Forces in Space

7/27/2019 Forces in Space

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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


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 vectorapproach


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Forces in Space


O

Y

Z

X

Fx = Fcos x

Fy = Fcos yFz = Fcos z


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(OE)2 = (OD)2 + (DE)2


O

Y

Z

X

D

A

B

= (OA)2 + (OC)2+ (DE)2


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(OE)2 = (OD)2 + (DE)2


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


O

Y

Z

X

D

A

B

(F)2 = (Fx)2 + (Fz)2+ (Fy)2

Fx = Fcos x ; Fy = Fcos y; Fz = Fcos z


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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


x = 30o ,y = 50o ,z = 100o

Fx = 4.33 kN, Fy = 3.21 kN, Fz = -0.868 kN


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Direction cosines


10

The cosines ofx, y & z are known asdirection cosines of the force F

Denoted by l, m & n

l = cos x

m = cos y

n = cos z l2 + m2 + n2 =1


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A force acts at the origin in a direction defined by the

angles y = 60o & z= 45o. Taking the component of

the force along X direction as -800N, (1) find the

value ofx. (2) find the other components and the

value of the force


Fcos x = -800 Ny = 60o & z= 45o

l = cos x , m = cos y & n = coszl2 + m2 +n2 =1 x =120

Fx = Fcos x F =1600 N

Fy= 800 N, Fz = 1131.37 N


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F passes through the origin and the co-ordinatesof a point along the line of action of F is known


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


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.


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


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


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 forceand its angles with the axes


dx = -1 , dy = -9 , dz = -1, d = 9.11

Fx = -10.976 N, Fy = -98.792 N, Fz = -10.976 N

x = 96.30, y = 171.085, z = 96.30


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Unit Vector


18

Unit VectorVector 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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the force is 100 N, find the components of the forceand its angles with the axes. Find the force vector.


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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the force is 100 N, find the components of the force

and its angles with the axes. Find the forcevector in terms of unit vectors


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


PracticeProblem

Vector joining A & B = 8i + 2 j + kUnit vector along AB = 8i + 2 j + k/ 8.30Force 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 is

100N. Length of the side of the cube is 2 cm.


A

B

(0, 2, 2)

(2, 0,0)

E h f P M i d f f


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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.


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

E h f P M i d f f


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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.


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 wire

is 2000N. Determine, (1) the components of theforce acting on the bolt (2) the angles definingthe direction of the force and (3) the force vector


8

0m

40 mA

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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80m

40 m A

B

The force in the bolt is the same as the tensionin the wire


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80m

40 mA

B

Hence the Force vector is-848i + 1696j +636k

Direction of forcex = 115.09o

y = 32o

z = 71.46o


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Vector addition


The sum of the vectors can be obtained by adding therespective components of the vectors

IfA= 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)


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


The horizontal component of R= Rx = Fx

Ry = Fy

Rz= Fz

R = sqrt ( Rx2+ Ry2 + Rz2)

Cos x = Rx/ R

Cos y = Ry/RCos z = Rz/R


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4 forces act on a particle as shown in fig.Determine the resultant of the forces in vector

form


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 in

vector form


PracticeProblem

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 is

2000 N in cable AB & 1500 N in the cable AC.Also calculate the angles that this resultantmakes with the X, Y & Z axes



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Z

Z

XO A

Y

B

C

62

5052


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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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Unit vector along AB = -0.63i + 0.606j + 0.485k

FAB = -1260i + 1212 j +970k

Unit vector along AC = -0.546i +0.6518j -0.5256kFAC= -819i +977.7j -788.4k

R = -2079i +2189.7j +181.6k

|R| = 3024.39

x = 133.42o

y = 43.61o

z = 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 the

cable AB is 40 N, calculate the required tension inAC& AD so that the resultant of the 3 forcesapplied at A is vertical



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Y

Z

X

X

Z

Y

DC

A

B


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Y

Z

X

X

Z

Y

DC

A

B


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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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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


42

For equilibrium, the components Rx, Ry & Rz mustbe 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 in

fig. Determine the magnitude of P and thetensions in the cables AB & AC



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Z

A

Y

X

Z

2m

1m

Mass -250 kgP

14m

B

C

( )


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Z

A

Y

X

Z

2m

1m

Mass =250 kgP

14m

B

C (0,14,-12)

(0,14,9)

(1,2,0)


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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


Z

A

Z

Y

X

D

C

B

1.2 m

1 m1.8m


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Z

A

Z

Y

X

D

C

B

1.2 m

1 m1.8m

A (1.2, 0,0) ; B (0, 0,1.2) ; C(-1, 0, -0.8);

D(0,1.8,0)

A (1 2 0 0) B (0 0 1 2) C( 1 0 0 8)


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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 F

AD

+ 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 kNFCD = 0.983 kN


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Dot Product


50

A. B = |A||B| cos , where is the angle betweenthe vectors

Dot product Scalar quantity

A= Axi + Ayj + Azk

B = Bxi + Byj + Bzk

A. B = AxBx + AyBy + AzBz Angle between A&B , cos = A.B/ (|A||B|)


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Dot Product


51

Application

Work done by a force F in causing a displacement sis given by

W= Fs cos

W = F . s

The point of application of a force F(5i+10j 15k)


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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


W = F.s

s = 2i-j-9k

W = 135 Nm


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Dot Product


53

Another Application to find the component orprojection of a force along a line

F = Fxi+ Fyj +Fzk

Projection ofF along a line AB = F. Unit vectoralong AB

A f F 3i 4j 12k t t i t A(1 2 3)


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A force F= 3i-4j+12k acts at a point A(1,-2,3).

Compute the component ofF along line AB if B

has co-ordinates (2,1,2)


Vector along AB = i+3j-k

Unit vector along AB = 0.301i+0.904j-0.301k

Component ofF 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


55

A x B = |A| |B| sin

A x B =

A x B = i(AyBz

AzBy)

j(AxBz

AzBx) + k( AxBy

AyBx)

i j k

Ax Ay AzBx By Bz


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Moment of a force in space


56

Moment of a force F about a point is given by M = r x Fwhere 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


57

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


58

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 about

a point B(2,-1,3).


A(1, -1, 2)

B(2, -1,3)


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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 pointA(4,-1,7) in the direction of vector 9i+6j-2k. Find

the moment of the force about the point O(1,-3,2)


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


O

L

Y

Z

X

(1) Calculate the Moment ofF

about O= Mo= r x F

(2) Calculate the component(projection) of Mo on the line OLMOL = Mo .Unit vector along OL

Find the moment of a force F(20i+30j) kN


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3 jacting at a point with position vector (6i+10j)about the line passing through the points

B(6i+10j-4k) and C(-4i+5j-2k)


(1) Calculate the Moment ofF about point B.

(2) Find the projection of this moment MB about theline BC

(1) Moment ofF about B

=Position vector of A wrt B x F = 4kx (20i+30j) = -120i +80j

( ) M f F b i B i 8 j


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(1) Moment ofF 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

Forces in Space

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