# Engg-Mechanics-Non-Coplanar-Force-system ```FS - 1
FORCES IN SPACE
(Noncoplanar System of Forces)
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Forces in space
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A Force in space: A Force is said to be in space if its line of
action makes an angle α, β and γ with respect to rectangular
co-ordinate axes X, Y and Z respectively as shown the Fig. 1.
F Fig. 1. A Force in space
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Forces in space
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Noncoplanar system of forces (Forces in Space) and
Their Classifications
System of forces which do not lie in a single plane is called
noncoplanar system of forces(Forces in space ). A typical
noncoplanar system of forces (forces in space) is shown in the
Fig. 2. below
Fig. 2 Forces in space (noncoplanar system of forces)
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Forces in space
Noncoplanar system of forces(Forces in space) can be
broadly classified into three categories. They are
1. Concurrent noncoplanar system of forces
2. Nonconcurrent noncoplanar system of forces
3. Noncoplanar parallel system of forces
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Forces in space
1. Concurrent noncoplanar system of forces: Forces which
meet at a point with their lines of action do not lie in a
plane are called “Concurrent noncoplanar system of
forces”. A typical system of Concurrent noncoplanar
system of forces is shown in the Fig.3.
Fig. 3. Concurrent noncoplanar system of forces
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Forces in space
2. Nonconcurrent noncoplanar system of forces: Forces
which do not meet at a point and their lines of action do not
lie in a plane, such forces are called “Nonconcurrent
noncoplanar system of forces”. A typical system of
nonconcurrent noncoplanar system of forces is shown in the
Fig.4.
Fig. 4. Nonconcurrent noncoplanar system of forces
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Forces in space
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3. Noncoplanar parallel system of forces: If lines of action of all
the forces in a system are parallel and they do not lie in a
plane such a system is called Non-coplanar parallel system
of forces. If all the forces are pointing in one direction then
they are called Like parallel forces otherwise they are called
unlike parallel forces as shown in the Fig.5.
Fig. 5 Noncoplanar parallel system of forces
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Forces in space
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Rectangular components of a force in space
Fig. 6. Resolving a force in space into rectangular components
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Forces in space
Rectangular components of a force in space
In the Fig.6(a) a force F is acting at the origin O of the system
of rectangular coordinate axes X,Y,Z. Consider OBAC plane
passing through the force F. This plane makes an angle  with
respect to XOY plane. Force F makes an angle θy with
respect to Y-axis.
Fig.6(a)
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Forces in space
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Rectangular components of a force in space
Fig.6(b)
In the Fig.6(b), the force F is resolved in the vertical (Y- axis) and
horizontal direction (X – axis) as
Fy = F Cosy and
Fh = F Siny respectively.
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Forces in space
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Rectangular components of a force in space
Fig.6(c)
In the Fig 6(c) the horizontal component Fh is again resolved in the X and
Z axes directions. These components are
Fx = Fh cos = F siny cos
Fz = Fh Sin = F siny sin
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Forces in space
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Now applying Pythagorean theorem to the triangles OAB
and OCD
F2 = (OA)2 = OB2 + BA2 = Fy2 +Fh2 ----------------(1)
Fh2 = OC2 = OD2 + DC2 = Fx2 +Fz2
----------------(2)
Substituting equation (2) into the equation (1), we get
F2 = Fx2 +Fy2 + Fz2
F =  Fx2 + Fy2 + Fz2
----------------(3)
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Forces in space
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The relationship existing between the force F and its three components Fx,
Fy, Fz is more easily visualized if a box having Fx, Fy, Fz for edges is drawn
as shown below. The force F is then represented by the original OA of this
box.
Fig. 7 Relationship between Force F and its components Fx, Fy and Fz
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Forces in space
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From the above Figure (Fig. 7)
Fx = F Cos x, Fy = F Cosy, Fz = F Cosz ------------(4)
Where x, y, z are the angles formed by the force F with X, Y, Z axes respectively.
Fx,Fy,Fz are the rectangular components of the force F in the directions of X, Y, Z
axes respectively.
Cos x = Fx/F; Cosy = Fy/F; Cosz = Fz/F
Substituting equation (4) into the equation (3), we get
F =  Fx2 + Fy2 + Fz2
F =  F2Cos2x + F2Cos2y + F2Cos2z
F2 = F2 ( Cos2x + Cos2y + Cos2z )
1 = Cos2x + Cos2y + Cos2z
-------------(5)
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Forces in space
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Force Defined by its magnitude and two points on its line of action
dx= d Cos x, dy = d Cosy,
dz = d Cosz
----(6)
d =  dx2 + dy2 + dz2
---(7)
Dividing member by
member the relations (4) and
(6), we obtain
Fx /dx = Fy/dy = Fz/dz = F/d
----------------------------(8)
Fig 8
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Forces in space
Resultant of concurrent forces in Space:Resolve all the forces into their rectangular components in X, Y and Z axes directions.
Adding algebraically all the horizontal components in the x direction gives
Rx =  Fx,
Similarly adding algebraically all the components in y and z directions yield the
following relations
Ry =  Fy,
Rz =  Fz
Thus magnitude of resultant
R =  Rx2 + Ry2 + Rz2
Angles x, y, z resultant forms with the axes of coordinates are obtained by
Cos x 
Rx
R
; Cos y 
Ry
R
; Cos z 
Rz
R
Forces in space
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Problems:
(1) A tower guy wire is anchored by means of a bolt at A is shown in the following
Figure. The tension in the wire is 6000N. Determine
(a) The components Fx, Fy, Fz of the forces acting on the bolt.
(b) The angles x, y, z defining the direction of the force.
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Forces in space
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Solution: (a) Here dx = 50m, dy = 200m, dz = -100m
Total distance A to B
d =  dx2 + dy2 + dz2
=  (50)2 + (200)2 + (-100)2
= 229.13 m
Using the equation (8)
Fx /dx = Fy/dy = Fz/dz = F/d
Fx = dx . (F/d) = (50 x 6000)/ 229.13 = 1309.3 N
Fy = dy . (F/d) = (200 x 6000)/ 229.13 = 5237.20 N
Fz = dz . (F/d) = (-100 x 6000)/ 229.13 = -2618.6 N
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Forces in space
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(b) Directions of the force:
Cos x = dx /d , x = Cos–1 (50/229.13) = 77.4
y = Cos–1 (dy /d) = Cos–1 (200/229.13) = 29.2
z = Cos–1 (dz /d) = Cos–1 (-100/229.13) = 115.88
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Forces in space
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Problem(2) Determine (a) the x , y and z components of the 250 N
force acting as shown below
(b) the angles x, y, z that the force forms with the coordinate
axes.
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Forces in space
Fh = 250 Cos60 = 125 N
Fx = 125 Cos25 = 113.29 N
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Fy = 250 Sin60 = 216.5 N
Fz = 125Sin25 = -52.83 N
x = Cos–1 ( Fx /F) = Cos–1 (113.29/250) = 63
y = Cos–1 ( Fy /F) = Cos–1 (216.5/250) = 30
z = Cos–1 ( Fz /F) = Cos–1 (-52.83/250) = 102.11o
Components of 250 N in the x, y, z axes directions are
Fx = 113.29N
Fy = 216.5 N Fz = -52.83N
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Forces in space
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Problem 3. Find the resultant of the system of forces as shown
below
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Forces in space
Solution
Components of Force F1 = 3000 N:
o
Fy1 = 3000 x Cos40 = 2298.13 N
o
Fh1 = 3000 x Sin40 = 1928.36 N
o
Fx1 = 1928.36 x Cos30 =1670 N
o
Fz1 = 1928.36 x Sin30 = 964.18 N
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 FS - 23
Forces in space
Components of Force F2 = 2000 N:
Fy2 = 2000 x Cos20 = 1879.39 N
Fh2 = 2000 x Sin20 = 684.04 N
Fx2 = 684.04 x Sin35 = 392.35 N
Fz2 = 684.04 x Cos35 = 560.33N
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Forces in space
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Rx =  Fx = Fx1 + Fx2 = 1670 – 392.35 = 1277.65
Ry =  Fy = Fy1 + Fy2 = 2298.13 + 1879.39 = 4177.52 N
Rz =  Fz = Fz1 + Fz2 = 964.18 + 560.33 = 1524.51 N
Resultant R = Rx2 + Ry2 +Rz2
= 1277.652 +4177.522 +1524.512
= 4626.9 N
Its inclinations with respect to x, y and z axes are
calculated as
x = Cos–1 (1277.65 /4626.9) = 73 58' 13.1&quot;
y = Cos–1 (4177.52/4626.9) = 25 27'
o
–1
z = Cos (1526.51/4626.9) = 70 45’36”
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Forces in space
FS - 26
Problem 4. In the Fig shown below, the forces in the cables AB and AC
are 100 kN and 150 kN respectively. At the joint ‘A’ loading is as
shown in the Fig. Find the resultant of system of forces in space and its
inclination with rectangular coordinates x,y and z axes.
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Forces in space
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Solution: Force in the cable AB = 100 kN
Force in the cable AC=150 kN
For the cable AB
dx = -20 m
dy = 15m
dz = 5m
dAB =  dx2 +dy2 + dz2
=  (-20)2 +(15)2 + (5)2 =  650 = 25.5 m
For the cable AC
dx = -20 m
dy = 25m
dz = -10m
2 +(25)
2 + | (-10)
2 = 1125 = 33.54 m
dAc |=Website
(-20)
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Forces in space
For the cable AB (1)
Fx1/dx1 = Fy1/dy1 = Fz1/dz1 = F/d
Fx1/-20 = Fy1/15 = Fz1/5 = 100/25.5
Fx1 = -78.41 kN
Fy1 = 58.82 kN
Fz1 = 19.61 kN
For the Cable AC (2)
Fx2/-20 = Fy2/25 = Fz2/-10 = 150/33.54
Fx2 = - 89.45 kN
Fy2 = 111.81 kN
Fz
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2 = - 44.72
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Forces in space
Component of force 60 kN
Fx3 = 60 x Cos(70) = 20.52 kN
Fy3 = 60 x Cos(30o) = 51.96 kN
Fz3 = 60 x Cos(11123)= - 21.88 kN
Component of the force 50KN
Fx4 = 50 kN
Fy4 = 0
Fz4 = 0
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Forces in space
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Algebraic summation of Rectangular Components in X, Y and Z
axes directions yield:
Rx =  Fx = Fx1 + Fx2 + Fx3 + Fx4
= -78.43 - 89.45 + 20.52 + 50
= -97.36 kN
Ry =  Fy = Fy1 + Fy2 + Fy3 + Fy4
= 58.82 + 111.81 + 51.96 + 0
= 222.59 kN
Rz =  Fz = Fz1 + Fz2 + Fz3 + Fz4
= 19.61 – 44.72 – 21.88 + 0
= -46.99 kN
 Resultant R =  Rx2 + Ry2 + Rz2
=  (-97.36)2 + (222.59)2 + (-46.99)2
= |247.45
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Forces in space
FS - 31
Inclinations of the resultant with X,Y and Z axes:
x = Cos-1(Rx / R) = Cos-1 (-97.36/247.45) = 113 10’
y = Cos-1(Ry / R) = Cos-1 (222.59/247.45) = 25 54’
z = Cos-1(Rz / R) = Cos-1 (-46.99/247.45) = 100 56’ 47”
Check:
Cos2x + Cos2y + Cos2z = 1
Cos2(11310’) + Cos2(2554’) + Cos2(10056) = 1
1 = 1
Hence OK
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Forces in space
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Problem 5. a) Forces F1, F2, and F3 pass through the origin and
points whose coordinates are given. Determine the
resultant of the system of forces.
F1 = 20 kN, (3,-2,1)
F2 = 35 kN, (-2,4,0)
F3= 25 kN, (1,2,-3)
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Forces in space
Solution:
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FS - 33
Forces in space
Force F1 = 20 kN
d =  32 + (-2)2 + 12 =  14 = 3.74
Cos x1 = dx / d = 3/3.74 = 0.802
Cos y1 = -2/3.74 = -0.535; Cos z1 = 1/3.74 = 0.267
Force F2 = 35 kN
d =  (-2)2 + 42 +0 = 20 = 4.47
Cos x2 = -2/4.47 = -0.45
Cos y2 = 4/4.47 = 0.9; Cos z2 = 0
Force F3 = 25 kN
d =  (1)2 + 22 +(-3)2 = 14 = 3.74
Cos x3 = 1/3.74 = 0.267
Cos y3 = 2/3.74 = 0.535; Cos z3 = -3/3.74 = -0.802
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Forces in space
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Summation of the rectangular components in X, Y and Z axes directions
yield:
Rx = Fx = 20 x 0.802 + 35 x (-0.45) +25 x 0.267 = 6.965 kN
Ry =  Fy = 20 x (-0.535) + 35 x 0.9 + 25 x 0.535 = 34.175 kN
Rz =  Fz = 20 x 0.267 + 35 x 0 + 25 x (-0.802) = -14.71 kN
Resultant R =  Rx2 + Ry2 + Rz2
=  6.9652 + 34.1752 + (-14.71)2
= 37.85 kN
Inclination of the resultant R with respect to X, Y and Z axes are
calculated as
x = Cos–1(Rx / R) = Cos-1(6.965/37.85 ) = 79.4
y = Cos–1(Ry /R) = Cos-1(34.175/37.85) = 25.46
-1(-14.71/37.85) = 112.87
z www.bookspar.com
= Cos–1(Rz |/R)
=
Cos
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Forces in space
FS - 36
Equilibrium of Concurrent non-coplanar system of forces:
When a rigid body subjected to concurrent noncoplanar system of forces
F1, F2…. ..FN as shown in the Fig. given below, is in equilibrium, then
algebraic summation of all the components of the forces in three mutually
perpendicular directions must be equal to zero.
i.e.
 Fx = 0
 Fy = 0
 Fz = 0
(1)
Above equations represent the static
conditions of equilibrium for concurrent
noncoplanar system of forces
Fig. A rigid body subjected to concurrent
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noncoplanar system of forces
Forces in
space
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Problem (1) Find the forces in the rods AB , AC and AO subjected to
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Forces in space
Solution:
For the cable AB:
dx1 = 0 - 4 = - 4
dy1= 8 - 0 = 8
dz1 = 15 – 0 =15
d1 =  (-4)2 +(8)2 +(15)2
= 17.46 m
Fx1/-4 = Fy1/8 = Fz1/15 = FAB/17.46
Fx1 = -0.23FAB,
Fy1 = 0.46FAB,
Fz1 = 0.86FAB
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Forces in space
For the Cable AC:
dx2 = 0 - 4 = -4m
dy2 = 8 - 0 = 8m
dz2 = -20 – 0 = -20m
d2 =  (-4)2 +(8)2 +(-20)2 = 480 = 21.91m
Fx2/-4 = Fy2/8 = Fz2/-20 = FAC/21.91
Fx2 = -0.18FAC, Fy2 = 0.365FAc, Fz2 = -0.91FAc
For the Force 120 N:
Fx3 = 120 N, Fy3 = 0, Fz3 = 0
For the force Fx4 = 300 N:
Fx4 = 0, Fy4 = -300N, Fz4 = 0
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Forces in space
Conditions of Equilibrium
 Fx = 0,  Fy = 0,  Fz = 0
Considering  Fx = 0
Fx1 + Fx2 + Fx3 + Fx4 + Fx5= 0
-FAD-0.23F – 0.18FAC + 120 + 0 = 0
FAD + 0.23FAB + 0.183FAC = 120 ----------------------(1)
 Fy = 0
0.46FAB + 0.365FAC + 0 +0 – 300 = 0
0.46FAB + 0.365FAC = 300 ------------------------------(2)
 Fz = 0
0.86 FAB – 0.91 FAC + 0 = 0
-----------------------(3)
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Forces in space
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Solving Equations (1), (2) and (3), we get,
FAB = 372.675 N; FAC = 352.25 N ; FAO = - 30.18 N
Force in the rod AB , FAB = 372.675 N (Tensile)
Force in the rod AC, FAC = 352.25 N (Tensile)
Force in the rod AO, FAO = 30.18 N (Compressive)
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Forces in space
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2) Three cables are connected at D and support, the 400
kN load as shown in the Fig given below. Determine the
tensions in each cable
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Forces in space
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Let FDA, FDB and FDC are the forces in the cables AD, BD,
and CD respectively.
For Cable DA:
dx1 = (0 – 3) = -3m
dy1 = (6 - 4) =2 m
dz1 = (6 – 0 = 6 m
dAD =  dx2 + dy2 + dz2
dAD =  (-3)2 + (2)2 + (6)2 = 9 + 4 + 36 = 7 m
Fx1/dx1 = Fy1/dy1 = Fz1/dz1 = FDA/dDA
Fx1/(-3) = Fy1/2 = Fz1/6 = FDA/7
 Fx1 = -0.43FDA, Fy1 =0.286FDA, Fz1 = 0.857FDA
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Forces in space
For Cable DB:
dx2= (0 – 3) = -3m
dy2 = (6 - 4) =2 m
dz2 = (-6 – 0 = -6 m
dBD = d2 = (-3)2 + (2)2 + (-6)2 =  49 = 7 m
Fx2/dx2 = Fy2/dy2 = Fz2/dz2 = FDB/dDB
Fx2/(-3) = Fy2/2 = Fz2/-6 = FDB/7
Fx2 = -0.43FDB,
Fy2 =0.286FDB,
Fz2 = -0.857FDB
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Forces in space
For Cable DC:
dx3 (0 – 3) = -3m
dy3= (0- 4) = -4m
dz3= (0 0 = 0m
dDC= d3=  (-3)2 + (-4) + (0)2
=  9 + 16
= 5m
Fx3/dx3 = Fy3/dy3 = Fz3/dz3 = FDC/d DC
Fx3/(-3) = Fy3/-4 = Fz3/0 = FDC/5
Fx3 = -0.6FDC,
Fy3 =-0.8FDC,
Fz3 = 0
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FS - 45
Forces in space
For the force 400KN
dx4 = 12m
dy4= -4m
dz4= 3m
d4= d400 =  (12)2 + (4) + (3)2
= 144 +16 + 9
= 69 = 13m
Fx4/dx4 = Fy4/dy4 = Fz4/dz4 = F400/d4
Fx4/(12) = Fy4/-4 = Fz4/3 = 400/13
Fx4 = 369.23 kN,
Fy4 = -123.08 kN,
Fz4 = +92.31 kN
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FS - 46
Forces in space
FS - 47
For Equilibrium, the algebraic summation of resolved
component in a particular direction is equal to zero.
i.e.  Fx = 0 -------------(1)
 Fy = 0 ------------- (2)
 Fz = 0 --------------(3)
(1)  Fx = Fx1 + Fx2 + Fx3 + Fx4 = 0
+ 0.43FDA + 0.43FDB + 0.6FDC = 369.23 ---------(1)
+ 0.286FDA + 0.286FDB – 0.8FDC = 123.08 ------(2)
+ 0.857FDA - 0.857FDB + 0 = 92.31 ------------(3)
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Forces in space
Solving equations (1), (2) and (3), we get
FDB = 304.1kN
FDA = 411.8 kN
FDC = 102 kN
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Forces in space
FS - 49
Practice Questions
1.
A tower guy wire is anchored by means of a bolt at A as shown below.
The tension in the wire is 2500 N. Determine (a) the components Fx, Fy
and Fz of the force acting on the bolt, (b) the angles θx, θy, θz defining
the direction of the force
(Ans: Fx= -1060 N , Fy = 2120 N, Fz= + 795 N
Θx = 115.1 o ; Θy= 32.0 o ; Θz= 71.5 o )
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Practice
2. Determine Questions
(a) x, y and z the components of the force 500 N in the below
Figure. (b) the angles θx, θy, θz that the force forms with the
coordinate axes
(Ans: Fx= + 278 N , Fy = + 383 N, Fz= + 160.7
N
Θx = 56.2 o ; Θy= 40.0 o ; Θz= 71.3 o )
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3.
In order to move
a wrecked truck, two cables are attached at A and
Practice
Questions
pulled by winches B and C as shown. Knowing that the tension in the
cable AB is 10 kN, determine the components of the force exerted by
the cable AB on the truck
(Ans: Fx= -6.30 kN , Fy = 6.06 kN, Fz= + 4.85
kN)
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4.
A 200 kg cylinder
is hung by means of two cables AB and AC, which are
Practice
Questions
attached to the top of a vertical wall. A horizontal force P perpendicular
to the wall holds the cylinder in the position shown. Determine the
magnitude of the force P and the tension in each cable
(Ans: P = 235 N , TAB = 1401 N, TAC= + 1236 N
)
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5.
Three cablesQuestions
are connected at A, where the forces P and Q are applied as
Practice
shown. Determine the tension in each cable when P = 0 and Q = 7.28
kN
(Ans: TAB= 2.88 kN , TAC = 5.76 kN, TAD= 3.6
kN)
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Forces in space
Practice Questions
6. A container of weight w = 400 N is supported by cables AB and AC which
are tied to ring A. Knowing that Q = 0, determine (a) the magnitude of
the force P which must be applied to the ring to maintain the container
in the position shown in figure below, (b) the corresponding the values
of the tension in cables AB and AC
(Ans: P = 138 N ,
TAB = 270N,
TAC = 196N )
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Practice Questions
7. A container supported by three cables as shown below. Determine the
weight of the container, knowing that the tension in the cable AB is 4 kN
 Ans: 9.32 kN
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Practice Questions
8. Determine the resultant of the two forces shown below.
(Ans: R = 498 N ,
Θx = 68.9 o ; Θy= 26.3 o ; Θz= 75.1 o )
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Forces in space
Practice Questions
9. A container of weight W = 1165 N is supported by three cables as shown
below. Determine the tension in each cable.
Ans: TAB = 500 N
TAC = 459 N