14. space forces

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INTRODUCTION
We have so far discussed and worked with a coplanar all system of forces,
wherein the forces in a system would lie in a single plane i.e. it was a two
dimensional force system. In this chapter we deal with" a system of forces
lying in different planes forming a three dimensional force system. Such a
system is also referred as a space Force System and a vector approach is
required to deal with these problems. The chapter begins with the study of
basic operations with forces using a vector approach. Later on we will learn to
find the resultant of a space force system and finally we shall deal with
equilibrium problems.
BASIC OPERATIONS USING VECTOR APPROACH
Since the forces have three dimensions, we employ a vector approach, which
simplifies the working. Here we will learn to represent a force vectorially, to
find vectorially moment of a force about a point and about a line, to find
magnitude and direction of a force given in vector form and other basic
operations useful in solution of a space force problem.
Force in vector form
Fig. shows a force of magnitude F in space passing through A (x1, y1, z1) and.
B (x2,y2, z2). The force in vector form is
AB
ˆF F.e
     
     
2 1 2 1 2 1
2 2 2
2 1 2 1 2 1
F F
x x i y y j z z k
x x y y z z
      
      
x y zF F i F j F k  
Note: i, j and k printed in bold type denote unit
vectors along the x, y and z axis respectively.
Space Forces

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MAGNITUDE AND DIRECTION OF FORCE
Fig. shows a force x y zF F i+F j+F k making angles
 x  y and  z, with the x, y
and z axis respectively.
Here Fx is the component of force in the x
direction. Similarly Fy and Fz are the force
components in the y and z direction.
The magnitude of the force is
2 2 2
x y zF F F F  
Also,
x x
y y
z z
F F cos θ
F Fcosθ
F Fcos θ



Here  x  y and  z are known as the force directions, the value of which lies
between 0 and 180. There is an important identity which relates them.
2 2 2
cos cos cos 1x y z    

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MOMENT OF A FORCE ABOUT A POINT
This is a very important operation while dealing with forces. For coplanar
forces, moment about g point was the product of the force and the 
distance. Here if the force is in space the moment calculation requires a
vector approach.
Fig. shows a force F in space passing through points A (x1, y1, z1) and B (x2,
y2, z2) on its line of action. Let C (x3, y3, z3) be the moment centre i.e. the point
about which we have to find the moment. The procedure of finding the
moment of the force about the point is as follows.
Step 1: Put the force in vector form i.e.
x y zF F i+F j+F k
Step 2: Find the position vector extending from the moment centre to any
point on the force i.e. x y zr r i+r j+r k .
Step 3: Perform the cross product of the position vector and the force vector
to get the moment vector i.e.
F
point
x y z
x y z
M
i j k
r r r
F F F
r F 


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EXERCISE 1
1. A force of 50 N acts parallel to the y axis in the -ve direction. Put the force
in vector form.
2. A 130 KN force acts at B (12, O, O) and passes through C (0, 3, 4). Put the
force in vector form.
3. A force F = (3i -4j 12 + k) N acts at a point A (1,-2, 3) m. Find
a. moment of the force about origin.
b. moment of the force about point B (2, l,2)m.
4. A force F = 80 i + 50 j - 60 k passes through a point A (6, 2, 6). Compute its
moment about a point B (8, 1, 4) (M. U. Dec12)
5. A 700 N force passes through two points A (-5,-1, 4) towards B (1, 2, 6) m.
Find moment of the force about a point C (2,-2, 1)m.
6. A force of magnitude 1200 N passes from A (2, -4, 6) to B (-3, 2, 3).
Calculate the moment of this force about the origin. Also determine the
component of this force along the line AC where C has the position vector
defined by cr i 2j 3kN    .
7. A rectangular plate is supported by
brackets to the wall at A and B by wire CD
as shown in figure. Knowing that tension in
wire is 2OO N determine the moment about
point A, of the force exerted by wire on
point C. All dimensions are in mm.
8. Find moment of force P = 500 N about point H.
9. Three forces act on a rectangular box as shown in figure. Determine the
moment of each force about the origin.

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RESULTANT OF CONCURRENT SPACE FORCE SYSTEM
Resultant of a concurrent space force system is a single force R , which acts
through the point of concurrence. Fig. (a) shows a concurrent system at point
P. The resultant of the system is shown 1n Fig. (b) and calculated as,
RESULTANT OF PARALLEL SPACE FORCE SYSTEM
The resultant of a parallel space force system is a single force R which acts
parallel to the force system. The location of the resultant can be found out
using Varignon's theorem.
Figure (a) shows a parallel system of three forces F1, F2 and F3. The resultant
of the system is shown in figure (b) and is calculated as,
1 2 3R F F F  
The resultant acts at P. The coordinates (x, y, z) of the point P can be
calculated by using Varignon's theorem, the moments for which can be taken
about any convenient point like point O. The equation of Varignon's theorem
for space forces is
F R
O OM M
0

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RESULTANT OF GENERAL SPACE FORCE SYSTEM
A general space force system is neither a concurrent nor a parallel system.
The resultant of such a system is a single force R and a moment M at any
desired point. Since the resultant contains one force and one moment, it is
also known as a Force Couple System. Fig. (a) shows a general system of four
forces F1, F2, F3 and F4. If it is desired to have the resultant at point O, then
as per figure (b).
The single force 41 2 3R F F F F   
and the single moment 31 2 4FF F F
O O O OM M M M M   

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EXERCISE 2
1. A force P1 = 10 N in magnitude acts along direction AB whose co-ordinates
of points A and B are (3, 2, -l) and (8, 5, 3) respectively. Another force P2 = 5
N in magnitude acts along BC where C has co-ordinates (-2, 11, -5).
Determine
a. The resultant of Pr and Pz.
b. The moment of the resultant about a point D (1, 1, 1).
2. Knowing that the tension in AC = 20 kN, determine the required values of
TAB and TAD so that the resultant of the three forces applied at A is
vertical. Also find the resultant.
3. A plate foundation is subjected to five vertical forces as shown. Replace
these five forces by means of a single vertical force and find the point of
replacement.
4. Determine the loads to be applied at A and F, if the resultant of all six loads
is to pass through the centre of the
foundation of hexagonal shape of side 3 m.

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5. Knowing that the resultant of the three parallel forces is 500 N acting in
positive y direction and passing through the centre of the rectangular plate,
determine the forces F1 , F2 and F3.
6. A square foundation mat supports four columns as shown in figure.
Determine magnitude and point of application of resultant of four loads.
7. A rectangular parallelepiped carries four
forces as shown in the figure. Reduce the
force system to a resultant force applied at
the origin and moment around the origin.
OA = 5 m, OB = 2m, OC = 4 m. (M. U
Dec 12)
8. Figure shows a rectangular
parallelepiped subjected to four
forces in the direction shown.
Reduce them to a resultant force at
the origin and a moment.

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9. A tetrahedron A B C D is loaded by forces
F1 = 100 N at A along DA, F2 = 2OO N at B
along CB and F3 = 300 N at C along DC as
shown in the figure. Replace the three
force system by a single resultant force R
at B and a single resultant moment vector
M at B. Take the co-ordinates in metre
units.
10. The following forces act on the block shown in
figure. F1 = 40 kN at point C along CD, F2 = 3O
kN at point D along DB and F3 = 1OO kN at point
D along DE. Find the resultant force and
resultant moment of these forces acting at O.
(SPCE Nov 12)

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EQUILIBRIUM OF SPACE FORCES
When the resultant of a system is zero, the system is said to be in
equilibrium. For the resultant to be zero, the resultant force F and the
resultant moment M should be zero.
This gives us six scalar equations of equilibrium viz.
0xF  0xM 
0yF  0yM 
0zF  0zM 
SUPPORTS FOR SPACE STRUCTURE
a) Ball and Socket Support
This support allows rotation in all the three
direction, unlike a hinge which allows rotation
about one axis only. A ball and socket support
does not allow any linear movement in any
direction. This results in a reaction force having
components along x, y and z axis. Fig. shows a
ball and socket support offering three components of reaction force.
b) Fixed End Support
A fixed end will not allow rotation in any
direction, nor will allow any linear movement.
This results in a reaction moment and a
reaction force having components along x, y
and z axis.
Refer Fig.

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Equilibrium of Concurrent Space Force System
For a concurrent system, only the resultant force needs to be zero. This is the
necessary and sufficient condition of equilibrium. This gives us three scalar
equations of equilibrium, viz.
0xF 
0yF 
0zF 
We can also use the other three scalar equations of equilibrium viz.
0xM 
0yM 
0zM 
EQUILIBRIUM OF NON-CONCURRENT SPACE FORCE SYSTEM
For equilibrium of both parallel and general space force systems, all the six
scalar equations of equilibrium are applicable viz.
0xF  0xM 
0yF  0yM 
0zF  0zM 

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EXERCISE 3
1. A boom AB supports a load of 1000 N as shown. Neglect weight of the
boom. Determine tension in each cable and the reaction at A.
2. A vertical load of 1100 N is supported by the three
rods shown in figure. Find the force in each rod.
Points C, O and D are in XZ plane while point B
lies 5 m above this plane.
3. A vertical load of 1OOO N is supported by three
bars as shown. Find the force in each bar. Point
C, O and D are in the x-z plane while B is 1.5 m
above this plane.
4. A vertical tower DC shown is subjected to a horizontal force P = 5O kN at its
top and is anchored by two similar guy wires BC and AC. Calculate
a. Tension in the guy wires.
b. Thrust in the tower pole.
Co-ordinates of the points are as below,
O (0, 0, 0), B (0, 0, -4), D (3, 0, 0), A (0, 0, 4), C (3, 20, 0)

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5. A uniform thin plate ABCD, square in shape is
suspended by strings GA (2a m length), GD (2a m
length), GE  3amlength and GH (a m length) so that it
remains in horizontal plane. If the size of the plate is
2a x 2a and E is the midpoint of BC, find the tension in
each of the string if the plate weighs 400 N.
6. A square steel Plate 2400 mm  2400 mm has a mass of 1800 kg with
mass centre at G. Calculate the tension in each of the three cables with
which the plate is lifted while remaining horizontal. Length DG = 2400 mm.
7. A crate is supported by three cables as shown. Determine the weight of the
crate, if the tension in the cable AB is 750 N.
8. Determine the tension in cable BC and BD and reactions at the ball socket
joint A for the mast as shown in figure.

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9. A vertical mast OD is having base ‘0’ with ball and socket. Three cables DA,
DB and DC keep the mast in equilibrium. If tension in the cable DA is 100
kN, find tensions in the cables DB and DC and force in the mast. Refer
figure.
10. A load of 500 N is held in equilibrium by means of two strings CA and CB
and by a force p as shown in figure. Determine tensions in strings and
magnitude of P.
11. A uniform equilateral triangular plate weighing 2.4
kN is held in horizontal plane by means of three
cables at A, B and C. An additional load of 1.2 kN
acts at midpoint of edge AC of the plate. Determine
the tensions in the three cables.
12. A plate ACED 1O mm thick weighs 76O0 kg/ms. It is held in horizontal
plane by three wires at A, B and C. find tensions in the wires.

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13. Figure shows steel Pipe ABCD of length 1.8 m having 900 bend at B. The
Pipe weighs 30 N/m and supported by three vertical wires attached to the
Pipe at A, C and D. Find tension in wires.
EXERCISE 4
THEORY QUESTIONS
Q.1 How is a space force represented in magnitude and direction.
Q.2 Write the equations of equilibrium to be satisfied for non co-planar
concurrent force system. (SPCE Nov 12)
Q.3 State conditions of equilibrium for non-concurrent forces in space.
UNIVERSITY QUESTIONS
1. A pole is held in place by three cables. If the force of each cable acting on
the pole is as shown in fig. determine the resultant. (6 Marks)
2. A force of 10 kN acts at a point P(2,3,5) m and has its line of action passing
through Q(10, -3, 4)m. Calculate moment of this force about a point
S(1,-10,3) m. (4 Marks)
3. Explain conditions for equilibrium for forces in spare. (6 Marks)

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4. Force F = (3i – 4j + 12k) N acts at point A (1,-2,3) Find -- (4 Marks)
(i) Moment of force about origin
(ii) Moment of force about point B (2,1,2)m.
5. A rectangular parallelepiped carries Three forces shown in fig. Reduce the
force system to a resultant force applied at the origin and a moment around
origin. (6 Marks)
6. The lines of action of three forces concurrent at origin ‘O’ pass respectively
through points A(-1,2,4), B (3,0,-3) and C(2,-2,4)m. The magnitude of
forces are 40N, 10N and 30N respectively. Determine the magnitude and
direction of their resultant. (6 Marks)
7. A force of 1200N acts along PQ, P(4,5,-2) and Q (-3,1,6)m. Calculate its
moment about a point A(3,2,0)m (4 Marks)
8. A force 80 50 60F i j k   passes through a point A Compute its moment
about point B (8,1,4). (6 Marks)

14. space forces

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14. space forces