cve 253 course outline

8/18/2019 CVE 253 course outline

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APPLIED MECHANICS I

1.0 Introduction to Mechanics

Mechanics is a science that describes and predicts the conditions of rest or motion of

bodies under the action of forces.

It is divided into three parts1. Mechanics of rigid bodies

2. Mechanics of Deformable bodies

3. Mechanics of Fluids

Mechanics of Rigid bodies is further divided into statics and dynamics. In this part of the

course, bodies are assumed to be perfectly rigid, although all bodies deform under the application

of load. But these deformations are usually small and do not adversely affect the conditions of

equilibrium or motion of the structure.

Mechanics is the foundation of most Engineering Science. The purpose is to explain and predict

physical phenomena and thus lay the foundations for Engineering applications

1.1 ForceForce is a push or pull on a body. It is generally characterized by its point of application,

its magnitude and its direction. The direction of a force is defined by the line of action and the

sense of the force.

1.1.1 Resultant of Concurrent Forces

Consider a body acted upon by co-planar forces as shown in Fig 1.1(a).

S

QP

P

SQ

Fig 1.1 (a) Fig 1.1 (b)

Since all the forces pass through point O, they are said to be concurrent forces.

The resultant R of the concurrent forces can be obtained by using the polygon rule which

is equivalent to the repeated use of the parallelogram law as shown in Fig 1.1(b)

1.1.2 Resolution of a Force

As two or more concurrent forces can be replaced by a single resultant force, similarly a

single force acting on a body may be resolved into two or more forces which togther have the

same effect on the body. These forces are referred to as components of the force.

Example 1.1

Determine the resultant force for the system of forces P and Q shown in Fig Q.1


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R

C

B = 40NA 20°

C = 60N

B

155° 25°

C = 60N

B = 40N

A25°

20°

Fig. E1.1

Solution

The triangle ABC is drawn

Using the triangle rule, and applying the Cosine law, we have,

R2 = B2 + C2 -2BC cos B

R= 97.73N

Applying Sine law

s in s in

s in s in

.

.

. .

A

C

B

R

A

A

=

=°

= °

= °+ ° = °

6 0

1 5 5

9 7 7 3

1 5 0 4

2 0 1 5 0 4 3 5 0 4α

1.1.3 Rectangular Components of a Force

A force can be resolved into two rectangular components by representing the orthogonal

forces as Fx and Fy in the x and y axes respectively.

If the unit vectors along x and y axes are

and , then vector F can be written as F = Fx

+ Fx

If F is the magnitude of force F and the angle between F and x axis is denoted by , then

Fx = F cos

Fy = F sin

Example 1.2

A man pulls with a force of 300N on a rope attached to a 6m tall building as shown in Fig

1.2. Compute the horizontal and vertical components of the force exerted on the rope at point A.


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Fig E1.2(a)

300N6m

8m

A

C

Solution

From Fig. E1.2(b)

AB = 240N

C = 300N

Fx = 300 cos

Fy = -300 sin

cos

cos

θ

θ

= =

= =

8

10

4

5

6

10

3

5

F = 240

- 180

1.1.4 Addition of Forces

Consider three forces P, Q and S acting on a body as shown in Fig 1.3.

A

S

P

Q

Fig 1.3 (a)

Their resultant R is defined by

R = P + Q + S

Resolving the forces into rectangular components, we have

Rx

+ Ry

= Px

+ Py

+ Qx

+ Qy

+ Sx

+ Sy

= (Px + Qx + Sx)

+ (Py + Qy + Sy)

This implies that


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Fig. 1.4

zx

x

z

y

Rx = Px + Qx + Sx

Ry = Py + Qy + Sy

or Rx = FxRy = fy

Thus the scalar components Rx and Ry of the resultant R of several forces acting on a

particle are obtained by adding the corresponding scalar components of the given forces

algebraically.

1.2 Three-Dimensional Force System

F F i F j F k

F i F j F k

F

F i j k

li m i n k

x y x

x y z

x y z

x y z

= + +

= + +

= = + +

= + +

= + + =

=

→

→

c o s c o s c o s

c o s c o s c o s

c o s c o s c o s

θ θ θ

λ θ θ θ

λ θ θ θ

λ

21

1

Example 1.3

A wooden plank is held by cable AB, AC and DB as shown


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8cm

27cm

16cm

11cm A

B C

Fig. E1.3(a)

If the tension is 840N in cable Ab and 1.2kN in cable AC, determine the magnitude and

direction of the resultant of the force exerted by cables AB and AC in state A.

Solution

F A B

F AC

y

z

x

Fig. E1.3(b)

If i,j and k are unit vector along x, y and z respectively

A B i j k A B c m

A C i j k A C cm

→

→

= − + + =

= − + − =

1 6 8 1 1 2 1

1 6 8 1 6 2 4

,

,

λ

λ

→

→ →

→

→

= =

=

A B

A B A B A B A B

uni t v e c tor a long A B

F F F A B

A B

A B

.

8 4 0

2 1

Substitute

∴ = − + +

= − + −

= + = − + −

= + + =

= = °

= ° = °

→

→

→ → →

F i j k N

S im ila r ly F i j k N

R F F i j k N

R R R R N

R

Sim ilarly

A B

A C

A B A C

x y z

x

x

x

y z

( )

, ( )

( )

c o s , .

, . , .

6 4 0 3 2 0 4 4 0

8 0 0 4 0 0 8 0 0

1 4 4 0 7 2 0 3 2 0

1 6 5 0

1 4 4 0

1 6 5 01 5 0 8

6 4 1 1 0 2 6

2 2 2

θ λ

θ

θ θ


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1.3 Moments

Moment is a tendency to rotate or twist a body about an axis which does not intersect the

line of action of the force.

The magnitude = F x Perpendicular Distance from its line of action to point of Moment arm.

Moment arm

for P

0

B

A

Y

Moment armfor B

X

P

Fig. 1.4

Most moments bends causing bending while some moments twist causing torque or torsional moment. It

is sometimes easier to work with the components of the force rather than with the force itself.

1'

d

25°

0 X

P=50kN

Y

Fig. 1.5(a)

2m

0

Y

Px=47

3m

X

Py=17.1kN

Fig. 1.5(b)

Moment arm with respect to 0 = dM = p x d

Counter Clockwise Clockwise

Also, M P x P y y x= +( ) ( )

Credit for this observation is given to Pierre Varignon (1654-1722), a French Mathematician.

Varignon’s theorem states that for coplanar forces, the moment of a force about any point is

equal to the sum of the moments of the components of the forces about the same point. Also referred to

as the Principle of Moments.

In the Fig1 .5b

M P P

kN m

M kN m C lockw is e

y x0

0

2 3

1 0 7

1 0 7

= + −

= −

=

( ) ( )

or determine the using triangle KLM


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3m

2m L

M

K 20°

d

20°

0

P

Fig. 1.5c

KM

m

=°

=

32 0

8 2 4

ta n

.

∴ =

∴ = °

=

K O m

d

m

6 2 4

6 2 4 2 0

2 1 4

.

. s in

.

M P d

k N m

0

1 0 7

= −

= −

Moment of P is independent of the location of the force along its line of action e.g. if P acts at point L,

then using Varignon’s theorem Mr = Px(On) - Py(6.24) = -107kNm. This technique of moving force until it

passes there, the derived moment centre can be a valuable shortcut for many moment calculations.

1.4 Moment of a Force about a Point

M r x F

rF

o

→ → →

=

= sin θ

O Fr

y

z

x

r x (F1 +

F2 .........) r x F1

+r x F2........

Varignon’s theorem

M

i j k

x y z

F F F

o

x y z

=


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A 200N r CA

C

k

i

j+

Example 1.4A rectangle plate is supported by brackets A and B and by a wire CD as shown in Fig. E 1.4(b). If

the tension in the wire is 200N, determine the moment about A of the same force extended by the wire on

point C.

B

y

Fig. E1.4(a)

A

C

2 4 0 m m 3 0 0 m

m 8 0

m m

8 0 m m

2 4 0 m

m

x

Solution

MA exerted by wire on point C is

M r x F

r C A i k m

F N d irec ted a lon g C D

C D C Di j k

m

F F i j k N

M r x F i j k N m

A CA

CA

CD

CD

A CA

=

= = +

=

= =

− + −

= = − + −

= = − + +

→ →

→ →

→

→

→ →

→ →

( . . )

/ ( . . . )

.

( )

( . . . )

0 3 0 08

20 0

0 3 0 24 0 32

0 5

1 2 0 96 12 8

7 6 8 2 8 8 28 8

λ

λ


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i j k

0 3 0 0 0 6

1 2 0 9 0 1 2 8

. .

− −

1.5 Equilibrium of Concurrent Forces

Equilibrium is a state of balance created by opposing forces which act on a body in such a manner

that their combined net effect is zero.

Newton’s first law states that:

When a body is at rest (or is moving with a constant velocity, with zero acceleration) in a straight line, the

resultant of the force system acting on the body is equal to zero.

For coplanar system

∑ =

∑ =

F

F

x

y

0

0

1.6 Couple:

Concept of a pair of equal and opposite parallel forces,

If produces only rotation, no translation, hence pure Moment.

Magnitude = Product of one of the forces and the perpendicular distance between the forces and its sense

in either clockwise or anticlockwise.

Equilibrium Couple consists of forces with identical magnitude but opposite sense to balance a couple. An

understanding of what a couple is enables us to work more closely at what happens inside a bending

member (beam).

Consider beam AB carrying a point load P as shown in Fig. 1.6(a). If the beam is cut at section X-

X, shear force V develops to balance reaction R at A as shown in Fig. 1.6(b)

V

R

P

x X

X

(a)

(b)

A B

A

R and V constitute a couple of magnitude Vx (or Rx) which can only be balanced by an opposing couple.

That is, the beam portion might be in force (translational) equilibrium ( F=0), but it is not in moment

(rotational) equilibrium.

The opposing couple can only be provided by the other beam part upon the cut face as shown in Fig.

1.6(c).


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∑ =

∑ =

∑ =

F

F

M

x

y

0

0

0

yQ

Q

V

R

Fig. 1.6(c)

Because the moment arm is limited by the beam depth the forces Q (Tensile/Compressive) developed will

be quite large.

For moment equilibrium ( M=0), Q times y must equal Vx.

This introduces the third equation of static equilibrium which joins the previous two equation as follows:


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2. Equilibrium of Simple Structures

2.1 Introduction

Engineering Structures are structures made up by connecting several members together in

order to support different loads under varying conditions. In this course three categories of

Engineering Structures would be considered. These includes:

1. Trusses

These are structures that are made up of straight members connected at joints only.

Generally, the members are usually slender and cannot support much lateral load. Hence, the

loads must be applied at the joints.

A simple truss is a truss that is formed from a triangle (basic truss) Fig. 2.1(a) and

subsequently connecting two members to two of the joints and joining their free ends. Fig. 2.1(b).

Fig. 2.1(c) is a truss formed by combining five basic

trusses.

(a) (b) (c) Warren TrussFig. 2.1

2.2 Analysis of Trusses by the Method of JointsThe following assumptions are made in the analysis of simple trusses.

1. All members are two force members, whether in tension or compression.

2. Weights of the connecting members are negligible.

3. All joints are perfectly pinned, hence, allowing for force rotation at joints.

4. All loads are applied at the joints only.

Since the entire truss is in equilibrium , each pinned joint must be in equilibrium. Since

each member is acted upon by two axial forces (tensile or compressive), there will be 2n equations

available for a simple truss with n pins. Since we have three unknown eternal reactions, and

representing the number of member by m, then 2n = m + 3 for trusses that are in equilibrium and

statically determinate required to compute the member forces. This truss is said to be staticallyindeterminate.

If m + 3 < 2n, then the truss would collapse under load. This truss is said to be unstable.

The following steps could be followed when analyzing a simple truss using the joint

method.

1. Draw the free-body diagram of the entire truss system with loadings.

2. Start with a joint at which only two forces are unknown.

3. Proceed to a joint where the force on one of the joints is known and its left with only two


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unknown joints.

4. Follow step 2 until all the joint forces are computed.

Example 2.1Find the forces in the members of the truss shown in

Fig.. 2.2(a) using the joint method.

B

4@3m

AC

2kN

D

6kNE

4m

F G H J K2kN

3

4 5

Fig 2.2(a)

Solution

Reactions

RA x 12 = 2 x 9 + 2 x 6 + 6 x 3

RA = 4kN

RB = 6kN

Joint A

AC

AF

A


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CG

CF

CDC

Fx: AC = 0

Fv: AF = RA = 4kN

Joint F:

FGF

Fv : 4/5 x FC = AF = 4kN

FC = 5kN

FH : FG = 3/5 x FC = 3kN

Joint C :

FV : GC = 4/5 x FC – 2

= 4 – 2 = 2kN

FH : CD = 3/5 x FC = 3kN

Joint G :

GHG

GD

GC

GF

FV : 4/5 x GD = 2GD/ (5/2)

FH : GH = 3/5 x GD + FC

= 9/2

Joint H :


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FV

: HD = 2

FH

: HJ = HG = 9/2

HJH

HD

HG

2kN

Joint D :

DEDC

DH

DJ

DG

FV

: 4/5 x DJ = 2 – 4/5 x GD

= 2 – 2 = 0

FH

: DE = DC + 3/5 x GD

= 9/2

Joint J :

JKJ

JH

JEJD

FV : JE = 4/5

X DJ = 0

FH

: JK = JH = 9/2

Joint E :

EBE

ED

EK

EJ

FV : 4/5 x EK = 6

EK = 15/2

FH

: BE = ED – 3/5 x EK = 9/2 – 3/5 X 15/2

= 0

Joint K :


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KKJ

KBKE

FV : KB = 4/5 x KE

= 6KN

Joint B :

KB

BE

6kN

Check KB = RB6kN = 6kN

2.3 Analysis of Trusses by the Method of Section

All the forces i n the proceeding members should be determined before the forces in a

member when using the method of joint. However, if only forces in few of the members is to be

determined, the method of section is non desirable.

The method of section allows the non concurrent forces acting on the cut section to be

determined by using the third equilibrium equation, that is,

M = 0.

To analyse the forces in a truss by using the method of section, the following section could

be followed:

1. Draw the free-body diagram of the truss system.

2. Pass a section through three members of the truss system one of which is the member

whose axial force is to be determined.

3. Draw the free-body diagram of one of the two portions.

4. Write out the equilibrium equations for the three non-concurrent forces and solve for the

unknown forces.


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Example 2.2

Find the forces in the members of the truss shown in Fig. 2.3(a) using the method of

section.

3@3m = 9m

FAY

FAX

L0

5kNU3GU2

FU1

E 1110

h

60°60°L3L2L1

987654

321 DB

10kN

A

10kN

Fig. 2.3(a)

Solution

Note that,

tan 6 0 =2 h

3 =

03 , =

3 3

2h m ,

Reactions

RB x 9 = 5 x + 10 x 3 + 10 x 63 3

2RB = 11.44kN

3@3m = 9m

FAYFAX

L0

5kNU3

GU2

FU1

E 1110

h

60°60°L3L2L1

987654

321 DB

10kN

A

10kN

1

1

2

2

3

34

45

5

Fi . 2.3(b)

Section 1-1

Section 1-1 is passed through the truss so that it intersects member GB and member BC.

The right portion is then chosen as a free-body diagram.


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L3

U3

U3

L2

Section 2-2

5kN

U3

Section 1-1L2 L3

60°

11.44

FV : U

3L

3 Sin 60 + 11.44 = 0

3L3 = - 13.21 KN (Compression)

FH :

3L3 Cos 600 + L2L3 = 0

L2L3 = 6.61 KN (Tension)

Section 2 – 2

Section 2-2 is passed through the truss so that it intersects members GF, GC and BC. The

right portion is then chosen as a free-body diagram.

ML2 : 11.44 x 3 + U2U3 h – 5h = 0

2U3 = - 8.21kN (Comp.)

FV :

3L2 Sin 600 – 11.44 = 0

3L2 = 13.21KN (t)

Section 3 – 3

Section 3-3 is passed through the truss so that it intersects members GF, FC and CD. The


Section 3-3

L3

11.44kN

U3

U2

L2L1

5kN

MU2 : 11.44 x (3 + 3/2) – 10 x 3/2 –L1L2 x h = 0


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L1L2 = 14.04kN (t)

FV : 11.44 – 10 + 2L2 Sin 600 = 0

U2L2 = - 1.66t KN (Comp.)

Section 4 – 4

Section 4-4 is passed through the truss so that it intersects members FE, FD and DC. Theright portion is then chosen as a free-body diagram.

Section 4-4

10kN

2 3

8 9

L2 L360°

11FU2

GU3 5kN

11.44kN

ML1 : 11.44 x 6 – 10 x 3 –5h + U1U2 x h = 0

U1U2 = -9.87kN (Comp.)

FV : 11.44 – 10 – U2L1 Sin 60 = 0

U2L1 = 1.66KN (t)

Section 5 – 5Section 5-5 is passed through the truss so that it intersects members FE, BE and AB. The


Section 5-5

10kN 10kN

B2 3

6 7 8 9

L1 L2 L360°

10 11FU2

GU3

5kN

MU1 : 11.44 x ( 6 + 3/2) – 10 x ( 3+ 3/2 ) – 10 x 3/2 – L0L1 x h = 0L0L1 = 9.93kN (T)

FV : 11.44 – 10 – 10 + U1L1 Sin 600 = 0

U1L1 = 9.88KN (t)

Joint U1

FV : U1L1 Sin 600 x U1L0 Sin 60

0 = 0


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U1L0 = - U1L1 = -9.88KN (Comp.)

2.4 Analysis of Truss using the Graphical (Maxwell) MethodThe graphical method of analysis is carried out by using the Maxwell’s diagram. The

diagram could be drawn by using the following steps to solve the truss in Fig. 2.4(a)

Qb

Qa

Qf QdFig. 2.4(a)

1. Using the Bow’s notation, label every region between loads and reactions, moving

clockwise around the truss using uppercase letters. Similarly, label every area inside the

truss in a clockwise direction using numerals as shown in Fig. 2.4(b).

f e d

a

Qb

b

BA

Qa

A

B

C

6

3 4

521

VBVA Qf Qd

DEFG

HA

2. Compute the external reactions at the supports using the free-body diagram for the whole

truss.

3. Choosing a suitable scale, draw the force polygon for the external forces and reactions

going in clockwise direction in order of encountering, that is, FG = VA, GA = HA and so

on as shown in Fig. 2.4(c).


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B

1

2

A

E

D

Fig. 2.4(c)

F

C

G

4. Select a joint with only two unknown forces and draw the force polygon of forces acting

on the joint. Begin at joint A and draw polygon FGA1F as shown in Fig. 2.4(c).

5. Repeat step 4 for other joints until the forces in all the members of the truss are drawn .as

shown in Fig. 2.4(c).

6. Scale out the magnitude of the force in each member by measuring the length of the

member on the Maxwell’s diagram. A1, 1F, B3, C4, C6 and so on.

7. Determine the direction of the force, that is, whether tensile or compressive by reading the

corresponding letters around a joint in clockwise order on the Maxwell’s diagram. A1 is

compressive because it is going into joint (A).

2.5 Frames and Machines

Frames and machines contain multiforce members. Frames are structure designed to

support loads and are usually stationary whereas machines are designed to transmit forces orcouples from input to output forces. They may or may not contain moving parts.

2.5.1 Analysis of Frames

To analyse a frame containing one or more multiforce members, the following steps

should be followed.

1. Draw a free-body diagram for the whole frame showing the external loads and machine at

supports.

2. Separate components making up the frame and draw the free-body diagram for each

component part.

3. Consider the two-force members, that is, member carrying only tensile or compressive

force, and draw the forces acting on these members.4. Consider the multiforce members and draw the free-body diagram for each member.

Where a multiforce member is joined to a two-force member, apply a force equal but

opposite in direction to the two-force member on the multiforce member.

Where a multiforce member is joined to another multiforce member, resolve the internal

force at that joint into a horizontal and vertical components.

5. Determine the internal forces at the joints and external reactions at supports by using the

three equilibrium equations on the free-body diagrams.


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6. The total number of equilibrium equations from the free-body diagram would be more

than the number of unknown forces and reactions. Use the extra equations to check the

correctness of the computed forces.

Example 2.4

Determine the reactions on the support of the frame shown in Fig.E.2.4(a).

30

20kN40kN

C40

4040

40

15

40

1525

Fig. E2.4(a)

B

Solution:The free-body diagram for the frame is as shown in

Fig. E.2.4(b).40

20kN

40

40

30

RAY15

15

C

Fig. E2.4(b)

40

40RCY

RCX

40kN

B

25

RAX

Equilibrium Equation:

A B

B

BC

B M o r M Σ Σ= =0 0

MA = 0, + ; (20) (15) + 40 (55) – 80 (RCY) + 15RCX = 0

RCY = 31.25 + RCX∴ 3

1 6

B C

Σ ΜΒ

= 0 ; (40) (15) − 40 ( ) + 45 = 0 R RC Y C X

RCX = RCY – 13.33∴ 8 9Substituting RCX = 17.33 kN , RCY = 34.5kN

Similarly,

Mc = 0, 80 RAY + 15 RAX – 20 (65) – 40 (25) = 0

RAY = 28.75 – RAX3

1 6


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A B

A Y A X R RΣ ΜΒ = 0 ; 40 − 30 − 20 (25) = 0

Solving, RAX = 17.33 KN , RAY = 25.5 KN

2.5.2 Analysis of Machines

To analyse a machine, the following steps may be followed1. Draw the free-body diagram for the entire machine showing all the forces and reactions on

the machine.

2. Separate the machine into components and draw the free-body diagram for each

component.

3. Consider the two-force members, that is, member carrying only tensile or compressive

force, and draw the forces acting on these members.

4. Consider the multiforce members and draw the free-body diagram for each member.

Where a multiforce member is joined to a two-force member, apply a force equal but

opposite in direction to the two-force member on the multiforce member.

Where a multiforce member is joined to another multiforce member, resolve the internal

force at that joint into a horizontal and vertical components.5. Determine the internal forces at the joints and external reactions at supports by using the

three equilibrium equations on the free-body diagrams.

Example 2.5

The mechanism shown in Fig. E2.5 has a lever AB pivoted at a fixed point A. The link CD

is pin-jointed to AB at C and pinned at D to a block sliding in horizontal guides which

compresses a spring. A downward force of 100N must be applied at B to maintain equilibrium.

Neglecting the effects of friction, determine

( i) The force exerted by the spring.

( ii) The vertical and horizontal components of the reaction at C.

( iii) The magnitude and direction of the reaction at A.

C

1 5 0 m m

A D

B

2 5 0 m

m

100N

Block

Spring

3 0 °

3 0 °

Fig. E2.5(a): Pivoted Lever

Solution:


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The free-body diagram is shown in Fig. E2.5(b).

Fig. E2.5(b)

Ax

Ay Dy

100N

B

C

Dx

Fig. E2.5(c)

Ax

Ay

Cy

100N

Cx

Cx

C y

Dx

Considering Fig. E2.5(c)

C yC

Cx


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C C

C C

C

C

C cx

C cx

C cx

C C

C xC

C C

x

y

y

x

y

y

y

x

y

x y

x y

=

=

∴ =

=

= °

=

=

=

=

cos

sin

ta n

ta n

ta n

.

ta n

.

.

θ

θ

θ

θ

θ

3 0

0 5774

1

0 5774

1 732

From free-body diagram Fig. E2.5(c)

Σ fy A C

A C i

y y

y y

:

( )

+ − =

+ =

1 0 0 0

1 0 0

Σ fx A C

A C ii

x x

x x

:

( )

− =

=

0

E M x C x C x

C C

C C

b u t C C

C C

C C

C

C

C N

A x y

x y

x y

x y

y y

y y

y

y

y

↓ + ° − ° − °

− − =

− =

=

∴

+ =

+ =

=

=

=

1 0 0 4 0 0 3 0 1 5 0 3 0 1 5 0 3 0

3 4 6 4 1 7 5 1 2 9 9 0

7 5 1 2 9 9 3 4 6 4 1

1 73 2

7 5 1 7 3 2 1 2 9 9 3 4 6 4 1

1 2 9 9 1 2 9 9 3 4 6 4 1

2 5 9 8 3 4 6 4 1

3 4 6 4 1

2 5 9 8

133 3

( c o s ) ( s in ) ( c o s )

.

.

.

( . ) .

. .

.

.

.


25/53

S u b s titu te C N o e qu a tio n i

A N

A N

A N d ow n lo a d

C on s id er in g C N

C N

C A N

y

y

y

y

x

x

x x

=

+ =

= −

∴ =

∴ =

=

∴ = =

133 3

1 3 3 3 1 0 0

3 3 3

3 3 3

1 7 3 2 1 3 3 3

2 3 0 8 8

2 3 0 8 8

. in t ( )

.

.

.

. ( . )

.

.

Σ

Σ

fy d c

d c N

fx c d

c d N

su l t fo rc e o n C x c

s ul t fo rc e o n C



su l t fo rc e o n C N

y y

y y

x x

x x

y

:

.

:

.

R e tan ( )

R e tan ( . ) .

R e tan . .

R e tan .

R e tan .

− =

= =

∴ =

= =

= +

= +

= +

=

=

0

133 3

2 3 0 8 8

2 30 8 8 1 33 3 3

5 3 3 0 3 5 4 1 7 7 6 8 8 9

7 1 1 3 2 4 3

266 7

2 2

2 2

Therefore, the force exerted on the spring is 266.7N

(ii) Cx=230.88N and Cy=133.3N

(iii)

R A A

R

R N

A x y

A

A

= + = +

=

=

( ) ( ) ( . ) ( . )

.

.

2 2 2 23 3 3 2 3 0 8 8

5 4 4 1 4 4 6

2 3 3 2 7


26/53

3. FRICTION

This is a tangential force generated between contacting surface. It is always in opposing

direction caused by interlocking of minute irregularities on surfaces.

Friction can be divided into two broad categories.Static Friction: where there is no relative motion between the surfaces.

Experiments have shown that the static frictional force is proportional to the normal component N

as shown in Fig. 3.1.

N

F

W

C

AFig. 3.1

That is,

Fs = µ sN

where µ s is a constant called the coefficient of static friction.

Dynamic (Kinetic) Friction: where there is relative motion between the contacting surfaces.

Similarly, experiments have shown that the kinetic frictional force is proportional to the normal

component N.

That is,

Fk = µ k N

where µ k is a constant called the coefficient of kinetic friction.

The kinetic friction µ k is always less than the static friction µ s. This is because there is less

interaction between the particles on the contacting surfaces when these surfaces move with

respect to each other.

The coefficients depend more on the nature of the surfaces. Typical values of coefficient of static

friction for various materials are given in Table 3.1

Table 3.1. Approximate Values of Coefficient of Static Friction for Dry Surfaces

Metal on metal 0.15-0.60

Metal on wood 0.20-0.60

Metal on stone 0.30-0.70

Metal on leather 0.30-0.60

Wood on wood 0.25-.050


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Wood on leather 0.40-0.70

Stone on stone 0.40-0.70

Earth on earth 0.20-1.00

Rubber on concrete 0.60-0.90

Source: Beer and Johnston (2)

Consider a block resting on an inclined plane as shown in Fig. 3.2. The forces acting on the body

is as shown in the figure.

Resolving the forces along an axis parallel to the surface:Σ

Σ

F n F W S in

N W S in

F y N W C o s

W N C o s

s

= − =

=

= − =

=

0 0

1

0 0

2

θ

µ θ

θ

θ

( )

( )

Substituting (1) and (2)

µ s = tan

That is, the coefficient of static friction is equal to the tangent of the angle of repose (angle of

inclination).

Example 3.1

A 250N block resting on a flat surface is being acted upon by a horizontal force P as

shown in Fig. E3.1(a). The coefficient of friction between block and plane are µ s=0.25.

Determine the minimum force P required to cause sliding?

s=0.25P

W=400N

Solution

The free-body diagram for the block is as shown in Fig. E3.1(b)

P

F

400N

N

Summing all the vertical forces we have,


28/53

Σ F v N

N N

B u t F N

F N

s

= − + =

∴ =

=

∴ = =

4 0 0 0

4 0 0

0 2 5 4 00 1 00

µ

. ( )

Summing all the horizontal forces we have,

Σ F

P F

P N

H =

− =

∴ =

0

0

10 0

Example 3.2

A force P acts on the 300N block placed on an inclined plane as shown Fig. E3.2(a). The

coefficient of friction between block and plane are µ s=0.25, µ k =0.20. Determine if the block is in

equilibrium and find the value of the frictional force if Pis (i) 100N (ii) 120N.

5

43

300N

P

Solution

The free-body diagram of the block is shown in Fig. E3.2(b)

x

y

P

300N

F

N

Summing all the forces along x axis we have,

Fx=0: P - 3/5(300)-F=0

F = P - 180 (i)

Summing all the forces along y axis we have,

Fy=0: N - 4/5(300) = 0

N = 240N


29/53

Case (i): For P = 100N

F = (100 - 180)N = -80N

F = 80N upwards

For Static equilibrium

Fmax = µ sN = 0.25(240)

= 60N

Since the value to maintain equilibrium is an 80N force directed up, the block tends to move

down.

Actual frictional force is,

Fk = µ k N

= 0.2(240) = 480

Case (ii) For P = 120NF = +120 -1 80 = -60N

Since F = Fmax , we have an impending motion down the plane.

Frictional force =, F = 60N.

Example 3.3

A horizontal force P acts on a 500N block placed on an inclined plane as shown Fig.

E3.3(a). The coefficient of friction between block and plane are µ s=0.25, µ k =0.20. What

horizontal force P would cause the block to slide upwards?

5 0 0 N

30°

P

SolutionThe free-body diagram of the block is shown in Fig. E3.3(b)


30/53

P S i n 3 0 °

W S i n 3 0

°

W C o s 3 0

°

N

P C o s 3 0 °

Fig. E3.3(b)

P

3 0 °

Summing all the forces parallel to the plane we have,

Fx = PCos300 - F - 500Sin300 = 0 (i)

Similarly, summing all the forces normal to the plane we have,

Fy = N - 500Cos300 - PSin300 = 0 (ii)

But the frictional force F = µ sN

Hence, the first equation becomes

PCos300 - µ sN - 500Sin300 = 0 (iii)

Solving the equations (ii) and (iii) simultaneously we have,

0.866P - 0.25N = 250

-0.5P + N = 433

N = 674.75N

P = 483.5N

The horizontal force P needed to start sliding up is 483.5N.


31/53

4. Structural Equilibrium

4.1 Introduction

In statics, it becomes convenient to ignore the small deformation and displacement. We

pretend that the materials used are rigid, having the property or infinite stiffness.The necessary and sufficient condition for a body lying in a plane to be in structural

equilibrium is by satisfying the static equilibrium equations as follows

∑ =

∑ =

∑ =

F

F

M

x

y

0

0

0

4.2 Supports

Bodies are usually supported by different types of supports. There are three types of

supports:

1. Supports which resist forces in one direction only e.g. Roller, Rocker and Frictionless support

Frictionless Surface

Rocker

Roller

Fig. 4.1

2. Supports which resist forces in two directions

Rough SurfaceHinge

Fig. 4.2


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y

x

ydA

z

x

y xC.G

x

3. Support which resist forces in all directions e.g. fixed or rigid, or built-in-support

Fy

MFx

Fig. 4.3

4.3 Free Body Diagram

In order to analyse a rigid body in equilibrium, it is necessary to idealize the two or three

dimensional body to a line diagram showing the forces acting on it. This diagram is referred to as

Free Body Diagram. The first major step in solving a problem in static equilibrium is to draw an

accurate free body diagram of the body together with the forces and reactions acting on it.

The following guideline should be followed in drawing the free body diagram:

Separate the body to be analysed

Draw the schematic diagram of the body by showing its eternal boundary.

Indicate all the external forces acting on the body.

Indicate all the reactions acting on the body.

Indicate all the dimensions and choice of coordinates axes (where necessary) on the

diagram.

4.4 First Moment of Area

First moment or area of an element about any axis in the plane of the area is given by the

product of the area of the 2nd element and the perpendicular distance between the element and the

axis.

Centroid

Centroids:

The center of mass or centre of gravity may visualized as the location of the resultant of a

set of parallel forces.

First Moment of Area about x-axis (Statical Moment) is mathematically given as:


33/53

Q xd A y A

= ∫ 0

Total Area = d A∫ The position for the centroid for the total area is given as :

−

−

= ∫

∫ =

= ∫

∫ =

x x d A

d A

Q

A

y y d A

d A

Q

A

A

A

y

A

A

x

0

0

0

0

;

Since for most structural applications the areas involved are regular geometric shapes, the integral

becomes

x x A

A

y y A

A

i i

i

i i

i

−

−

= ∑

∑

= ∑

∑

( )

( )

Example4.1Determine the centroid location of the T shape in the figure below. All dimensions are in

mm.

301530

y1=7060

20

y2=30

A2

A1

Solution:

The shape consists of two rectangles with their centroids as shown in Fig. E4.1(a)

Fig. E4.1

By symmetry, is =0 x−

y y A y A

A A

m m

−

= +

+

= +

+

=

1 1 2 2

1 2

7 0 2 0 7 5 3 0 1 5 6 0

2 0 7 5 1 5 1 0

5 5

( )( ) ( ) ( )

( ) ( )

The centroidal location is shown in Fig. E4.1(b)


34/53

x

Centroidal axes

y=55

y

x

y

Fig. E4.1(b)

Example 4.2

A rectangular plane section 70mm x 90mm is having a square hole 30mm x 30mm in it as

shown in Fig. E4.2. Locate the centroid of the plane area.

hole 3 0

2 0

4 0

303010

y

x

Fig. E4.2(a)

Solution:

The centroid along the x and y axes are located by subtracting the square hole from the

rectangular section.

x x A x A

A A

m m

y

m m

−

−

=

−

−

=

−

−

=

=

−

−

=

1 1 2 2

1 2

3 5 7 0 9 0 2 5 3 0 3 0

7 0 9 0 3 0 3 0

3 7

4 5 9 0 7 0 5 5 3 0 3 0

9 0 7 0 3 0 3 0

4 3

( )( ) ( )( )

( ) ( )

( )( ) ( )( )

( ) ( )

The centroidal location is shown in Fig. E4.2(b)


35/53

hole 3 0

2 0

y=43

303010

y

x

Fig. E4.2(b)

C.G

x=37

3 0

30

30

4.2 Second Moment of Area

Moment of Inertia is a mathematical concept that is used to quantify the resistance of

various sections to bending or buckling.

A beam section with large Moment of Inertia, I value, will have smaller stress and deflections

under a given load than one with a lesser I value. A long, slender column will not buckle laterally

if I of its cross-section is sufficient.Moment of inertia of Area of an element shown in Fig. 4.5 about any axis is defined as the

product of the area of the element and the square of the perpendicular distance between the

element and the axis.

I y d A

I x d A

x

A

y

A

= ∫

= ∫

2

0

2

0

y

x

y

dA

x

An I value has units of length to the 4th power. Hence, it is sometimes called Second Moment of

Area.

It may help to understand the concept of moment of inertia if we draws an analogy based upon

real inertia due to motion and mass with equal area.

Imagine the two shapes in the Fig.4.6 below to be cut out of heavy sheet material and placed on

an axle (xx) so they could spin about it.

x x


36/53

It would be much harder to start the first shape Fig. 4.6(a) and once moving much harder to stop

than the second shape in Fig. 4.6(b)

The same principle is involved when a skater spins on ice. With arms held close in, the skater will

rotate rapidly, and with arms outstretched (creating more moment of inertia, I), the skater slows

down.

Example 4.3

Find the Moment of Inertia,. Ixx of a rectangle

Solution:

The rectangle is shown in Fig. E4.3. Dividing the rectangle into strips parallel to the x-

axis, we obtain

h/2

h/2

dyy

x x

bFig. E4.3

I y d A

y bd y

b y b h h

I b h

xx

A

h

h

h

h

xx

= ∫

= ∫

=

= +

=

−

−

2

0

2

2

2

3

2

2

3

3 3

3

8

3

8

12

Example 4.4

Find the moment of inertia about the ax and y axes for the shape shown in Fig. E4.4. All

dimensions are in mm.


37/53

y

8

16

16

x

100

x 220

Solution:

The shapes could be divided into three rectangles to determine the Iyy. To compute the Ixx,

a rectangle could be used for the entire area and then subtract two rectangular spaces from it as

follows.

I I va lu es o f th e rec g le s

I I rec g le I rec g ular sp ace s

yy

xx

= ∑= −

3

2

ta n

tan tan

I I I

b h b h

x m m

I I I

x m m

yy yy We b yy Fl ange

W eb F lange

xx xx gr os s xx s p ac e

= +

=

+

= +

=

= −

= −

=

2

1 22

1 2

2 2 0

1 22

1 6 1 0 0

1 22 6 8 1 0

2

1 0 0 2 5 2

1 22

4 6 2 2 0

1 25 1 7 1 0

3 3

3 3

6 4

3 36 4

(8) ( ).

( ) ( ).

4.3 Parallel Axis Theorem

This provides a simple way to compute moment of inertia I about any axis parallel to

centroid axis.

Consider the moment of inertia I of a small elemental area dA with respect to x’ axis as shown in

Fig. 4.7 which could be written as

I y d A x A

' = ∫ 20


38/53

y

d

xC.G

y'

x'

y'

dA

Fig. 4.7

Since y = y’ + d, this could be expressed as

I y d d A

y d A d y d A d d A

x

A

A

' ( ' )

' '

= +∫

= ∫ + + ∫ ∫

2

0

2 2

02

The second integral represents the first moment of the area with respect to its own centroidal axiswhich must be zero. Hence,

I I A d x x'

= + 2

This expression is known as the parallel axis theorem.

Parallel axis theorem :The parallel axis theorem for the second moment states that the second

moment of area about any axis is equal to the second moment area about a parallel axis through

the centroid of the area plus the product of the area and the square of the perpendicular distance

between the two axes.

Example 4.5

Determine the moment of inertia Ixx of the T-shape in Example 4.1 with respect to its

horizontal centroidal axis as shown in Fig. E4.5(a).

301530

y=55

20

x

60

x

Solution

The T-shape could be divided into two rectangles with their centroids as shown in Fig


39/53

E.4.5(b)

2

x x

1

d1=15

25

Fig. E4.5(b)

Since neither rectangle has it centroid coincident in the centroid of the entire section, as shown in

Fig. E4.5(b) we have

[ ] [ ] I I A d I A d

b h A d

b h A d

x m m

xx xx xx= + + +

= +

+ +

= +

+ +

=

1 21 1

2

2 2

2

1 1

3

1 1

2

3

2 2

2

3

2

3

2

6 4

1 2 1 2

7 5 2 0

1 2 7 5 2 0 1 5

1 5 6 0

1 2 1 5 6 0 25

1 2 2 1 0

( )( )( )

( )( )( )

.

Example 4.6

Determining the moment of inertia, Iyy for the retaining wall shown in Fig. E4.6.

x x

1m

1m 1m

9m

y

Solution

The retaining wall could be divided into a rectangle and two triangles. The moment of

inertia with respect to the y axes, Iyy is the computed as follows:

[ ] [ ] I I I A d

m

yy yyrect

yytriangle

= + +

= + +

=

2

9 1

1 2 2

9 1 5

3 6

1

2 9 1 5 1

1 5 9

2

3 3

2

4

( ) ( . )( ) ( . ) ( )

.


40/53

Example 4.7

The section of a steel is as shown in Fig. E4.7. The section is symmetrical about y-axis.

Determine the moment of inertia of the cross-section with respect to y-axis.

1 0 m m

7 .5 m m

1 0 m m

y

7. 5 m m

y

8mm 8mm

Fig. E4.7

Solution

Moment of inertia of

(i) Outer Box, Iy = 20 x 163

= (20 x 163)/12

= 6826mm4

(ii) Inner Box, Iy = 15 x 163 /12

= 5120mm4

( iii) Outer Half Circle, Iy = R R

4

2 2

8 2

4

38

Π ΠΠ

+ +

= 27476.24 mm4

( iv) Inner Half Circle, Iy = r r

4

2 2

8 2

4

38

Π ΠΠ

+ +

= 12286.5 mm4

( v) Small hatched rectangle,

Iy= bh3 /12 + Ad2


41/53

= 1710mm4

Moment of Inertia about y, Iy= (6826 -5120) + 2(27476.24-12286.8) +2(1710)

= 28665.5mm4


42/53

5. Newton’s Laws of Motion

The basic laws for expressing the relationship between the external forces on a body and its

motion are the Newton's law of motion. These were proposed in Newton's famous work, theprincipia published in 1687.

These laws are usually expressed as:

First Law : A particle free of any external influence continues in its state of rest or uniform

motion unless acted upon by an external force .

If F

0 , then F

d/dt (mv)

Second Law : The product of mass of a particle and the acceleration imparted to it by a unit

force is proportional to the force acting on it .

F = ma

Third Law : Two particles exert on each other forces equal in magnitude and acting in opposite

directions along the straight line connecting the two particles .

= 60°

x

y

Example 5.1

The block shown in Fig. E5.1 weighs 20kg and is released from rest .

(i) Check if the block moves (ii)Find its speed after it has descended a distance d=5m down theplane. The coefficients

of friction are:

s(static) = 0.30

k (kinematic) = 0.25

Solution


43/53

F

mg

x

y

NFig. E5.1(b)

Referring to the FBD

F = ma

Nj + mg (Sin i - Cos j)- Fi = mai

or

fx = 0 : mg Sin - F = ma

fy = 0 : N - mg Cos = 0

First determine if the block will moveFor equillibrium

F = mg Sin

F is limited by 0 F Fmax = s N

Hence ,

mg sin

s N

mg sin

s mg cos

∴

Ta n s

But tan60 = 1.73 >

s = 0.3,

hence, the block moves and,as it does so ,is acted upon by acted upon by k N up the plane.

Having checked the statics we now solve the equation of motion:

mg sin -

k N = ma

Substitute for N = mg cos

a = g(sin - k cos )

a = constant

V = = g(sin-

k cos )t + c1ad t ∫

Using boundary conditions that v=0 when t=0c1=0

Integrating v

x = g/2(Sin -

k Cos )t2 + c2

If we choose x = 0, when t = 0,

c2=0

x = g/2(Sin -

k Cos ) t2


44/53

Using = 60,

k = 0.25, x = 5m, g = 9.81 m/s2

t=1.17s

Velocity , v=8.50 m/s


45/53

6. Kinematics of Particles and Rigid Bodies

6.1 Introduction

Dynamics is the study of the motion of bodies and forces that cause those motions while

kinematics is an aspect of dynamics that deals with consideration of space and time , that is ,

geometry of motion .

6.2 Kinematics of a Point in Rectilinear Motion

In which a point P moves along a straight line.

P ositio n V ec to r r X i

V elocity V r X i

a r X i

O P

P O P

P O P

,^

. . ^

.. .. ^

=

= =

= =

Three cases of interest are:

(i) Acceleration is a known function of time , f(t)

a X f t

X f t d t C

X f t d t C t C

= =

= +∫

= +∫∫ +

..

.

( )

( )

( )

1

1 2

We can also use the V – t diagram , although limited in application.

(ii) Acceleration is a known function of velocity g(v)

a X d v

d t g v

d v

d t g v

t C d v

g v p v

= = =

=

+ = ∫ =

..

( )

( )

( ) ( )

1

1

We can then solve for velocity as

v q t d x

d t

X q t d t C

= =

= ∫ +

( )

( )2

Example 6.1

If acceleration of a point P is proportional to velocity on a = = –2v m/s2 with initial X ..

conditions (0) = 2 m/s and X(0) = –7 m . Find the position of the particle x(t) . X .

Solution


46/53

X d v

d t v

d v

d t v

t C d v

v

..

= = −

= −

+ = − ∫

2

1

2

21

Integrating ,

t C I n V

I n V t C

V e t C

+ = −

= − −

= − −

1

1

2 2

2

2 2

1

Using (0) = 2 , X .

C

In

d x

d t V e e m s

X e d t C

e d t C

t I n t

t

t

1

2 2 2

2

2

2

2

2

2

2

2

=

−

= = =

∴ = +∫

= +

− + −

−

−

/

Using x(0) = –7, we have

C2 = –6

Hence,

x(t) = (–6 – e-2 t) m

(iii) Acceleration is a function of position, that is:

a = = h(x)...

X

We may combine to obtaina V d v

d t a n d V X

d x

d t = = = =

. .

ad x

d t V

d v

d t =

If a function r(x) exits, such that

h xd r x

d x a a nd r x h x d x( )

( )( ) ( )= = = ∫

We obtain,

d r

d x

d x

d t V

d v

d t

d r

d t V

d v

d t

. =

=


47/53

Integrating ,

r xV

C ( ) = +

2

12

Thus, V r x C 2 12 2= −( )

Example 6.2

Let a = = -4x, find velocity using initial conditions (0) = 2, X(0) = -7 X ..

X

.

Solution

h(x) = –4x

r(x) = –

4x dx = –2x 2

Substituting,

− = +22

2

2

1 X

V C

Using the boundary conditions,

V X

V X m s

2 2

2

2 0 0 4

2 0 0 4

= −

= ± − /

Example 6.3

The system of pulleys shown in Fig. E6.3 shows a block A traveling downwards with VA

= 3t2 m/s. Find the velocity of block B when t = 4 s.

K

A

C

D

B

yC

yA

j

Solution

The length L of the rope that passes around both pulleys is a constant .

L = yc +

rc + (yc -k) +

rD + (yA-k)

Differentiate with respect to time


48/53

0 2

2

3

2

2

= +

= = −

y y

y y

t

C A

C

A

. .

.

.

VB = Vc since both move in the same pattern

V t j m s

V j m s o r m s U p w a rd s

B

B t

= −

= −=

3

2

2 4 2 4

2

4

^

^

/

/ /

6.3 Kinematics in 3-dimensional Cartesian Coordinates

r o p

ji

o

kP

Position vector of position P shown in Fig. 6.1 is

r x i y j z k

v x i y j z k

a x i y j z k

o p

o p

p

= + +

= + +

= + +

− − −

^ ^ ^

. . .

.. .. ..

Example 6.4

A point Q has the acceleration vector

aQ = (4i - 6tj + sin 0.2tk) m/s2

At t = 0, roQ = (1,3,5) and VQ = 2i - 7j + 3.4k m/s.

Find the speed and its distance from starting point when t=3s.

Solution

Integrating aQ, we have VQ = 4ti - 3t2 j - 5cos0.2k + C1Using initial conditions

2i - 7j + 3.4k = -5k + C1Therefore,

C1 = 2i - 7j +8.4k

Integrating

rOQ = (2t2+2t)i - (t3+7t)j + (8.4t-25sin0.2t)k + C2


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Using initial conditions at t=0

rOQ|t = 0 = i +3j -5k = C2Substituting the required time t=3s

VQ|t=3 = 14i - 34j + 4.27 m/s

Thus speed V m sQ t = = + − + =32 2 2

1 4 3 4 4 2 7 3 7 0( ) ( . ) . /

rOQ|t=3 = 25i - 45j + 6.08k m

The distance d between Q and its starting point is

d = |rOQ(3)-rOQ (0)|

= ( ) ( ) ( . ( ))2 5 1 4 5 3 1 6 0 8 52 2 2− + − − + − −

= 54.8 m


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7. Kinetic Energy and Momentum

7.1 Kinetics

This deals with the consideration of the manner in which the motion of a body is related to

external mechanical actions including forces and couples.

7.1.1 Work and Kinetic Energy

F = ma

Forming the dot product with velocity V,

F. v = ma.v

= m dv/dt.v

= m/2 d/dt (v.v)

= d/dt (m/2 v2)

Integrating,

= m/2[ |v(t2)|2-|v(t1)|

2 ]F v d t t

t .

1

2∫

The RHS is called the Kinetic Energy while the LHS is called the work done on the

particle

Work done on the particle = Change in the particle's K.E

W = T∆

7.2 MomentumMomentum = mv

(momentum) = ma =d

d t F

d L

d t −

−

= change in Momentum = DLF d t L t L t t

t

− − −

∫ = −2 2 1( ) ( )

= Impulse

If during some time interval, the sum of the external forces varnishes , i.e, thed L

d t

− = 0

Hence ,the momentum is a constant, or is conserved,during that interval:

It is possible by conservation of Momentum ,to obtain limited quantitative information about the

motions of the colliding bodies

7.2.1 Colliding Bodies

In the analysis of collision (or impact) , the Momentum of a body is the sum of the

Momenta of the individual parts

L = L1 + L2 + ... + Ln


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Example 7.1

A wooden block of mass m1=50kg is at rest on a smooth horizontal surface. When it is

struck by a bullet of mass m2=30kg travelling at a speed v=10m/s, the block slides to the right.

Find the speed v after the bullet becomes embedded in the block

4

3m 2

m 1

Fig. E7.1

Solution

L t m m v

L t m v

f −

−

= +

=

( ) ( ) )

( ) ( . )

2 1 2

1 2 20 8

Equation

( ) .

.

m m v m v

vm v

m m f

1 2 2 2

2 2

1 2

0 8

0 8

+ =

=+

Substituting

v m s f = 3 /

Coefficient of Restitution provide measure for colliding bodies to rebound off each other.

eF d t

F d t

t

t

t

t = ∫

∫ 2

3

1

2

− ∫ = −

− ∫ = −

− ∫ = −

∫ = −

F d t m v v

F d t m v v

F d t m v v

a n d F d t m v v

t t

A c A i

t

t

B c B i

t

t

B A f c

t

t

B B f c

1

2

1

2

2

3

2

3

( )

( )

( )

( )

Substituting

ev V

V V

v V

V V

A F C

C Ai

B F C

C B i

= −

−

= −

−

Eliminating,

ev v

v v

B F A F

A i B i

= −

−

Quotient of the “relative velocity of seperation” and the “relative velocity of approach”.


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i

j

43 5

0 m / s

4 0 m / s

3 0 °

Line of impact

Fig. E7.2

e = 0

disk sticks together

e = 1

no energy dissipation

Example 7.2

Two identical blocky pucks collide as shown. If e =0.8. Find the velocities of the pucks

after collision and impulse of the force of interaction.

Solution

Neglecting friction and assuming non-deformation,

Let m = mass of last puck

VAF x Vai = final and initial velocities of ball A and B respectively.

V S in i C o s j m s

i jm s

V V j

V i j

i jm s

V V jm s

A i

A f i

B i

B f i

− −

−

−

−

−

= ° + °

= +

= +

= − −

= − −

= −

4 0 3 0 3 0

2 0 3 4 6

3 4 6

5 04

5

3

5

4 0 3 0

3 0

1

2

( ) /

. /

.

/

/

Since there are no external forces on the system of two pucks,

m V V m V V A F B F A i B i( ) ( )− = +

Taking the i - direction along the line of impact


53/53

V V

B y d e fin i ti o n

eV V

V V

S o lv in g V m s V m s

V i j

V i j

AF

BF

1 2

2 1

2 1

1 2

2 0

0 82 0 4 0

48

3 4 1 4

34 3 4 6

14 3 0

+ = −

= =

−

− −⇒ − =

= − =

= − +

= +

−

−

.( )

/ , /

. ,

as shown in Fig. E7.2(b)

14

30

17.3

17

Fig. E7.2(b)

cve 253 course outline

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