engineering mechanics: module 1 - vidyarthiplus
TRANSCRIPT
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A N A N T H I T H A
D E P T . O F C I V I L E N G I N E E R I N G
F I S A T
Engineering Mechanics: Module 1
1/8/2013
1
Engineering Mechanics, Module 1
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Pre read
1/8/2013Basic Civil Engineering, Module 4
2
Dear Student,
Please use this document only as an aid tosupplement the lectures & the lecture notes.
Please note that the format conversion has resultedin the removal of animations .
Happy & Fun learning !!!
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1/8/2013Engineering Mechanics, Module 1
X
Y
Z
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Learning Objectives
1/8/2013Engineering Mechanics, Module 1
4
Force systems in Space
Introduction to Vector approach
Elements of vector algebra
Position Vector
Moment of a force about a point & axis
Resultant of forces
Equilibrium of forces in space using vector approach
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Forces in Space
1/8/2013Engineering Mechanics, Module 1
O
Y
Z
X
Fx = Fcos ƟxFy = Fcos ƟyFz = Fcos Ɵz
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(OE)2 = (OD)2 + (DE)2
1/8/2013Engineering Mechanics, Module 1
O
Y
Z
X
D
A
B
= (OA)2 + (OC)2+ (DE)2
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(OE)2 = (OD)2 + (DE)2
1/8/2013Engineering Mechanics, Module 1
O
Y
Z
X
D
A
B
= (OA)2 + (OC)2+ (DE)2
= (Fx)2 + (Fz)2+ (Fy)2
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(OE)2 = (F)2
1/8/2013Engineering Mechanics, Module 1
O
Y
Z
X
D
A
B
(F)2 = (Fx)2 + (Fz)2+ (Fy)2
Fx = Fcos Ɵx ; Fy = Fcos Ɵy; Fz = Fcos Ɵz
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A force of magnitude 5 kN makes 30o, 50o &
100o with X, Y & Z axes respectively. Find
the magnitude of its components along X, Y
& Z axes
1/8/2013Engineering Mechanics, Module 1
Ɵx = 30o ,Ɵy = 50o ,Ɵz = 100o
Fx = 4.33 kN, Fy = 3.21 kN, Fz = -0.868 kN
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Direction cosines
1/8/2013Engineering Mechanics, Module 1
10
The cosines of Ɵx, Ɵy & Ɵz are known as direction cosines of the force F
Denoted by l, m & n
l = cos Ɵx
m = cos Ɵy
n = cos Ɵz
l2 + m2 + n2 =1
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A force acts at the origin in a direction defined by the
angles Ɵy = 60o & Ɵz= 45o. Taking the component of
the force along X direction as -800N, (1) find the
value of Ɵx. (2) find the other components and the
value of the force
1/8/2013Engineering Mechanics, Module 1
Fcos Ɵx = -800 N
Ɵy = 60o & Ɵz= 45o
l = cos Ɵx , m = cos Ɵy & n = cosƟz
l2 + m2 +n2 =1 Ɵx =120
Fx = Fcos Ɵx F =1600 N
Fy= 800 N, Fz = 1131.37 N
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F passes through the origin and the co-ordinates of a point along the line of action of F is known
1/8/2013Engineering Mechanics, Module 1
O
Y
Z
X
D
B
dydx
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d2 = (dx)2 + (dy)2+ (dz)2
Fx = (F X dx) /d ; Fy = (F X dy)/ d ; Fz= (F X dz)/d
1/8/2013Engineering Mechanics, Module 1
O
Y
Z
X
D
B
dydx
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Calculate the components of force 1000 Nshown in figure along X, Y & Z co-ordinates.
1/8/2013Engineering Mechanics, Module 1
O
Y
Z
X
40 mm
120 mm
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Co-ordinate of A is (120, 40,50) dx =120, dy
=40, dz = 50
1/8/2013Engineering Mechanics, Module 1
O
Y
Z
X
40 mm
120 mm
A
Distance of A from the origin, d = sqrt ( 1202+ 402+ 502)
= 136.01 mm
Fx = 882.29 N, Fy = 294.1 N, Fz = 367.62 N
d2 = (dx)2 + (dy)2+ (dz)2
Fx = (F X dx) /d ; Fy = (F X dy)/ d ; Fz= (F X dz)/d
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When A & B are 2 points on the line of actionof F, neither of which is at the origin
1/8/2013Engineering Mechanics, Module 1
Y
Z
X
A (xa,ya,za)
B (xb,yb,zb)
dx = (xb- xa)dy = (yb-ya)dz = (zb- za)d2 = dx2 + dy2 +dz2
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A force has a line of action that goes through the
coordinates (2,4,3) & (1, -5,2). If the magnitude of
the force is 100 N, find the components of the force
and its angles with the axes
1/8/2013Engineering Mechanics, Module 1
dx = -1 , dy = -9 , dz = -1, d = 9.11
Fx = -10.976 N, Fy = -98.792 N, Fz = -10.976 N
Ɵx = 96.30, Ɵy = 171.085, Ɵz = 96.30
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Unit Vector
1/8/2013Engineering Mechanics, Module 1
18
Unit Vector Vector having unit length
Unit Vector = (Force Vector)/ Magnitude of force vector
= F/ |F|
Unit vectors along X, Y &Z i, j, k
Vector F= Fxi + Fyj + Fzk
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A force has a line of action that goes through the
coordinates (2,4,3) & (1, -5,2). If the magnitude of
the force is 100 N, find the components of the force
and its angles with the axes. Find the force vector.
1/8/2013Engineering Mechanics, Module 1
Fx = -10.976 N, Fy = -98.792 N, Fz = -10.976 N
Force Vector is F = Fxi + Fyj + Fzk
F = -10.976 i – 98.792 j -10.976k
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A force has a line of action that goes through the
coordinates (2,4,3) & (1, -5,2). If the magnitude of
the force is 100 N, find the components of the force
and its angles with the axes. Find the force
vector in terms of unit vectors
1/8/2013Engineering Mechanics, Module 1
The vector joining the points =
(1-2)i + (-5-4)j + (2-3)k
Unit vector along this vector = (-i -9j-k)/ 9.11
Unit vector = F/|F|Hence F = Unit Vector X |F|
= ((-i -9j-k)/ 9.11 ) X 100
= -10.976i – 98.79j – 10.976k
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A force has a line of action that goes throughpoints A(4,2,5) & point B(12,4,6). If themagnitude of the force in the direction of AB is100N, find the force vector in terms of unit vector
1/8/2013Engineering Mechanics, Module 1
Practice Problem
Vector joining A & B = 8i + 2 j + kUnit vector along AB = 8i + 2 j + k/ 8.30
Force vector = unit vector X magnitude of
the force
= (8i + 2 j + k/ 8.30) X 100
=96.38i + 24.096j + 12.04k
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A force P acts as shown in fig. Express theforce P as a vector. Magnitude of force P is100N. Length of the side of the cube is 2 cm.
1/8/2013Engineering Mechanics, Module 1
A
B
(0, 2, 2)
(2, 0,0)
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Express the force P as a vector. Magnitude of forceP is 100N. Length of the side of the cube is 2 cm.
1/8/2013Engineering Mechanics, Module 1
A
B
(0, 2, 2)
(2, 0,0)
Vector joining A&B is 2i-2j-2k, |AB| = 3.46Unit vector along AB = 0.577i -0.577j-0.577k
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Express the force P as a vector. Magnitude of forceP is 100N. Length of the side of the cube is 2 cm.
1/8/2013Engineering Mechanics, Module 1
A
B
(0, 2, 2)
(2, 0,0)
Force Vector = 57.7i -57.7j – 57.7k
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A vertical post guy wire is anchored by means of abolt at A as shown in fig. The tension in the wireis 2000N. Determine, (1) the components of theforce acting on the bolt (2) the angles definingthe direction of the force and (3) the force vector
1/8/2013Engineering Mechanics, Module 1
80
m
40 m A
B
Coordinates of A (40, 0, -30)Coordinates of B (0,80,0)Vector joining A&B is -40i+80j+30k|AB| = 94.33Unit vector in the direction of AB = -.424i + 0.848 j +0.318 k
Z
X
Y
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1/8/2013Engineering Mechanics, Module 1
80
m
40 mA
B
The force in the bolt is the same as the tension in the wire
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1/8/2013Engineering Mechanics, Module 1
80
m
40 mA
B
Hence the Force vector is-848i + 1696j +636kDirection of forceƟx = 115.09o
Ɵy = 32o
Ɵz = 71.46o
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Vector addition
1/8/2013Engineering Mechanics, Module 1
The sum of the vectors can be obtained by adding the respective components of the vectors
If A= Axi + Ayj+ Azk &
B = Bxi + Byj+ Bzk, then,
A+B = B+A = (Ax+Bx)i + (Ay+By)j + (Az+Bz)k
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Find the unit vector in the direction of theresultant of vectors A = (2i-j+k), B= (i+j+2k) ,C= (2i+2j+4k)
1/8/2013Engineering Mechanics, Module 1
Resultant R = A+B+C
= 5i + 2j + 7k
Unit Vector = R/|R|
= 0.57i + 0.23j + 0.79k
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Resultant of forces in space
1/8/2013Engineering Mechanics, Module 1
The horizontal component of R= Rx = ∑Fx
Ry = ∑Fy
Rz= ∑Fz
R = sqrt ( Rx2+ Ry2 + Rz2)
Cos Ɵx = Rx/ RCos Ɵy = Ry/RCos Ɵz = Rz/R
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4 forces act on a particle as shown in fig.Determine the resultant of the forces in vectorform
1/8/2013Engineering Mechanics, Module 1
30o70o
140 N
120 N100 N
100 N
R= (140 cos 30-100cos70+100cos15)i + (140sin30+100sin70-120-100sin15)j=183.63i +18.09j
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4 forces 30 kN, 20 kN, 25 kN & 100kN arerespectively directed through the points whose co-ordinates are A(2,1,5), B(3,-1,4) C(-3,-2,1) &D(4,1,-2). If these forces are concurrent at theorigin, calculate the resultant of the forces invector form
1/8/2013Engineering Mechanics, Module 1
Practice
Problem
90.01i +10.01j + 6.68 k
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2 cables ‘AB’ & ‘AC’ are attached at A as shownin fig. Determine the resultant of the forcesexerted at A by the 2 cables, if the tension is2000 N in cable AB & 1500 N in the cable AC.Also calculate the angles that this resultantmakes with the X, Y & Z axes
1/8/2013Engineering Mechanics, Module 1
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1/8/2013Engineering Mechanics, Module 1
Z
Z
XO A
Y
B
C
62
5052
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1/8/2013Engineering Mechanics, Module 1
Z
Z
XO A
Y
B
C
62
5052
Co-ordinates of A (52,0,0)Coordinates of B (0, 50,40)Co-ordinates of C (0, 62,-50)
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1/8/2013Engineering Mechanics, Module 1
Unit vector along AB = -0.63i + 0.606j + 0.485k
FAB = -1260i + 1212 j +970k
Unit vector along AC = -0.546i +0.6518j -0.5256k
FAC= -819i +977.7j -788.4k
R = -2079i +2189.7j +181.6k
|R| = 3024.39
Ɵx = 133.42o
Ɵy = 43.61o
Ɵz = 86.56o
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A post is held in vertical position by 3 cables AB,AC &AD as shown the fig. If the tension in thecable AB is 40 N, calculate the required tension inAC& AD so that the resultant of the 3 forcesapplied at A is vertical
1/8/2013Engineering Mechanics, Module 1
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1/8/2013Engineering Mechanics, Module 1
Y
Z
X
X
Z
Y
DC
A
B
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1/8/2013Engineering Mechanics, Module 1
Y
Z
X
X
Z
Y
DC
A
B
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1/8/2013Engineering Mechanics, Module 1
A (0,48, 0)
B(16,0,12)
C(16,0,-24)
Vector along AB = 16i-48j+12k
Unit Vector along AB = 0.307i-0.923j+0.23k
Vector along AC = 16i-48j-24k
Unit Vector along AC = 0.2857i-0.857j-0.4257k
Vector along AD = -14i-48j
Unit Vector along AD= -0.28i-0.96j
FAB = 12.28i -36.92j +9.2k
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1/8/2013Engineering Mechanics, Module 1
Given , Rx = 0 & Rz= 0
Rx=0 12.28 +0.287 FAC – 0.28 FAD = 0
Rz= 0 9.2- 0.42587 FAC= 0
FAC = 21.47N
FAD = 65.86 N
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Equilibrium of forces in space
1/8/2013Engineering Mechanics, Module 1
42
For equilibrium, the components Rx, Ry & Rz must be zero
∑Fx= Rx = 0,
∑Fy = Ry= 0 &
∑Fz= Rz= 0
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A horizontal force ‘P’ normal to the wallholds the cylinder in the position shown infig. Determine the magnitude of P and thetensions in the cables ‘AB’ & ‘AC’
1/8/2013Engineering Mechanics, Module 1
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1/8/2013Engineering Mechanics, Module 1
Z
A
Y
X
Z
2m
1m
Mass -250 kg
P
14
m
B
C
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1/8/2013Engineering Mechanics, Module 1
Z
A
Y
X
Z
2m
1m
Mass =250 kgP
14
m
B
C (0,14,-12)
(0,14,9)
(1,2,0)
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1/8/2013Engineering Mechanics, Module 1
Unit vector along AB= -0.06i +0.798j +0.5988k
Unit vector along AC = -0.0588i+ 0.705j-0.705k
Since the cylinder is in equilibrium, ∑Fx= 0, ∑Fy=0 &
∑Fz = 0
∑Fy=0 0.798 FAB + 0.705 FAC -250X9.81= 0
∑Fz=0 0.598 FAB - 0.705 FAC = 0
∑Fx=0 P- 0.06 FAB - 0.058 FAC = 0
P = 193.02 N, FAB = 1755.47 N, FAC=1491.48 N
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A tripod supports a load of 2 kN as shown in fig.
A, B & C are in XZ plane. Find the forces in the 3
legs of the tripod
1/8/2013Engineering Mechanics, Module 1
Z
A
Z
Y
X
D
C
B
1.2 m
1 m
1.8
m
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1/8/2013Engineering Mechanics, Module 1
Z
A
Z
Y
X
D
C
B
1.2 m
1 m
1.8
m
A (1.2, 0,0) ; B (0, 0,1.2) ; C(-1, 0, -0.8);
D(0,1.8,0)
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1/8/2013Engineering Mechanics, Module 1
A (1.2, 0,0) ; B (0, 0,1.2) ; C(-1, 0, -0.8);
D(0,1.8,0)
FAD= FAD( -0.55i + 0.833j)
FBD= FBD(+0.833j – 0.55k)
FCD = FCD(+0.452i + 0.815j + 0.362k)
∑Fx= 0 , ∑Fy= 0 & ∑Fz= 0-0.556 FAD + 0 FBD + 0.453 FCD = 0
0.333 FAD + 0.833 FBD + 0.815 FCD = 0
0 FAD – 0.556 FBD + 0.362 FCD = 0
FAD = 0.801 kN
FBD = 0.641 kN
FCD = 0.983 kN
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Dot Product
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50
A . B = |A||B| cos Ɵ, where Ɵ is the angle between
the vectors
Dot product Scalar quantity
A = Axi + Ayj + Azk
B = Bxi + Byj + Bzk
A . B = AxBx + AyBy + AzBz
Angle between A&B , cosƟ = A.B/ (|A||B|)
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Dot Product
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Application
Work done by a force F in causing a displacement ‘s’ is given by
W = Fs cos Ɵ
W = F . s
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The point of application of a force F(5i+10j-15k)is displaced from point A(i+oj+3k) to the point(3i-j-6k). Find the work done by the force
1/8/2013Engineering Mechanics, Module 1
W = F.s
s = 2i-j-9k
W = 135 Nm
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Dot Product
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Another Application to find the component or projection of a force along a line
F = Fxi+ Fyj +Fzk
Projection of F along a line AB = F. Unit vector along AB
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A force F= 3i-4j+12k acts at a point A(1,-2,3).
Compute the component of F along line AB if B
has co-ordinates (2,1,2)
1/8/2013Engineering Mechanics, Module 1
Vector along AB = i+3j-k
Unit vector along AB = 0.301i+0.904j-0.301k
Component of F along AB = (3i-4j+12k) .
(.301i+0.904j-0.301k)
= -6.325.
To convert this in vector form, -6.325(0.301i+0.904j-
0.301k)
= -1.895i-5.717j+1.903k
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Cross Product
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A x B = |A| |B| sin Ɵ
A x B =
A x B = i(AyBz – AzBy) – j(AxBz – AzBx) + k( AxBy –AyBx)
i j k
Ax Ay Az
Bx By Bz
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Moment of a force in space
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Moment of a force F about a point is given by
M = r x F where r = xi + yj + zk; r is the position vector
M = r x F =
M = i(yFz-zFy) – j(xFz- zFx) + k (xFy- yFx)
Mx = yFz-zFy ; My= -xFz+zFx ; Mz = xFy- yFx
Mx,My& Mz are the moments about the X,Y & Z axes
i j k
x y z
Fx Fy Fz
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Position Vector
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Position Vector of a point passing through the origin
A (x, y, z)
X
Y
Z
x
y
r = xi + yj + zkMagnitude of position vector r = sqrt (x2+ y2+z2)
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Position Vector of a point wrt another point
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Position vector of point A, wrt point B
rAB is given by the line BA
If A (xa,ya,za) & B(xb,yb,zb) then
rAB is given by line BA (xa-xb)i + (ya-yb)j+ (za-zb)k
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A force F= 2i+ 3j-4k is applied at the pointA(1, -1,2).Find the moment of the force abouta point B(2,-1,3).
1/8/2013Engineering Mechanics, Module 1
A(1, -1, 2) B(2, -1,3)
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1/8/2013Engineering Mechanics, Module 1
M = r x F
r = Position vector of the point A wrt point B = (1-2)i + (-1- -1)j + (2-3)k
= -i + 0j –k
Moment of force at A about B = r x F
M = 3i -6j -3k
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A force of magnitude 44 N acts through the point A(4,-1,7) in the direction of vector 9i+6j-2k. Find the moment of the force about the point O(1,-3,2)
1/8/2013Engineering Mechanics, Module 1
Unit vector in the direction of 9i+6j-2k is 0.818i + 0.545j + 0.181k
Force vector, F = 36i +24j -8k
Position vector of A wrt point O is 3i+2j+5k
M = -136i + 204j
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Moment of a force about an axis
1/8/2013Engineering Mechanics, Module 1
O
L
Y
Z
X
(1) Calculate the Moment of F
about O= Mo= r x F
(2) Calculate the component (projection) of Mo on the line OLMOL = Mo .Unit vector along OL
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Find the moment of a force F(20i+30j) kNacting at a point with position vector (6i+10j)about the line passing through the pointsB(6i+10j-4k) and C(-4i+5j-2k)
1/8/2013Engineering Mechanics, Module 1
(1) Calculate the Moment of F about point B.
(2) Find the projection of this moment MB about the line BC
(1) Moment of F about B
=Position vector of A wrt B x F
= 4k x (20i+30j) = -120i +80j
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1/8/2013Engineering Mechanics, Module 1
(1) Moment of F about point B = -120i +80j
(2) Find the projection of this moment MB about the line BC
Vector along BC = -10i-5j+2k
Unit vector along BC = -0.88i-0.44j+0.176k
MBC = (-120i +80j). (-0.88i-0.44j+0.176k)
= 70.4 kNm