space forces

7/28/2019 Space Forces

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Engineering Mechanics (Statics) 10-2 Space Forces

Ex. 10. 9 : The cord exerts a force of 30 N on the hook. If the length of cord is 8m find co-ordinates x and y of

point B. The x-component of force is 25 N at A. Refer Fig. Ex. 10.9.

Fig. Ex. 10.9

Soln. : The position vector from A to B is,

rAB

= 2 i 4k + y j + x i = (x 2) i + (y) j (4) (k)

(Follow the path A C D E B) to reach from A to B,A C = 2 i, C D = 4 k, D E = y j, E B = x i

Fig. Ex. 10.9(a)

The magnitude of position vector represents length of cord

r = (x 2)2

+ (y)2

+ (4)2

= 8 (1)


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(x 2)2 + (y)2 + 16 = (8)2

(x 2)2 = 48 y2 (2)

The force vectorF

AB= F

ABe

AB= 30

(x 2) i + y (j ) 4 (k)

(x 2)2

+ (y)2

+ (4)2

Here from Equation (1) denominator represents length.

F

AB=

30

8[(x 2) i + (y) j (4) k]

x component of force Fx =30

8(x 2)

25 =30

8(x 2)

200

30 = x 2

x = 8.67 m Ans.

Substituting in Equation (1)

[(8.67) 2]2 = 48 y2

y2 = 48 (6.67)2

y = 1.873 m Ans.

TYPE II : EXAMPLES BASED ON RESULTANT OF CONCURRENT AND

PARALLEL FORCES IN SPACE

Ex. 10.11 : Determine the resultant of the two forces as shown in

Fig. Ex. 10.11.Soln. : Force vectors :

1. Finding components of 250 N along co-ordinate axis.

Resolving 250 N into two components along directions oa

and y-axis using parallelogram OABC (Ref. Fig. 10.11(a))

Fa = 250 cos 60 = 125 N

and Fy = 250 sin 60 = 216.51 N

Now resolving Fa = 125 N into two directions along x and

z axis using parallelogram ODCE

(Ref. Fig. Ex. 10.11(b)). Fig. Ex. 10.11

Fx = 125 cos 25 = 113.29 N

Fz = 125 sin 25 = 52.83 N

Force vector

F = Fx i + Fy j + Fz k

Fig. Ex. 10.11(a) Fig. Ex. 10.11(b)


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F = (113.29) i + (216.51) j + ( 52.83) k

2. Finding components of 300N force along co-ordinate axis.

Resolving 300N into directions y and ob, using parallelogram OAFG (Ref. Fig. Ex. 10.11(c))

Fy = 300 cos 40 = 229.81 N

Fb = 300 sin 40 = 192.84 N

Resolving Fb into two components Fx and Fz using parallelogram OHIJ (Ref. Fig. Ex. 10.11(d))

Fz = 192.84 cos 20 = 181.21 N

Fx = 192.84 sin 20 = 65.955 N

Fig. Ex. 10.11(c) Fig. Ex. 10.11(d)

Force vector of 300N force will beF = Fx i + Fy j + Fz k

F = (65.955) i + (229.81) j + (181.21) k Resultant of two forces by adding , co-efficients of i, j and k

Rx = (113.29 + 65.955) i = 179.245 i

Ry = (216.51 + 229.81) j = 446.32 j

Rz = ( 52.83 + 181.21) k = 128.38 k

R = 179.245 i + 446.32 j + 128.39 k Magnitude R = R

2

x + R2

y + R2

z R = 497.81 N Ans.

Directions,

x = cos 1

Rx

R y = cos

1

Ry

R z = cos

1

Rz

R

x = 68.89 Ans. y = 26.29 Ans. z = 75.05 Ans.


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Ex. 10.14 : Three parallel forces act on a circular plate.

If FA = 200 N determine FB and FC so

that resultant of three forces has a line of

action along y-axis. Refer Fig. Ex. 10.14.

Soln. : The force vectors

F

A= 200 j

F

B= F

Bj (All forces are parallel to y-axis)

F

C= F

Cj

Co-ordinates of points of application of forces.

A ( r cos 45, 0, r sin 45)

A ( 1.06, 0, 1.06)B (+ 1.5 cos 30, 0, 1.5 sin 30) Fig. Ex. 10.14

B (+ 1.3, 0, 0.75)and C (0, 0, 1.5)

Since,R is acting along y-axis, it is passing through point O.

MO = 0

Fig. Ex. 10.14(a)

i

1.06

0

j

0

200

k

1.06

0 M

FA

+

i

+ 1.3

0

j

0

FB

k

0.75

0 M

FB

+

i

0

0

j

0

FC

k

1.5

0 M

FC

= 0

i ( 212) + k (+ 212) + i ( 0.75 FB) + k ( 1.3 F

B) + i (1. 5 F

C) + k (0) = 0

212 0.75 FB

+ 1.5 FC

= 0 and 212 1.3 FB

= 0

FB

=2121.3

= 163.07 N Ans.

and FC =212 + 0.75 (163.07)

1.5 FC = 222.87 N Ans.


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TYPE III : EXAMPLES BASED ON EQUILIBRIUM OF CONCURRENTFORCES AND PARALLEL FORCES

Ex. 10.18 : Three cables are connected at point A.

Find the value of forceP = Pi for which

tension in cable AD is 300N. Refer

Fig. Ex. 10.18.

Soln. : The concurrent forces acting at A are, P,T

AB,

T

ACand

T

AD.

Force vectors :

1. P = P i

2.T

AB= T

ABe

AB

= TAB

240 j 950 i + 375 k

( 240)2

+ ( 950)2

+ (375)2

Follow the path A E O Bto reach from A to B

A E = 240 jE O = 950 iO B = 375 k

T

AB= T

AB[ 0.905 i 0.229 j + 0.357 k ]

3.T

AC= T

ACe

AC Fig. Ex. 10.18

= TAC

240 j 950 i 310 k

( 240)2

+ ( 950)2

+ ( 310)2

Follow the path A E O C to reach from A to C

A E = 240 j, E O = 950 i. O C = 310 k T

AC= T

AC[ 0.924 i 0.233 j 0.302 k ]

4.T

AD= T

AD e

AD= 300

240 j 950 i + 950 j 200 k

(710)2

+ ( 950)2

+ ( 200)2

= 3001202.75

[ 950 i + 710 j 200 k ]

TAD

= 236.96 i + 177.094 j 49.88 k

Now apply conditions of equilibrium for concurrent forces,

Fx = 0 P 0.905 TAB 0.924 TAC 236.96 = 0 (1)

Fy = 0 0.229 TAB 0.233 TAC + 177.094 = 0 (2)

Fz = 0 0.357 TAB 0.302 TAC 49.88 = 0 (3)


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Solve above equations,

We get , P = 937.91 N Ans.

TAB

= 427.36 N

TAC

= 340.03 N

TYPE IV : EXAMPLES BASED ON EQUILIBRIUM OF PARALLEL FORCES IN SPACE

Ex. 10.23 : A homogeneous semi circular plate of weight 10 kN and radius 1m is

supported in horizontal plane by three vertical wires as shown in

Fig. Ex. 10.23. Determine the tensions in the wires.

Fig. Ex. 10.23

Soln. : Force vectors and co-ordinates of points of application.

Fig. Ex. 10.23(a) Fig. Ex. 10.23(b)

Now, Force vectors, Co-ordinates B (0, 0, 1)

T

A= T

Aj D (1 cos 30, 0, 1 sin 30)

T

B= T

Bj D (0.87, 0, 0.5)

T

D= T

Dj A (0, 0, + 1)

W = 10 j G (

4r

3 , 0, 0)

G (0.424 r, 0, 0)

G (0.424, 0, 0) (r = 1m)

Now, apply conditions of equilibrium,

Fy = 0 TA+TB + TD 10 = 0

TA +TB + TD = 10 (1)

M = 0 (Taking moment about origin)


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i0

0

j0

TB

k 1

0

M

TB

+

i0.87

0

j0

TD

k 0.5

0 M

TD

+

i0

0

j0

TA

k1

0 M

TA

+

i0.424

0

j0

10

k0

0 M

W

= 0

expanding,

i (TB) + i (0.5T

D) + k (0.87 T

D) + i ( T

A) + k ( 4.24) = 0

TB

+ 0.5 TD

TA

= 0 (2)

and 0.87 TD

4.24 = 0 (3)

Solving above Equations (1), (2) and (3)

We get, TA

= 3.782 kN Ans.

TB

= 1.345 kN Ans.

TD = 4.873 kN Ans.

TYPE V : EXAMPLES BASED ON MOMENTS

Ex. 10.28 : A force of 500N passes through points whose position vectors arer1 = 10 i 3j + 12 k and

r2

= 3 i 2j + 5 k

What is the moment of this force about a line in x y plane, passing through the origin and inclined at

30 with x-axis ?

Soln. : The magnitude of force F = 500 N

unit vector in the direction of force e =(3 10) i + ( 2 + 3) j + (5 12) k

(3 10)2

+ ( 2 + 3)2

+ (5 12)2

= 7 i + j 7k9.95

= 0.703 i + 0.1 j 0.703 k

Force vectorF = F e = 500 [ 0.703 i + 0.1 j 0.703 k]F = 351.5 i + 50 j 351.5 k

Now line OL is in x y plane making 30 with x-axis as shown in Fig. Ex. 10.28.The direction cosines of line OL

x = l = cos x = cos 30 = 0.866

y = m = cos y = cos 60 = 0.5

z = n = cos z = 0 (Line in x-y plane) Fig. Ex. 10.28

Now, Moment of a force F about line OL is given by,

MOL =

x

x

Fx

yy

Fy

zz

Fz

D.C.S. of line OLCo-ordinates of point of application of force F

Components of force F

MOL

=

0.866

10

351.5

0.5

3

50

0

12

351.5

= 0.866 (1054.5 600) 0.5 ( 3515 + 4218) + 0 = 42.097 N.m Ans.


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Ex. 10.30 : A box of size 3 4 2m is subjected to three forces as shown in Fig. Ex. 10.30. Find in vector form

the sum of moments of the three forces about diagonal OB.

Fig. Ex. 10.30

Soln. : Co-ordinates of different points are,

O (0, 0, 0)

C (4, 2, 0) (path O G C)G (4, 0, 0) (path O G)F (4, 0, 3) (path O G F)B (4, 2, 3) (path O G C B)

Force Vectors :

P1

= 60 j

P

2= 20 i

P

3= 20 i

Moment of forces about diagonal OB.

Important Observation :

(a) Line of action of P2

is passing through

O so moment of P2

about line OB

must be zero.

(b) Line of action of P1

is passing through

point B so moment of P1

about line

OB must be zero.

Now taking moment of P3

about line OB.

Finding first D.C.S. of line OB : (OB

) Fig. Ex. 10.30(a)

OB

=(4 0) i + (2 0) j + (3 0) k

(4)2

+ (2)2

+ (3)2

=4i + 2j + 3k

29

x =4

29, y =

2

29, z =

3

29

Now, Force vectorF = F e


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P3 = 20 i

MOB

=

x

x

Fx

y

y

Fy

z

z

Fz

D.C.S. of line OB

Co-ordinates of point of application of force

Components of force.

=1

29

4

4

20

2

2

0

3

0

0co-ordinates of point of application C

=1

293 (0 40) = 22.28 kN.m.

In vector formM

OB= M

OBe

OB(Product of magnitude and unit vector in direction of OB)

M

OB= 22.28

1

29(4i + 2j+ 3k)

MOB

= 16.548 i + 8.274 j + 12.411 k Ans.

Ex. 10.33 : To lift a heavy block a man uses

block and tackle system as

shown in Fig. Ex. 10.33. If

moments about y and z axis of

the force acting at B by portion

AB of the rope are respectively,

120 N.m and 460 N.m.

Determine distance b.

Fig. Ex. 10.33

Soln. : Since tension is acting from B to A at B the tension vector

T = T e

BA= T

bk 4.8j + 2.1 i + 1.5j

(2.1)2

+ ( 3.3)2

+ ( b)2

[Follow the path B C O D A]B C = bk, C O = 4.8 j, O D = 2.1 i , D A = 1.5 j ]


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ArrangingT = T

(2.1)i + ( 3.3)j + ( b) k

15.3 + b2

Now, Moment of T about axis oy

(y axis) is 120 N.m

Moy =T

15.3 + b2

0

0

2.1

1

4.8

3.3

0

b

b

120 =T

15.3 + b2 [2.1 b]

2.1 T.b

15.3 + b2

= 120

T.b = 1202.1

15.3 + b2

T. b = 57.14 15.3 + b2

(1) Fig. Ex. 10.33(a)

Moment of T about z-axis (oz) is 460 N.m

Moz =T

15.3 + b2

0

0

2.1

0

4.8

3.3

1

b

b= 460

10.08 T

(15.3 + b2)

= 460

T

15.3 + b2 = 45.63 (2)

Substituting in Equation (2)

45.63 15.3 + b2b = 57.14 15.3 + b2 b =

57.14

45.63 b = 1.252 m Ans.

Ex. 10.34 : To lift the table without tilting, the combined

moment of four parallel forces must be zero

about x and y axis passing through centre of

table. Determine the magnitude of force and

distance a. Refer Fig. Ex. 10.34.

Soln. : Since forces are parallel to z-z axis

The moments of forces about x-x and y-y axis can be

found directly.

Mxx = 0 (given)

(20 0.6) + (18 0.6) (30 0.6) F (0.6 d) = 0

F (0.6 d) = 4.8 (1)

Myy = 0 (given) Fig. Ex. 10.34

(18 0.5) + (F 0.5) = (20 0.5) + (30 0.5)

18 + F = 20 + 30 F = 32 N Ans.Substituting this value in Equation (1)

32 (0.6 d) = 4.8 d = 0.45 m Ans.


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TYPE VI : GENERAL FORCE SYSTEM AND SOME MISCELLANEOUS EXAMPLES

Ex. 10.40 : A circular plate of radius 150 mm is supported by three

wires each of length 600 mm each. Determine tension

in each wire. Weight of plate is 10kg. Refer

Fig. Ex. 10.40.

Soln. : Consider F.B.D. of plate we see that the forces acting on

plate are,

TBA

, TDA

, TCA

, weight of plate.

The co-ordinates of different points are,

D (150, 0, 0)

C [ 129.9, 0, 75]

B [ 129.9, 0, 75]

A [0, 580.95, 0] Fig. Ex. 10.40

Fig. Ex. 10.40(a) Fig. Ex. 10.40(b)

Force Vectors :

T

BA= T

BAe

BA= T

BA

(0 + 129.9) i + (580.95 0) j + (0 + 75)k

(129.9)2

+ (580.95)2

+ (75)2

TBA

= TBA

600[129. 9 i + 580.95 j + 75 k]

T

CA= T

CAe

CA

= TCA

(0 + 129.9) i + (580.95 0) j + (0 75)k

(129.9)2

+ (580.95)2

+ (75)2

T

CA=

TCA

600[129. 9 i + 580.95 j 75 k]


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TDA = TDAeDA = TDA

(0 150) i + (580.95 0) j

( 150)2 + (580.95)2

T

DA=

TDA

600[ ( 150) i + (580.95 j) ]

W = mg ( j )

W = 98.1 ( j )

Apply conditions of equilibrium,

Fx = 0

129.9

600 TBA +

129.9

600 TCA

150

600 TDA = 0 (1)

Fy = 0580.95

600T

BA+

580.95

600 T

CA+

580.95

600T

DA 98.1 = 0 (2)

Fz = 075

600T

BA

75

600T

CA= 0 (3)

Solving Equations (1), (2) and (3)

We get, TBA

= 27.15 N Ans.

TCA

= 27.15 N Ans.

TDA

= 47.02 N Ans.

Space ForcesEnds.

space forces

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