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Equilibrium Conditions

2-1

2
...

Distinguish between a vector and a scalar quantity
...
"



Understand the Addition of Forces




Resolve a force into 2 components analytically
...




Understand the Moment of a Force



Define the moment of a force about a point
...

Sum up the moments due to several forces acting on body about a point
...
0

2-2

Equilibrium Conditions



Understand the concept of Free Body Diagrams



Define a Free Body Diagram (FBD)
...
g
...

Know the reactions present at various supports, e
...
rollers, pin-joints, fixed-ends
...











State the conditions of equilibrium for concurrent forces
...

Solve Statics problems involving concurrent force systems
...

Determine if a body is in equilibrium
...


Version 1
...
1

2-3

Vector Quantity

A force is a vector quantity that has magnitude(size) and direction
...

5N

Fig
...
1
Fig
...
1 shows a force vector lying in the plane of the paper with a magnitude of
5 N acting at an angle of counter-clockwise from the positive x-axis
...
Fig
...
2
shows three sets of co-planar forces, each on a different plane
...
2
...
0

2-4

2
...
It is the reverse process of adding two given forces together
...


F

F

F2
F2




F1
F1



F1 & F2 form the sides of the rectangle

F
F

F2





F1

F2

F1

Applying simple trigonometry to the right angle triangle,

&

Rectangular components,

F2 = F sin

&

F1 = F cos

Example 2
...

 Draw a rectangle with F as the diagonal
...

 Indicate the angle to be used for resolving
...
60 N 

x

Component Fy = F sin = 100 sin 30°
= 50 N 

Fy


Fx
Version 1
...
2
Resolve the 100 N force along the x- and y- axes
...
1
y
F = 100 N

30°

Horizontal

x

TUTORIAL
2
...
(Follow the steps as in
Example 2
...
2

y

(b)

Resolve the 300 N pulling force (tension) into rectangular components in
(a)
the x and y directions
...
1 (a) Fn = 433 N
Ft = 250 N
2
...
71 N
Fy = 241
...
2 (a) Fx = 281
...
6 N
2
...
1 N 
Fh = 245
...
0

2-6

Equilibrium Conditions

2
...

F1

To determine the resultant of the two concurrent
forces, F1 and F2 (Fig
...
4a),

F2


Steps to follow:
1
...
2
...
2
...



Fig
...
4 a

F1
F2
F1 sin 

F2 sin 





F1 cos 
Fig
...
4 b

2
...
2
...
e
...
If  direction is assigned as positive,
Fx = F2 cos  - F1 cos 
...


Sum up all y-components (i
...
Fy )
...


= F1 sin  + F2 sin 
...
2
...


Its direction from the x-axis is calculated using
the tangent function:

 Fy
tan θ 
 Fx

FR


Fx
Fig
...
4 d

Version 1
...
Each force exerts in the direction that it acts, but if several forces
are present, the net effect will be in the direction of the resultant force
...
3
Two perpendicular forces of 30 N and 40 N are acting on a particle as shown below
...


30 N

Applying Pythagoras’ Theorem and simple
trigonometry to the right-angle triangle,
40 N

FR =

402  302 ;

tan θ 

30
40



30 N

FR
Resultant force

FR = 50 N

Direction:

 = 37º

40 N

Example 2
...

 Fy = 50 + 60 sin30
 Fx = 60 sin30

30°
60 N

= __________
= __________

50 N

60cos30

60sin30

50

Mechanics I (ME1101/0101)

Version 1
...
5
Find the resultant force for the given co-planar concurrent forces using analytical
method
...
6
Find the resultant force for the given co-planar forces
...
0

Mechanics I (ME1101/0101)

Equilibrium Conditions

2-9

TUTORIAL
2
...
Determine the resultant force
acting on the bracket
...
4)
2
...
2 kN

(Ans : 6
...
4

16
...
99 N

70
...
108 kN

Mechanics I (ME1101/0101)

85
...
0

2 - 10

Equilibrium Conditions

(c)
6 kN
10 kN
75°

45°
30°

30°

20 kN

16 kN

(Ans : 19
...
5

75
...

If two forces, each of magnitude 250 N, can be applied by means of cables as
shown below, determine analytically what the angle  should be
...
9 o )
vertical


250 N

Version 1
...
4

2 - 11

Moment of a Force

The moment of a force produces or tends to produce rotation about an axis normal to
the plane in which the force acts
...
There may be no
motion, but we can still determine the moment if the force tends to cause rotation
...

(Since we are dealing with coplanar forces, a point actually represents an axis
normal to the plane in which the forces are acting
...
d

Moment of F about O,

d

Moments are indicated as clockwise () or anticlockwise () sense, depending on
the direction in which the force tends to rotate the body about the given point or axis
...
7
Determine the moment of the 100 N force about points A, B, C and D
...
a

MD 

A

2m

100 N

MA 

30°
0

100cos30

D

2 mC

B 2m D

MC 
100sin30’s line of action
passes through A, B, C &
D, hence no moment
about these points
...
b

100 N
30°
A

2m

MB 

dC

B 2m D
dB

C

2m

dA

MD 

Alternative solution for Fig
...


line of action of the
100 N force

Mechanics I (ME1101/0101)

Version 1
...
5
Addition of Moments
If more than one force or its components act on a body, the resultant moment ( M ),
is the algebraic sum of all the moments acting about the same point
...

Example 2
...
(Resolve the force where necessary)
...
a

25 N

MA ( for 25N force)

= _________________



= _________________



MA

A

4m

B

25 N

Fig
...
5 m
D

D

MD ( for 100N force)

MD ( for 25N force)

30° 100 N

1
...
c



MD

MD ( for 100N force)

= _________________

MD ( for 25N force)

= _________________



= _________________



MD



MA

MC

F
100 N
3m
C

Fig
...
0

2m

3m

D

= _____________________________
= _____________________________

ME

= _____________________________

MF

= _____________________________

E

Mechanics I (ME1101/0101)

Equilibrium Conditions

2 - 13

Example 2
...

Given : AB = BC = 3 m; CD = 1
...

D
C

100 N
B

500 N
80°
A

Example 2
...
AB = 100 mm; BC = 160 mm
...
6 Determine the sum of moments of the given forces about the fixed end A of the
cantilever
...
2 m
BC = 0
...
0

2 - 14

2
...

Indicate the resolved components
...
99 Nm  , 953
...
6 m
BC = 0
...
2 m
BC = 0
...
a
2
...
b

Couples

Couples are commonly encountered in engineering
...
We often represent a couple with
a curved arrow, i
...
() or ()
A couple has the following characteristics:
1
...

2
...

3
...


b

a/2

F

a/2

For the die-holder shown, calculate
the moment of the couple about points
A, B, C & D
...
11
Calculate the moment of the 3 N
couple applied to the steering
wheel shown
...


Version 1
...
8

For the piping system shown below, calculate the moment
(a) due to the couple
...

( Ans : 17
...
18 Nm  )
A

AB = 0
...
8 m
CD = 0
...
9

Determine the sum of the moments about fulcrum B of the bell-crank due to
the applied forces
...
603 Nm  )
C

75°

AB = 0
...
18 m
130 N
100 N
kg
A

Mechanics I (ME1101/0101)

B

60°

Version 1
...
10 For the framework shown below, calculate the sum of moments of the 3 given
forces about point A
...

( Ans : 13
...
11 Determine the sum of moments about point A due to all forces and couple
acting on the 10 kg plate, which has its centre of gravity at G
...
8 m by 1
...

( Ans : 217
...
0

Mechanics I (ME1101/0101)

Equilibrium Conditions

2
...
The first
important step to analyzing a system is to isolate it or its component parts
...
Applied loads (forces & couple/moment) and
reactions (support forces/moments) are then put in
...
See illustrations below
...


Free body diagram of the
combined body is used when
forces at its supports need to be
analyzed, for e
...
R, the reaction
at the wheels
...

The correct sense of the reactions
can be confirmed through
calculations
...
g
...


F
(Hydraulic piston force)
Mechanics I (ME1101/0101)

Version 1
...
of unknowns

1
...

R

Direction is normal to
the supporting surface
...
Smooth surfaces

R
R

3
...


F

4
...


R
50o

rigid link

R

Version 1
...


50o

Mechanics I (ME1101/0101)

Equilibrium Conditions

Type of Support
5
...
of unknowns

X
Y

2 unknown rectangular

X

components of reaction
...
Rough Surfaces
X

2 unknown rectangular

Y

components of reaction
...


X

Y

2 unknown rectangular

7
...


X
M

Mechanics I (ME1101/0101)

Y

1 unknown fixing
(resisting) moment M

Version 1
...
12

Complete the free body diagram of the body to be isolated:

Body to isolate

a)
Bell
crank

Mechanical system

FBD of body

C

A
30 N

110°

B

30°

C

A
20 N

B

C

C

B

b)
Structure
ABCD

B
D

70°

D

38°

A

A
800 N

AB = 0
...
4 m; CD = 0
...
2 m; BC = 0
...
0

Mechanics I (ME1101/0101)

Equilibrium Conditions

2
...
In other words, resultant forces as well as moments equal zero
...

For a 2-dimensional or coplanar force system, the following conditions are
necessary and sufficient for complete equilibrium
...


Note : Only the first 2 conditions need to be used when considering the equilibrium
of a point where all forces meet i
...
concurrent force system
...
12 ( Equilibrium of coplanar concurrent forces )
Figure shows concurrent forces acting on a truss joint
...

( Hint: Resolve all forces along the directions A & B )
2 kN
5 kN

5cos30

30°

2
10

10 kN
5sin30
B
A
B
A

Mechanics I (ME1101/0101)

Version 1
...
13
Three cables supporting a weight D form a
concurrent force system at B
...


Equilibrium Conditions

A

C

45°

B

30°

D

5 kg

FABsin45

FBCsin30
FBCcos30

FABcos45
B
FBD

FBD
D
mg

Version 1
...
14
For the plane truss shown, find
the forces in the members AB & BE as assumed in the magnified view of joint B,
given the forces in members BC & BF
...

a)

B

5
...
125 m
FE = 6
...
0

2 - 24

Equilibrium Conditions

TUTORIAL (Equilibrium of Coplanar, Concurrent Forces)
2
...
5 kN and B = 2 kN are applied to the gusset plate shown below
...

( 1
...
5 kN )

B = 2 kN
C

30 °

A = 3
...
14

D

O

The members of a plane truss are connected to a gusset plate shown below
...

(See Example 2
...
3 kN , 10
...
0

5 kN

Mechanics I (ME1101/0101)

Equilibrium Conditions

2
...
If the length of the cable is 20 m and the deflection BD at
mid-point is 0
...

( 3
...
16 Shown below is a plane truss consisting of light metal rods pinned together at
their ends
...
Calculate
a)
the forces in members AD & AB pinned at the joint A
...
(See Example 2
...
5 N ; – 292
...

(292
...
5 N; 40º  )
C

D

40°
A
B

Mechanics I (ME1101/0101)

Version 1
...
17

Equilibrium Conditions

A 100 N force pulls at point B of a cable ABC as shown
...
(See example
2
...
25 N, 96
...
0

Mechanics I (ME1101/0101)

Equilibrium Conditions

2 - 27

Worked example ( Equilibrium of non-concurrent forces)
For the beam shown,
a)

b)

40 N

sketch a free body diagram of the
beam ABC, given that its weight is
negligible
...


B

YB

A

a)

Indicate the dimensions of the body
for computing the components and
moments of forces
...
2 m

40 N
C

XB
0
...


* Summing moments about the point where an unknown acts will
enable the other unknown to be found
...
3) – RA(0
...
96 N 

YB – RA – 40sin50
YB – 45
...
64
YB

=0
=0
= 76
...
71 N 

Fx = 0

RB =

2
YB  X 2
B

B = tan-1(

Mechanics I (ME1101/0101)

YB
)
XB

=

25
...
602 = 80
...
60
) = 71
...
71

Version 1
...
15
The 500 mm x 400 mm plate is subjected to forces as shown
...
Neglect the weight of the plate
...
16
Determine the reactions at A and B for the light beam shown below
...
0

3m

A

C

D

2m

Mechanics I (ME1101/0101)

B

Equilibrium Conditions

2 - 29

Example 2
...

b) calculate reactions at the fixed end A
...
5 m
BC = 1
...
9 m

20 N

A
C
B
FBD
D

Mechanics I (ME1101/0101)

Version 1
...
18
An awning structure is supported as shown
...
A load of 100 N hangs at C
...

the reaction at the pin support A
...
5 m
BC = 1
...
0

Mechanics I (ME1101/0101)

Equilibrium Conditions

2 - 31

Example 2
...
Determine
a)
b)

the reaction acting at the centre B of the pulley
...

1
...
5 m

C

B
A

30°
600 N

D

Solution: Pulley has to be isolated from the pin support at B
...
Note the opposing sense of the component
reactions at B on both the bodies
...
0

2 - 32

Equilibrium Conditions

TUTORIAL (Equilibrium of Coplanar, Non-concurrent, Forces)
2
...
Neglect
the mass of the beams
...

19 kN

i)
2m

2m

2m
B

5 kN

Ans ( 11 kN  , 3 kN  )

8 kN

2 kN

ii)

40 °
D

B

C

AB = CD = 0
...
3 m

4 kN

Ans ( 5
...
8 o

iii)

A

D

; 557 Nm  )

C

B
100 N
50 N
AB = 4
...
5 m
CD = 1
...
7 N 88
...
0

; 54
...
19 Calculate the tension in the chain at B and
the reaction at the pin support A
...
4 N, 1083
...
779°

)
B
D

70°

A

38°

800 N

AB = 0
...
4 m; CD = 0
...
20

An advertising sign of mass 10 kg is supported on a light beam AB as shown
...

magnitude and direction of the reaction at A
...
3o

30°

B
0
...
4 m

10 kg

Mechanics I (ME1101/0101)

Version 1
...
Determine
with the aid of a free body diagram,
a)
b)

the reaction at point A of the crate
...


AB = 5 m
BC = 2 m

A
G

Smooth wall

2
...
8782 kN ; 4
...
62°

)

Rough ground

B

2
...
It is supported
by a hinge at A and a pin-ended bar at B
which is anchored at the wall
...

( 219 N, 36
...
0

o

, 274
...
2 m
BC = 0
...
23 A car park gate control has a beam CB of negligible mass and a counter weight
C of 200 kg
...

(435
...
7o )

B
60°
3
...
0

2 - 36

2
...

Determine the reactions acting on the bracket at:
( 1
...
358 kN 58o

a) A
...


)

(128
...
1 N, 11
...
25

2 kN

For the bell crank lever shown, determine the:
a)
b)

reaction at R
...

80 N
P

75°

45o

Q

R

Version 1
...
3 m
QR = 0
...
26 A load of 3 kN acts on a light beam 2
...
The part of the cable between the pulley and the beam is vertical
...

Find the tension F in the cable required to hold the beam
in equilibrium
...
76 kN)
Determine the reaction at O on the beam
...
91 kN 83
...

(16
...
7 m

Mechanics I (ME1101/0101)

2
...
0

2 - 38

Equilibrium Conditions

2
...
Assuming that all the forces are coplanar,
a)
b)
c)
d)
e)

determine the tension in the cable;
sketch the free body diagram of the pulley alone;
sketch the free body diagram of the angle bracket ABC alone;
determine the magnitude and direction of the force exerted on the bracket
by the pin at C;
determine the reaction at A on the bracket
...
5 N , 849
...
0 N  )
Pulley 75 mm
Cable
C

500 mm

60O

50 kg

Angle
Roller
A

B
300 mm

Version 1
...
28 Figure shows a portable mobile crane lifting a cargo weighing half a tonne
...
Find:
(a) force in the hydraulic cylinder BD which acts at point B of the boom
...

C
Given: AC = 2
...
2 m
G
30°
AG = 0
...
80 kN ; 10
...
08o)

60°
D

Mechanics I (ME1101/0101)

Version 1
...
29 A mass of 10 kg is supported by two cables DB and DE as shown in the figure
...

a) Sketch a free body diagram of the point D and show that the tensions in
the cables DB and DE are 67
...
2 N respectively
...

( (i) Rc = 204
...
4 N,

17
...
6 m

A
60°

G

frictionless
roller

B
E

0
...
doc

Version 1

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