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