e-learningcourse on engineering mechanics” –concurrent
TRANSCRIPT
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E-Learningcourse on“Engineering Mechanics” –Concurrent
coplanar forces - Forces on a plane-Continuation& Concurrent Non-coplanar
Forces – Forces in spacePPT-3
Dr. Vela Murali,Ph.D.,Head& Professor i/c – Engineering Design Div.,
Mechanical Engineering Department,College of Engineering, Guindy,
Anna University, Chennai – 600 0251
By
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1. Quick Review of PPT – 1,2
2. Concurrent coplanar forces - Forces on aplane - Equilibrium conditions
3. Concurrent Non-coplanar forces – Forcesin space
2
CONTENTS
Course on “Engineering Mechanics” by Dr. Vela Murali
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3
•Overview about Engineering Mechanics
•Resolving of ‘n’ no. of concurrent coplanar
forces – Forces on plane – Straight Quadrant
and Inclined Quadrant approaches
•Example Problems
Review of PPT -1,2
Course on “Engineering Mechanics” by Dr. Vela Murali
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• Equilibrium of a Particle
i j( Fx) +( Fy) = 0
Fx = 0 and Fy = 0
Example 7
Course on “Engineering Mechanics” by Dr. Vela Murali
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Fx = 0, ( , +ve) 100 Cos (30) – F1 Cos (45) = 0
F1 = 122.7 N
Applying the second equilibrium condition
Fy = 0, ( , +ve) 100 Sin (30) + F1 Sin (45) – F2 = 0
Substituting F1,
F2 = 136.76 N
The three forces, 100 N, F1 and F2 are applied on the
particle such that the particle is under equilibrium.
Course on “Engineering Mechanics” by Dr. Vela Murali
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6
• Equilibrium of a Particle on an Inclined plane
For equilibrium of a particle on an inclined plane,the resultant R = 0
Falong the plane = 0 and FPerpendicular to the plane = 0
Example 8
Course on “Engineering Mechanics” by Dr. Vela Murali
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30
30500
Cos
SinF
30o
Falong the plane = 0 , ( , +ve )
F Cos(30) – 500 Sin(30) = 0
= 288.7 N
Applying second equilibrium equation
FPerpendicular to the plane = 0, ( , +ve)
R- F Sin(30) – 500 Cos(30) = 0
R = 577.4 N
30o
Course on “Engineering Mechanics” by Dr. Vela Murali
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Equilibrium of a Particle by force polygon
Three or more concurrent coplanar forces, which are acting on the particle, are such that the particle is being under equilibrium.
For this condition the force polygon, which is to be drawn to the scale according to the direction and magnitude of the system of the forces one after the other and is a closed one.
Applicability of Newton’s I – law - Equilibrium
Course on “Engineering Mechanics” by Dr. Vela Murali
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• Applicability of equilibrium of a particle –different engineering problems
• Many of engineering problems are subjected to concurrent coplanar forces and satisfy the conditions of static equilibrium
Either they straight plane problems (or) inclined plane problems.
A free body diagram is to be drawn at a point in the body, where the lines of actions of the concurrent forces pass through (called as particle) and representing the direction and magnitude of these forces.
Course on “Engineering Mechanics” by Dr. Vela Murali
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Body weight of ‘W’
C
W
Centroid of
the body
W
Centroid of
the body
Weight acting on a
horizontal plane
Weight acting on an Inclined
plane
Course on “Engineering Mechanics” by Dr. Vela Murali
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W
θ
θ
•Resolution and components of the force along edges of the Inclined quadrant
– Inclined Plane Problems
Dr. Vela Murali
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Newton’s III law – Reaction force - Equilibrium
Fig. Reactive force „R‟
at the Contact surface Fig. Reactive forces R1 &
R2 at two contact surfaces
W
RW
W1
R1=W1
W2 W2
W1
R2=W1+W2
Course on “Engineering Mechanics” by Dr. Vela Murali
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Reactive force R from an inclined surface due to
the body weight
Course on “Engineering Mechanics” by Dr. Vela Murali
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• Space diagram and Free Body Diagrams (FBD)
A
BC
O5O6
O3O2
36 cms
O1O4
Cylinders A, B and C have equal diameter and weight of 16 cm and 100 N respectively
Space diagram showing the physical conditions of the problem
Course on “Engineering Mechanics” by Dr. Vela Murali
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WB
RB/A
RB/2
RB/1
45o
B
Free Body Diagram (FBD) drawn at point ‘B’
Course on “Engineering Mechanics” by Dr. Vela Murali
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RA/BWA
45o
A
45o
RA/C
Free body diagram drawn at point „A‟
Course on “Engineering Mechanics” by Dr. Vela Murali
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WC
45o
C
Free Body Diagram drawn at point „A‟
RC/4
RC/A
RC/3
Course on “Engineering Mechanics” by Dr. Vela Murali
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Dr. Vela Murali
Problem 9Find the resultant of the a system of concurrent coplanar forces shown in Fig. 2.32 by using polygon law of graphical approach.
45o F1 = 500 NO
30o
F2 = 300 N
F3 = 400 N
F4 = 200 N60o
F5 = 100 N
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F2=300 N
F5=100 N
F4=200 N
200cos(30)F3=400 N 45o
30o
60o
300cos(45)
100cos(60)
F1=100 N
300sin(45)
200sin(30)
100sin(60)
Dr. Vela Murali
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20
Course on “Engineering Mechanics” by Dr. Vela Murali
For solution refer the Book on “Engineering Mechanics” By Vela MuraliPublished by Oxford University Press (2010)
Practice your self from the free body diagram
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Problem 10
A stone of weight 500 N as shown is supported against
to plane surface which is perpendicular to the 45o
inclined surface. What is the magnitude of the force that
is supporting the stone in the direction of the inclined
plane and the reaction force from the surface.
Dr. Vela Murali
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500N
45o
Dr. Vela Murali
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500 N500cos(45)
45o
Dr. Vela Murali
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45o
Falong the plane = 0 , ( , +ve )
RV - 500 Sin(45) = 0RV = 500 Sin(45) = 353.55 N
Applying second equilibrium equation
FPerpendicular to the plane = 0, ( , +ve)
RP- 500 Cos(45) = 0
45o
Dr. Vela Murali
45oRP = 353.5 N,
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25
Problem 11
Determine the magnitudes of F1 and F2 which
are holding the body of weight 6 kN
suspended from the string at point ‘O’ as
shown. Assume the pulley over which the
string passes is smooth.
Dr. Vela Murali
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6 kN
45o
30oF2
F1
O
Dr. Vela Murali
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F2
30o
6 kN
F1 cos(45)F1
F2 cos(30)F1 sin(45)
45o
F2 sin(30)
O
Dr. Vela Murali
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Fx ( , +ve) = 0, F2 cos(30) – F1 sin(45) = 0
F2= 0.816 F1 (1)
Fy ( , +ve) = 0, F2 sin(30) + F1 cos (45) - 6= 0 Substituting (1)
(0.816F1)sin(30) + F1 cos (45) = 6
F1= 5.379 kN (2)Substituting (2) in (1)
F2= 0.816 (5.379) = 4.39 kN
Dr. Vela Murali
30
4512Cos
SinFF
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Problem 12
Two cables are tied together at point ‘O’ and
loaded as shown. Determine the tension in
OO1 and OO2.
Dr. Vela Murali
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30
150 kg
80o
10oO
O1
O2
Dr. Vela Murali
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TOO2
10o
TOO1
80o
TOO1 sin(80)
OTOO2 cos(10)
TOO2 sin(10)
150 X 9.81 N
TOO1 cos(80)
Dr. Vela Murali
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Fx ( , +ve) = 0, TOO2 cos(10) – TOO1 cos(80) = 0
TOO2= 0.176 TOO1 (1)
Fy ( , +ve) = 0, TOO1 sin(80) – 150 (9.81) – TOO2 sin (10) = 0
Substituting (1)
TOO1= 1542.45 N, (2)Substituting (2) in (1)
TOO2= (0.176) (1542.45) = 271.47 N,
Dr. Vela Murali
10
8012Cos
CosTT OOOO
5.1471954.01OOT
80o
10o
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Problem 12
A string of length 20 cms is attached to a
point A on a smooth vertical wall and to a
point C on the surface of the sphere of radius
10 cms. The sphere whose weight is 300 N
hangs in equilibrium against the wall. Find
the tension in the string and the reaction of
the wall.
Dr. Vela Murali
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A
O
300 N
20 cms
C 10 cms
Dr. Vela Murali
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TCA
70.53o
TCA sin(70.53)
ORV
300 N
TCA cos(70.53)
Dr. Vela Murali
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36
Course on “Engineering Mechanics” by Dr. Vela Murali
For solution refer the Book on “Engineering Mechanics” By Vela MuraliPublished by Oxford University Press (2010)
Practice your self
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Problem 13
A spherical ball of weight ‘W’ rest in ‘V’ shaped surface whose sides are inclined at angles and to the horizontal. Find the pressure on each side of the ‘V’ shaped surface
Dr. Vela Murali
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21O
Dr. Vela Murali
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39
R2
O R2 sinβ
W
Drawing FBD @ Point ‘O’
βR1
R1 sin
R2 cos β
R1 cos
Dr. Vela Murali
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Fx ( , +ve) = 0, R1 sin( ) – R2 sin(β) = 0
(1)
Fy ( , +ve) = 0, R1 cos( ) – W + R2 cos(β) = 0 Substituting (1)
, (2)
Substituting (2) in (1)
,
Dr. Vela Murali
Sin
SinRR 21
Sin
WSinR2
β
Sin
WSinR1
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Problem 14
Determine the mass that must be
supported at ‘P’ and the angle ‘α’ of
the cord in order to hold the system in
equilibrium.
Dr. Vela Murali
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40 kg
80 kgP
45o
Dr. Vela Murali
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43
Course on “Engineering Mechanics” by Dr. Vela Murali
For solution refer the Book on “Engineering Mechanics” By Vela MuraliPublished by Oxford University Press (2010)
Practice your self
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44
Determine the required length of cord PR in Fig. so that the10 kg block suspended in the position shown. The un-deformed length of the spring PQ is 0.5 m, and the has the stiffness 250 N/m.
Problem 15
P
10 kg
45O
1.5 m
kPQ= 250 N/m
Q
R
P
Dr. Vela Murali
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45
Drawing FBD at point ‘P’
TPR
TPRcos(45)
TPRsin(45)
TPQ
10x9.81 N
45o
Dr. Vela Murali
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46
Applying equilibrium conditions
,1.98
)1(
0)45cos(
,,0
,7.138
01.98)45sin(
,,0
NT
ngSubstituti
TT
veF
NT
T
veF
PQ
PQPR
x
PR
PR
y
θ=45O )1(
Dr. Vela Murali
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47
nDeformatiolengthUndeformedlengthDeformed
m
KStiffness
TforceTensionPQinnDeformatio
PQ
PQ
PQ
3924.0250
1.98
,
m8924.03924.05.0
Dr. Vela Murali
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48
45o
R
P0.6076m
m8593.0PR
PR
6076.0
PR
8924.05.145cos
Dr. Vela Murali
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Particle Mechanics -Concurrent Non-coplanar forces - Forces in space
49
Dr. Vela Murali
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• Concurrent Non-coplanar forces
x
y
O
F1
F3
F2
z
F1 = Fx1i + Fy1 j + Fz1 k
F2 = Fx2i + Fy2 j + Fz2 k
F3 = Fx3i + Fy3 j + Fz3 k
R
θx1
θy1
θz1
Dr. Vela Murali
Fx1 =F1 cos(θx1), Fy1 =F1 cos(θy1), Fz1 =F1 cos(θz1) etc……
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51
Dr. Vela Murali
1222
zyx CosCosCos
321
321
321
zzzz
yyyy
xxxx
FFFF
FFFF
FFFF
222
zyx FFFR
R
F
R
F
R
F z
z
y
y
x
x
111 coscos,cos
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• Problem 16
52
A force of 200 kN is acting at a point makingan angle of 120o and 60o with x- and y- axesrespectively. Find the components of theforce and express the force as a vector.
Dr. Vela Murali
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53
Dr. Vela Murali
Solution
o
z
z
z
z
valuepositiveTaking
Cos
Cos
CosCosCos
45,
7071.0
5.0
160120
2
222
1222
zyx CosCosCos
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54
Dr. Vela Murali
kjiF
NCosFCosF
NCosFCosF
NCosFCosF
zz
yy
xx
4.141100100
4.141)45(200
100)60(200
100)120(200
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A force acts at a point ‘O. Find the magnitude and the direction of the force.
kjiF
50100200
• Problem 17
Dr. Vela Murali
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56
•Solution
Fx = 200 units, Fy = -100 units
Fz = -50 units
Magnitude of the force =
UnitsFFF zyx 13.229
222
o
z
o
y
o
x 6.102,87.115,2.29
Dr. Vela Murali
R
F
R
F
R
F z
z
y
y
x
x
111 coscos,cos
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57
Dr. Vela Murali
• Problem 18
A force of 300 kN acts along a line joining two points A (1,2,3) and B (2, 4,-5). Determine its components and express it as vector.
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58
Course on “Engineering Mechanics” by Dr. Vela Murali
For solution refer the Book on “Engineering Mechanics” By Vela MuraliPublished by Oxford University Press (2010)
Practice your self
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59
Dr. Vela Murali
Problem 19
A force of 250 kN acts along a line joining two points P (-3,4,6) and Q(5,-5,8). Determine the component of the force along the line joining two points A (3,2,1) and B (4,3,5).
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60
•Solution
2.12
2
9
8
222
PQPQPQPQ
PQPQ
PQPQ
PQPQ
dzdydxr
zzdz
yydy
xxdx
To find out the inclination of the line ‘AB’ with respect to x, y& z axis
Dr. Vela Murali
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61
Dr. Vela Murali
o
PQ
PQ
PQz
o
PQ
PQ
PQy
o
PQ
PQ
PQx
r
dz
r
dy
r
dx
56.80cos
5.137cos
02.49cos
1
1
1
To find out the inclination of the line ‘AB’ Whose coordinates are A(3,2,1)& B(4,3,5) with respect to x, y& z axis
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62
24.4
4
1
1
222
ABABABAB
ABAB
ABAB
ABAB
dzdydxr
zzdz
yydy
xxdx
o
AB
ABABx
r
dx36.76cos 1
Dr. Vela Murali
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63
xP
F=250 kN
z
B Q
A
Component of force along the line ‘AB’
kNFF
FF
ABzPQzAByPQy
ABxPQxAB
85.33))cos()cos(())cos()cos((
))cos()cos((
Dr. Vela Murali
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64
kNkjiF
3020601
kNkjiF
1040202
kNjiF
25153
The following forces act at a point.
Find the resultant and its direction.
Problem 20
Dr. Vela Murali
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65
Course on “Engineering Mechanics” by Dr. Vela Murali
For solution refer the Book on “Engineering Mechanics” By Vela MuraliPublished by Oxford University Press (2010)
Practice your self
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66
F1 = 400 N is acting along the line joining the two points of (0,0,0) & (x1, y1, z1) = (4,-1,5) units; F2 = 300 N is acting along the line joining the two points of (0,0,0) & (x2, y2, z2) = (5,-3,-5) units; F3 = 500 N is acting along the line joining the two points of (0,0,0) & (x3, y3, z3) = (-6, -6, -5) units. Find the resultant and its direction.
Problem 21
Dr. Vela Murali
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67
Course on “Engineering Mechanics” by Dr. Vela Murali
For solution refer the Book on “Engineering Mechanics” By Vela MuraliPublished by Oxford University Press (2010)
Practice your self
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68
Problem 22
A force F acts at the origin of a coordinatesystem in a direction defined by the anglesθy = 70o and θz = 60o. If the component ofthe force F along ‘x’ direction equals to– 180 N.Determine (i) the angle θx (ii) themagnitude of the force ‘F’ (iii) thecomponents of the force ‘F’ along ‘y’ and ‘z’directions (iv) The components of the force‘F’ along a line through the origin and thepoint (2,2,2).
Dr. Vela Murali
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•Solution
o
x
x CosCosCos
7.142
16070 222
θy = 70o , θz = 60o
X- component of the force = -180 N
NF
FCosF xx
23.226
180
Dr. Vela Murali
1222
zyx CosCosCos
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70
NFCosF
NFCosF
zz
yy
12.113
37.77
To find out the inclination of the line ‘OP’ Whose coordinates are O(0,0,0)& P(2,2,2) with respect to x, y& z axis
unitsdzdydxr
zzdz
yydy
xxdx
OPOPOPOP
OPOP
OPOP
OPOP
46.3
2
2
2
222
Dr. Vela Murali
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71
o
OP
OPOPz
o
OP
OPOPy
o
OP
OPOPx
r
dz
r
dy
r
dx
7.54cos
7.54cos
7.54cos
1
1
1
Component of force along the line ‘OP’
Dr. Vela Murali
NFF
FF
OPzzOPyy
OPxxOP
11.6))cos()cos(())cos()cos((
))cos()cos((
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72
Equilibrium of the Particle applied with a system of ConcurrentNon-coplanar forces
Fx = 0 ( , +ve)
Fz = 0 ( , +ve) and
( , +ve) Fy = 0
Dr. Vela Murali
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73
650 mm
300 mm
200 mm
P
Q
z
x
SR
O
OS=500 mmOQ=600mm
A container of weight W=1500 N is supportedby three cables as shown in Fig. Determine the tension in each cable.
Problem 23
Dr. Vela Murali
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74
To find out the inclination of the line ‘PR’ whose coordinates are P(0,-650,0)& R(200,0,-300) with respect to x, y& z axis
mmdzdydxr
mmzzdz
mmyydy
mmxxdx
PRPRPRPR
PRPR
PRPR
PRPR
3.743
300
650
200
222
•Solution
Dr. Vela Murali
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75
o
PR
PRPRz
o
PR
PRPRy
o
PR
PRPRx
r
dz
r
dy
r
dx
8.113cos
02.29cos
39.74cos
1
1
1
)1(
403.0cos
87.0cos
27.0cos
PRPRzPRPRz
PRPRyPRPRy
PRPRxPRPRx
TTF
TTF
TTF
Dr. Vela Murali
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76
To find out the inclination of the line ‘PS’ whose coordinates are P(0,-650,0)& S(-500,0,0) with respect to x, y& z axis
mmdzdydxr
mmzzdz
mmyydy
mmxxdx
PSPSPSPS
PSPS
PSPS
PSPS
1.820
0
650
500
222
Dr. Vela Murali
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77
o
PS
PSPSz
o
PS
PSPSy
o
PS
PSPSx
r
dz
r
dy
r
dx
90cos
56.37cos
6.127cos
1
1
1
)2(
0cos
79.0cos
61.0cos
PSzPSPSz
PSPSyPSPSy
PSPSxPSPSx
TF
TTF
TTF
Dr. Vela Murali
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78
To find out the inclination of the line ‘PQ’ whose coordinates are P(0,-650,0)& Q(0,0,600) with respect to x, y& z axis
mmdzdydxr
mmzzdz
mmyydy
mmxxdx
PQPQPQPQ
PQPQ
PQPQ
PQPQ
59.884
600
650
0
222
Dr. Vela Murali
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79
o
PQ
PQ
PQz
o
PQ
PQ
PQy
o
PQ
PQ
PQx
r
dz
r
dy
r
dx
3.47cos
3.137cos
90cos
1
1
1
)3(
678.0cos
735.0cos
0cos
PQPQzPQPQz
PQPQyPQPQy
PQxPQPQx
TTF
TTF
TF
Dr. Vela Murali
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80
Components of force acting along the line PO
)4(
0
1500
0
POz
POy
POx
F
NF
F
For equilibrium it must satisfy the following Equations,
i) Fx = 0, ( , +ve) ii) Fy = 0, ( , +ve)
iii) Fz = 0, ( , +ve)
Dr. Vela Murali
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81
From equations (1), (2), (3) and (4)
)9(75.668
)6.()7.(
)8(11.1125
)5.()7.(
)7(497
)6.(&)5.(
00678.00403.00
)6(756.28.2040
)5.(
01500735.079.087.00
500061.027.00
NT
EqinEqngSubstituti
NT
EqinEqngSubstituti
NT
EqEqngSubstituti
TTF
TT
EqngSubstituti
TTTF
TTF
PQ
PR
PS
PQPRz
PSPQ
PQPSPRy
PSPRx
Dr. Vela Murali
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82
ooo
zyxPQ
ooo
zyxPR
ooo
zyxPS
NT
NT
NT
3.47,3.137,90,,,75.668
8.113,02.29,39.74,,,11.1125
90,56.37,6.127,,,84.497
Tensions are
Dr. Vela Murali
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83
R
5 m
5 m
7 m hP
Q
S
5 m
4 m
2.5 m
3 m
3 m4 m
z
xy
Determine the tension developed in the three cables required to support the traffic light, which has a mass of 30 kg. Take h = 4 m
Problem 24
Dr. Vela Murali
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84
Course on “Engineering Mechanics” by Dr. Vela Murali
For solution refer the Book on “Engineering Mechanics” By Vela MuraliPublished by Oxford University Press (2010)
Page 111, Example 3.17
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Reference:
“Engineering Mechanics”
by
Vela MuraliPublished by
Oxford University
Press (2010)