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# Engineering Mechanics Part II: Dynamics . Lectures 7 - 9

## 1. Engineering Mechanics Part II: Dynamics Dr. Bahaa Saleh

## 2. Course Topics

• Chapter 1: Introduction to dynamics• Chapter 2: Kinematics of a Particle:

Topic # 1: Particle motion along a straight line

Topic # 2: Particle motion along a curved path

Topic # 3: Dependent motion of connected particles

Topic # 4: Relative motion of two particles

• Chapter 3: Kinetics of a Particle:

Topic # 1: Force and Acceleration

Topic # 2: Work and energy

Topic # 3: Impulse and momentum

## 3. Course Topics – Cont.

• Chapter 4: Planer Kinematics of a RigidBody.

• Chapter 5: Planar Kinetics of a Rigid

Body: Force and Acceleration.

• Chapter 6: Introduction to Mechanical

Vibration.

3

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## 4. Chapter 3: Kinetics of a Particle Topic # 1: Force and Acceleration

## 5. First Law

• Law of inertia - a body in motion will stay inmotion and a body at rest will stay at rest

unless acted upon by a net external force.

• A particle originally at rest, or moving in a

straight line with a constant velocity, will

remain in this state provided the particle is

not subjected to an unbalanced force.

• Static law

## 6. Second Law

• A particle acted upon by an unbalanced forceF experiences an acceleration that has the

same direction as the force and a magnitude

that is directly proportional to the force

F a

F ma

a2

F2

a3

F3

## 7. Third Law

• Law of Action-Reaction - For everyaction, there is an equal and opposite

reaction.

• The mutual forces of

action and reaction

between two particles

are equal, opposite,

and collinear.

F12 F21

## 8. Summary of Newton’s laws

1- Law of inertia a body in motion will stay in motion and a body atrest will stay at rest unless acted upon by a net external force.

2- Law of force-acceleration A particle acted upon by an unbalanced

force F experiences an acceleration a that has the same direction

as the force and a magnitude that is directly proportional to the

force

F ma

3- Law of action-reaction for every action, there is an equal and

opposite reaction

mg = FN

## 9. The Equation of Motion Free Body and Kinetic Diagram

When more than one force acts on a particle, the resultant force isdetermined by a vector summation of all the forces. The equation

of motion may be written as

FR F m a

• Free-Body diagram (Force Diagram)

• Kinetic diagram (acceleration Diagram)

## 10. Equations of Motion: Rectangular Coordinates

• When the net force isprojected to separate

coordinate axes the

Newton’s second law

still holds

F ma

F

F

F

x

max

y

ma y

z

maz

## 11.

Free Body Diagram Method•Draw each object separately

•Draw all the forces acting on that object

•Get x and y components of all the forces to

calculate the net force

•Apply Newton’s second law to get acceleration

•Use the acceleration in any motion analysis and

establish a Kinetic Diagram

ma

## 12. Normal & Frictional Force

Normal & Frictional Forcemg = FN

F

mg

Ff

FN

## 13.

Static Friction ( ms )• Static friction – parallel force on the

surface when there is no relative

motion between the 2 objects

• Static friction force can vary from

zero to Maximum

Static Dynamic

Ff = msFN Ff = mkFN

F f μ s FN

Friction

The coefficient of static friction

is material dependent.

Applied external force

## 14.

Kinetic Friction ( mk )• Kinetic friction – parallel force on the

surface when there is relative motion

between the 2 objects

• Kinetic friction force is constant

F f μ k FN

• is material dependent.

Friction

• The coefficient of Kinetic friction

Static

Ff = msFN

Dynamic

Ff = mkFN

Applied external force

## 15. Spring Force

• Spring forceFs ks

• k : spring stiffness (N/m)

• s : stretched or compressed

length

s

s l lo

lo

lo

ll

## 16.

Problemmg

y

ma

x

N

mk N

a=?

N mg cos 0

F

y

0;

Fx max

N 40 9.81 cos 20 0

N 369 N

mk N mg sin ma

0.25(369) 40 9.81 sin 20 40ax

ax 5.66 m / s 2

## 17.

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## 18. Chapter 3: Kinetics of a Particle Topic # 2: Work and Energy

## 19.

Work of external Force W1 2• A force does work when it moves

through a

displacement in the direction of the force.

•Work is positive when the force component is in the

same direction as its displacement, otherwise it is

negative

• Forces that are functions of displacement must be

integrated to obtain the work. Graphically, the work is

equal to the area under the force-displacement curve.

## 20.

Work of external Force W1 2Work = Force F, (N) * displacement S, (m) = Joule

F

1

2

F cos θ

S

W1 2 S * F cos

## 21.

Work of a Weight• The work of a weight is the product of the

weight magnitude (W) and the vertical height

from reference plane (y).

UW W y

• The work is positive when the weight moves

downwards (the reference plane under the body).

• The work is negative when the weight moves

upwards (the reference plane over the body).

• The work is zero when the reference plane

pass with the body).

## 22.

Work of a WeightUU1 12 2 W ( yy) )

magnitude of the particle' s weight times

its vertical displaceme nt.

magnitude of the particle' s weight times

its vertical displacement.

## 23.

Work of a Spring ForceThe work of a spring is of the form

US

1

2

ks

2

where k is the spring stiffness (N/m)

S is the stretch or compression of the spring, m

## 24.

Work of a Spring ForceA mistake in sign can be avoided when applying this equation

if one simply notes the direction of the spring force acting on

the particle and compares it with the sense of direction of

displacement of the particle if both are in the same sense,

positive work results; if they are opposite to one another, the

work is negative.

11 2 2 1 1 2 2

UU

(( ks

ks2 2 ksks

) )

1 1

1 2s

22

22

1

2

U s k s

2

## 25.

The kinetic energyThe kinetic energy (T), Joule at the initial and final

points is always positive, since it involves the speed

squared

1

2

T mV

2

Where m is the particle mass, kg

V is the particle speed in m/s

## 26.

Principle of Work and EnergyW1 2 T2 T1

whereW1 2 is the external force work

T1and T2 are the initial and final kinetic energy

## 27.

Principle of Work and EnergyW1 2 T1 U1 T2 U 2

whereW1 2 is the external force work

T1and T2 are the initial and final kinetic energy

U UW U S

UW is the weigth work

U S is the spring work

## 28.

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## 29.

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## 30. Chapter 3: Kinetics of a Particle Topic # 3: Impulse and Momentum

## 31.

Principle of Linear Impulse and MomentumF ma m dv

dt will integrate the equation of

In this section we

t2

v2

t1

v1

Fdt respect

m dv

motion

to time and thereby obtain

with

the principle of impulse and momentum.

The resulting equation will be useful for solving

problems involving force, velocity, and time.

Principle of linear impulse and momentum equation

## 32. Newton’s Second Law

dvF madv m dt

t2

v2 dt

t1

v1

Fdt m d v

t2

Fdt mv

2

m v1

t1

Impulse

change in momentum

This equation is referred to as the principle of linear impulse and

momentum

## 33. Principle of Linear Impulse and Momentum

t2Fdt mv

2

m v1

t1

For problem solving, pervious equation will be rewritten in the

form

tt22

mv

v11

Fdt m v 2

m

tt11

• which states that the initial momentum of the particle

at time t1 plus the sum of all the impulses applied to

the particle from t1 to t2 is equivalent to the final

momentum of the particle at time t2.

## 34. Principle of Linear Impulse and Momentum

tt22m

Fdt

dt

m

mvv22

mv 1 F

tt11

If each of the vectors in above Eq. is resolved into its x, y, z

components, we can write the following three scalar equations

of linear impulse and momentum.

t2

m (v x )1 Fx dt m (v x ) 2

t1

t2

m (v y )1 Fy dt m (v y ) 2

t1

t2

m (v z )1 Fz dt m (v z ) 2

t1

## 35.

The 100-kg stone shown in Fig. is originally at rest on the smooth horizontal surface. Ifa towing force of 200 N, acting at an angle of 45°, is applied to the stone for 10 s,

determine the final velocity and the normal force which the surface exerts on the stone

during this time interval.

m=100 kg

t2

( ) m( x )1 Fx dt m( x ) 2

t1

0 200 N cos 45o (10s) (100 kg) 2

2 14.1 m / s

m(

t2

) Fy dt m( y ) 2

y 1

t1

0 N c (10s) 981N (10s) 200 N (10s) sin 45o 0

N c 840 N

At rest smooth

t=10 s

V2=?

N=?

## 36.

m = 28 MgAt rest

V=?

t=4s

F = 4 - 0.01 t2

t2

m( x )1 Fx dt m( x ) 2

t1

4

0 ( 4 0.01t ) (10 ) dt 28(10 ) 2

2

0

2 0.564 m / s

3

3

## 37.

The 15-Mg boxcar A is coasting at 1 .5 mls on the horizontal trackwhen it encounters a 12-Mg tank car B coasting at 0.75 mls

toward it as shown in Fig. 15-Sa. If the cars collide and couple

together, determine (a) the speed of both cars just after the

coupling, and (b) the average force between them if the coupling

takes place in O.S s.

m A ( A )1 mB ( B )1 (m A mB ) 2

(15000)(1.5) (12000)(0.75) (27000) 2

2 0.5 m / s

mA=15 Mg

mB=12 Mg

Couple together

V2=? After coupling

Favg = ? In 0.8 s

( ) m ( p )1 Favg (t ) m ( p ) 2

(15000)(1.5) Favg (0.8) (15000)(0.5)

Favg 18.8 kN

## 38.

mA=15 MgmB=12 Mg

Couple together

V2=? After coupling

Favg = ? In 0.8 s

m A ( A )1 mB ( B )1 (m A mB ) 2

(15000)(1.5) (12000)(0.75) (27000) 2

2 0.5 m / s

( ) m ( p )1 Favg (t ) m ( p ) 2

(15000)(1.5) Favg (0.8) (15000)(0.5)

Favg 18.8 kN

## 39.

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