along a fixed shaft are examples
of more fully constrained motion.of more fully constrained motion.
The choice of an appropriate coordinate system is frequently
indicated by the number and geometry of the constraints.
Thus, if a particle is free to move in space, the particle is said
to have three degrees of freedom since three independent
coordinates are required to specify its position at any instant.
The marble sliding on the surface
has two degrees of freedom.
Collar sliding along a fixed shaft
has only one degree of freedom.
Free-Body Diagram (FBD)
Serbest Cisim Diyagramı (SCD)
When applying any of the force-mass-acceleration equations of
motion, we must account correctly for all forces acting on the
particle.
The best way to do this is to draw the particle’s free body diagram The best way to do this is to draw the particle’s free body diagram
(FBD).
In statics the resultant equals zero
whereas in dynamics it is equated to the
product of mass and acceleration .
0=∑Fr
amFrr
=∑product of mass and acceleration .amFrr
=∑
If we choose the x-direction, for example, as thedirection of the rectilinear motion of a particle, theacceleration in the y- and z-direction will be zero .
=Σ xx maF
0
0
=Σ
=Σ
z
y
F
F
xx maF =∑
xva xx&&& ==
yy maF =∑
yva yy&&& ==
( ) ( )22∑∑∑ += yx FFF
22yx aaa +=
1) 2)
tt maF =∑ nn maF =∑1) 2)
sva t &&& == ( )ρ
=ρ
=22
n
sva
&
( ) ( )22∑∑∑ += nt FFF
22nt aaa +=
rr maF =∑ θθ =∑ maF1) 2)
2
r rra θ−= &&& θ+θ=θ&&&& r2ra
( ) ( )22∑∑∑ θ+= FFF r
22 22θ+= aaa r
xx maF =∑
xva xx&&& ==
yy maF =∑
yva yy&&& ==
1) 2) zz maF =∑zva zz&&& ==
3)
( ) ( ) ( )222∑∑∑∑ ++= zyx FFFF
222zyx aaaa ++=
rr maF =∑ θθ =∑ maF1) 2)
2
r rra θ−= &&& θ+θ=θ&&&& r2ra
( ) ( ) ( )2z
22
r FFFF ∑∑∑∑ ++= θ
2
z
22
r aaaa ++= θ
zz maF =∑
zva zz&&& ==
3)
RR maF =∑ θθ maF =∑ φφ maF =∑222cos φθφ &&&& RRRaR −−=
θφφθφθφθ&&&&&& sin2cos2cos RRRa −+=
2sincos2 θφφφφφ&&&&& RRRa ++=
( ) ( ) ( )222
∑∑∑∑
1) 2) 3)
( ) ( ) ( )222
∑∑∑∑ ++= φθ FFFF R
222φθ aaaa R ++=
1. The block shown is observed to have a velocity v1=20 m/s as it passes
point A and a velocity v2=10 m/s as it passes point B on the incline.
Calculate the coefficient of kinetic friction µk between the block and
incline if x=75 m and θ=15o.
RectilinearRectilinear MotionMotion
2. A force P is applied to the initially stationary cart. Determine the
velocity and displacement at time t=5 s for each of the force histories P1
and P2. Neglect friction.
RectilinearRectilinear MotionMotion
3. The sliders A and B are connected by a light rigid bar and move with
negligible friction in the slots, both of which lie in a horizontal plane. For
the position shown, the velocity of A is 0.4 m/s to the right. Determine
the acceleration of each slider and the force in the bar at this instant.
RectilinearRectilinear MotionMotion
4. A 0.8-kg slider is propelled upward at A along the fixed curved bar which
lies in a vertical plane. If the slider is observed to have a speed of 4 m/s as it
passes position B, determine, (a) the magnitude N of the force exerted by
the fixed rod on the slider and (b) the rate at which the speed of the slider
is increasing. Assume that friction is negligible.
CurvilinearCurvilinear MotionMotion
5. The collar has a mass of 5 kg and is confined to move along the smooth
circular rod which lies in the horizontal plane. The attached spring has an
unstretched length of 200 mm. If, at the instant β=30o, the collar has a
speed v=2 m/s, determine the magnitude of normal force of the rod on the
collar and the collar’s acceleration.
CurvilinearCurvilinear MotionMotion
CurvilinearCurvilinear MotionMotion
6. The 1 kg collar slides along the smooth parabolic rod in the vertical plane
towards point O. The spring whose stiffness is k=600 N/m has an
unstretched length of 1 m. If, at the position shown in the figure, velocity
of the collar is 3.5 m/s, determine the force acting on the collar by the
parabolic rod for this instant. Neglect the friction.
2x9
32y =
3/4 m
1 m
k
B
O
m
x
y
2x9
32y =
0.375 m
CurvilinearCurvilinear MotionMotion
7. The slotted arm revolves in the horizontal plane about the fixed vertical
axis through point O. The 2 kg slider C is drawn toward O at the constant
rate of 50 mm/s by pulling the cord S. At the instant for which r=225 mm,
the arm has a counterclockwise angular velocity ω=6 rad/s and is slowing down
at the rate of 2 rad/s2. For this instant, determine the tension T in the cord
and the magnitude N of the force exerted on the slider by the sides of the
smooth radial slot. Indicate which side, A or B, of the slot contacts the slider.smooth radial slot. Indicate which side, A or B, of the slot contacts the slider.
CurvilinearCurvilinear MotionMotion
8. The slotted arm OB rotates in a horizontal plane about point O of the
fixed circular cam with constant angular velocity 15 rad/s. The spring has a
stiffness of 5 kN/m and is uncompressed when θ=0. the smooth roller A has a
mass of 0.5 kg. determine the normal force N which cam exerts on A and also
the force R exerted on A by the sides of the slot when θ=45o. All surfaces are
smooth. Neglect the small diameter of the roller.
CurvilinearCurvilinear MotionMotion
9. Pin B has a weight of 1.2 N and is moving both in slotted arm OC and
circular slot DE. Determine the radial and transverse forces acting on B.
Take 15 rad/s, 250 rad/s2, θ=20o. Neglect the friction.=θ& =θ&&
D
r
CB
b
E
θO