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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Engineering Mechanics (Statics) 10-3 Space Forces
(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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Engineering Mechanics (Statics) 10-4 Space Forces
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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Engineering Mechanics (Statics) 10-5 Space Forces
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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Engineering Mechanics (Statics) 10-6 Space Forces
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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Engineering Mechanics (Statics) 10-7 Space Forces
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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Engineering Mechanics (Statics) 10-8 Space Forces
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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Engineering Mechanics (Statics) 10-9 Space Forces
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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Engineering Mechanics (Statics) 10-10 Space Forces
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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Engineering Mechanics (Statics) 10-11 Space Forces
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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Engineering Mechanics (Statics) 10-12 Space Forces
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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Engineering Mechanics (Statics) 10-13 Space Forces
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.
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