# 1.13 FORCES IN SPACE

When a force is represented in three dimensions, then the force is known as ‘Force in space’. Forces in space are generally represented in vector form.    In Fig. 1.80, the force F is represented by vector OA. Through point A, draw planes parallel to co-ordinate planes i.e., Coordinate planes are xy, yz and zx as shown in Fig. 1.80).

These planes along with co-ordinate planes make a rectangular box. The force F is then represented by the diagonal of the box and its three components Fx ,Fy and Fz by its edges.

Let θx= Angle made by force F with x-axis
θy= Angle made by force F with y-axis
θz= Angle made by force F with z-axis

Now    Fx  = F cos θx
F= F cos θy                                        … (1.26)
Fz = F cos θz

The cosines of θx, θy and θz are known as the direction cosines of the force F and denoted by

l = cos θx,           m = cos θy        and      n = cos θz                                                            …(1.27 A)

The three angles” are related by

cos2 θx,+ cos2 θy + cos2 θz = 1                                                                                                          …(1.28)

or         l2 + m2 + n2 = 1

Let       i = vector of unit length in the positive x-direction,

j = vector of unit length in the positive y-direction,

and      k = vector of unit length in the positive z-direction the force vector F is represented by

F = Fx i + Fy j + Fz k                                                                             …(1.29)

Magnitude of the vector F is given by

F = | F | =

But Fx = F cos θx, Fy cos θy and Fz cos θz. Substituting these values in equation (1.29), we get

F = (F cos θx)i + (F cos θy) j + and (F cos θz) k                                               …(1.30)

1.13.1. Unit Vector. The vector having a unit length (or unit magnitude) is known as unit vector. Hence unit vector corresponding to force vector F is equal to F + Magnitude of vector F

*For x, y co-ordinates cos2 θx + cos2 θy = 1. This is proved as, θy = 90 – θx Hence cos θy= cos (90
– θx) = sin θx Or cos2 θy= sin2 θx. Therefore, cos2 θx + cos2 θy = cos2 θx + sin2 θx which is equal to 1.

Note. Force vector is represented by THICK, BOLD FACE and magnitude is represented by

capital letters (without thickness) only.

1.13.2. Components of a Force when Two Points on its Line of Action are Given.

Fig. 1.81 shows a force F, which is defined by two points A(x1, y1, z1) and B(x2, y2, z2)

Let d = Distance between A and B

θx = Angle made by force F with x-axis

θy = Angle made by force F with y-axis

θz = Angle made by force F with z-axis

dx = Distance along x-axis between A and B

dy = Distance along y-axis between A and B

dz = Distance along z-axis between A and B.

Then                dx = (x2 – x1), dy = (y2y1)

and                  dz = (z2 z1)                                        … (1.32)

Also                 dx = d cos θx, dy = d cos θy

and                  dz = d cos θ2

Also we know that

Fx = F cos θx, Fy = F cos θy

and                              Fz = F cos θz

From equation (1.34A) the components of the force can be obtained, provided we know ‘F’ and ‘d’.

1.13.3. Position Vector of a Given Point. The position vector of a point can be given with respect to origin or with respect to another point.

(i) The position vector of a point A with respect to origin O is the vector OA. It is represented by r. It is given by (see Fig. 1.82)

r = xi + yj + zk

(ii) The position vector of a point A with co-ordinates (x2, y2, z2) with respect to point
B(x1, y1, z1) is given by (see Fig. 1.82a)

r = (x2 x1) i+ (y2 y1) j + (z2 z1) k.
Problem 1.44.A force vector is represented by a line AB. The co-ordinates of point A are (2, 4, 3) and of point B(1, – 5,2) respectively. If the magnitude of force= 10 N, then determine:

(i) the components of the force along X, y and z-axis,
(ii) angles with the X, y and z-axis, and
(iii) specify the force vector.

Sol. Given:
Co-ordinates of point A are (2,4,3)
∴ x1 = 2,
y1 = 4
and z1 = 3
Co-ordinates of point B are (1, – 5,2)
∴ x2 = 1,
y2 = -5
and z2 = 2
Magnitude of force, F = 10 N

Here the magnitude of the force and co-ordinates of two points through which force passes, are given. The force components can be obtained if we know ‘d’ (i.e., distance between two points). But d = √(〖dx〗^2+ 〖dy〗^2 + 〖dz〗^2 )
From equation (1.32),

dx= x2 – x1 = 1- 2 = -1
dy= y2 – y1 = – 5 – 4 = – 9
de = z2 – z1= 2 – 3 = – 1
and from equation (1.33),
d = √(〖dx〗^2+ 〖dy〗^2 + 〖dz〗^2 )= √(〖(-1)〗^2+〖(-9)〗^2 + 〖(-1)〗^2 )= √(1+ 81 + 1)= √83=9.11

(i) Components of the force along the axes
Let Fx = Component of the force F along x-axis.
Fy = Component of the force F along y-axis.
Fz = Component of the force F along z-axis.
Now using equation (1.34), we get
F_x/dx=F_y/dy=F_z/dz=F/d
Substituting the values of dx, dy, dz, F and d, we get
F_x/(-1)=F_y/(-9)=F_z/(-1)=10/9.11
∴ F_x=(-1)× (-10)/9.11=(-10)/9.11=-1.1 N. Ans.
F_y=(-9)× 10/9.11=(-90)/9.11=-9.88 N. Ans.
F_z=(-1)× 10/9.11=(-10)/9.11=-1.1 N. Ans.

(ii) Angles with x, y and z-axes
Let θx = The angle that the force F makes with z-axis,
θy = The angle made by F with y-axis, and
θz = The angle made by F with z-axis.
Using equation (1.26),
Fx = F cos θx
∴ 〖cos θ〗_x=F_x/F=(-1.1)/10=-0.11
θx = cos-1 (- 0.11) = 96.3°. Ans.
Fy =Fcos θy
∴ 〖cos θ〗_y=F_y/F=((-9.88))/10=-0.988
∴ θ_y=θ^(-1) (-0.988)=〖171.09〗^o Ans.
Also Fz =Fcos θz
∴ 〖cos θ〗_z=F_z/F=((-1.097))/10=-0.11 or=〖cos〗^(-1) (-0.11)=〖96.3〗^o. Ans

(iii) Specify the force vector
The force vector F is represented by equation (1.29) as
F =Fxi +Fyj +Fzk =-1.1 i + (- 9.88) j + (-1.1) k
(∴ Fx = -1.1, Fy = – 9:88 and Fz = -1.1)
= -1.1 i – 9.88 j -1.1 k. Ans.

IInd Method
The force vector F can also be obtained as given below:
The vector joining the points (2, 4, 3) and (1, – 5, 2) is given by

= dx i +dx j + dz k = (x2 –x1) i + (y2 –y1) j + (z2 –z1) k
= (1- 2) i + (- 5 – 4) j + (2 – 3) k = – 1 i – 9 j – 1 k.

Now find the unit vector in the direction of above vector i.e., in the direction of Vector (-1 i -9 j -1 k).

This unit vector is obtained from equation (1.31) and is equal to the given vector divided by the square root of the sum of squares of its components.

∴ Unit vector in the direction of the vector (- 1 i – 9 j – 1k) is
= ((- 1 i – 9 j – 1k))/√(〖(-1)〗^2+ 〖(-9)〗^2+ 〖(-1)〗^2 )= ((- 1 i – 9 j – 1k))/√(1+81+1)= ((- 1 i – 9 j – 1k))/9.11
The force vector F = Magnitude of force times unit vector in the direction of force
= (10(- 1 i – 9 j – 1k))/9.11= (10 ×1)/9.11 i-(10 ×9)/9.11 j-(10 ×1)/9.11 k

= – 1.1 i – 9.88 j -1.1 k. Ans.

Problem 1.45. With respect to origin (0, 0, 0), a point A is located at t- 5, 2, 14). Specify the position of the point:
in terms of orthogonal components,
in terms of directional cosine, and
in terms of unit vector.

Sol. Given:

Co-ordinates of origins are (0, 0, 0)
Co-ordinates of point A are (- 5, 2, 14).
(i) Position of the point in terms of orthogonal components
The components of the vector along x-axis, y-axis and a-axis are given as :

Along x-axis = (- 5 – 0) = – 5
Along z-axis = (2 – 0) = 2
Along z-axis = (14 – 0) = 14
∴ Position vector, r = – 5 i + 2 j + 14 k. Ans.

(ii) Position of the point in terms of direction cosine
The magnitude of vector is given by,
|r| = r = √(〖(- 5)〗^2+〖(2)〗^2+〖(14)〗^2 )= √(25+4+196) = √255 = 15.
The direction cosines are given by equation (1.27 A) as
1 = cos θx,
m = cos θy
n = cos θz
Also from Eq. (1.26), we have
cos θ_x = (Component of vector along x-axis )/(Magnitude of vector )= (-5)/15 = -0.333

cos⁡〖θ_y 〗= (Component of vector along y-axis )/(Magnitude of vector )= 2/15 = 0.0133

cos θ_z = (Component of vector along z-axis )/(Magnitude of vector )= 14/15 = 0.933
∴ Direction cosines are
l = cos θx = – 0.333
m = cos θy = 0.0133
n = cos θz = 0.933

(iii) In terms of unit vector

The unit vector in the direction of the vector r = – 5 i + 2 j + 14 k) is (- 5 i + 2 j + 14 k )/(|r|)

= (- 5 i + 2 j + 14 k )/15= -0.333 i+0.0133 j+0.933 k. Ans

Problem 1.46.A force P is acting as shown in Fig. 1.83. Express the force P as vector. Magnitude of force P = 500 N.

Sol. Given:
Length a = 2 cm.
Magnitude of P = 500 N.
The force P is acting from A to G.
Hence point A corresponds to 1 and point G corresponds to 2.

The point G is lying on x-axis.
Hence co-ordinates of point G are (a, 0, 0) or (2,0,0).
‘The point A is in y-z plane
∴ x2=2, y2=0 and z2=0
The co-ordinates of the point A are (0, a, a) or (0,2,2).
∴ x1=0, y1=2 and z1=2
The vector joining the points A (0,2,2) and G (2,0,0) is given by
= dx i+ dy j+ dz k
= (x2- x1) i + (y2 – y1) j + (z2 – z1) k
= (2 – 0) i + (0 – 2) j + (0 – 2) k
= 2 i – 2 j -2 k.

The unit vector in the direction of vector 2 i – 2 j -2 k is
=(2 i – 2 j -2 k)/√(〖(2)〗^2+〖(-2)〗^2+〖(-2)〗^2 )=(2 i – 2 j -2 k)/√12
The force vector P is given by
= Magnitude of force P times unit vector in the direction of force

=500((2 i – 2 j -2 k)/√12)=1000/√12 i-1000/√12 j-1000/√12 k

=288.683 i-288.683 j-288.683 k Ans.

IInd Method

The force vector P can also be obtained as
P = Pxi+ Pyi+ Pzk ….(i)
where P_x=P/d (x^2-x^1 )=P/d (dx)=2P/d (∵ dx=2)
P = 500 N (given)
d=√(〖dx〗^2+〖dy〗^2+〖dz〗^2 )= √(2^2+〖(-2)〗^2+〖(-2)〗^2 ) (∵dy=-2,dz=-2)
∴ P_x=P/d dx=2P/d=(2×500)/√12=1000/√12
P_y=P/d dy=500/√12×(-2)=(-1000)/√12
P_z=P/d dz=500/√12×(-2)=(-1000)/√12
∴Force vector P=P_x i+P_y j+P_z k
=1000/√12 i+(((-1000))/√12)j+(((-1000))/√12)k
=1000/√12 i-1000/√12 j-1000/√12 k
=288.683i-288.683j-288.683k Ans.

Problem 1.47.The lines of action of three forces concurrent at the origin 0 passes respectively through points A, B and C having coordinates:
xA=-1, yA = 2, zA =4 ; xB = 3, yB = 0, zB=-3 and xC=2,yC=-2 and zC =4
The magnitude of the forces are FA = 40 N, FB = 10 N and FC = 30 N. Find the magnitude direction of their resultant.

Sol. Given:
Co-ordinates of point A are (- 1, 2, 4)
Co-ordinates of point B are (3, 0, -3)
Co-ordinates of point C are (2, – 2, 4)

The forces passes through the origin and points A, B, C respectively.
Magnitude of forces are
FA=40 N, FB=10 N and FC=30 N.
The position vector of force FA

= (- 1 – 0) i + (2 – 0) j + (4 – 0) k = – 1 i + 2 j + 4 k
The unit vector in the direction of force FA
=((-1 i+ 2 j+4 k))/√(1^2+2^2-4^2 )=((-1 i+ 2 j+4 k))/√21=((-1 i+ 2 j+4 k))/4.5825
Magnitude of force FA = 40 N
∴ Force vector FA = Magnitude of FA times unit vector
=40((-i+2 j+4 k)/4.5825)=-40/4.5825 i+(40×2)/4.5825 j+(40×4)/4.5825 k
=-8.278 i+ 17.456 j+34.912 k …(i)
Similarly, the position vector of force FB
=3 i+0 j-3 k=3 i -3 k
The unit vector in the direction of force FB
=(3 i -3 k)/√(3^2+3^2 )=(3 i -3 k)/4.242=3/4.242 i-3/4.242 k=0.7072 i-0.7272 k
∴ Force vector FB = Magnitude of FB times unit vector in the direction of force FB
= 10 (0.7072 i – 0.7072 k)
= 7.072 i – 7.072 k … (ii)
Similarly force vector FC = Magnitude of FC times unit vector in the direction of FC
=30 ((2 i-2 j+4 k))/√(2^2+2^2+4^2 )=30 ((2 i- 2 j+4 k))/√24=((60 i+60 j+120 k))/4.899
=60/4.899 i-60/4.899 j+120/4.899 k
= 12.247 i – 12.247 j + 24.494 k … (iii)
The resultant force vector is obtained by adding force vectors FA, FB and FC [or by adding equations (i), (ii) and (iii)].
∴ Resultant force vector
= Force vector FA + Force vector FB + Force vector FC
= (- 8.728 i + 17.456 j + 34.912 k) + (7.072 i -7.072 k) + (12.247 i -12.247 j + 24.494 k)
= (- 8.728 i + 7.072 i + 12.247 i) + (17.456 j -12.247 j) + (34.912 k – 7.072 k + 24.494 k)
= 10.591 i + 5.209 j + 52.334 k
Force Vector FA
Let us find the force vector FA first. It is given as
FA = (FA)x i + (FA)y j + (FA)z k … (i)
The force FA passes through point A (- 1, 2, 4) and origin (0, 0, 0)
∴ dx = (x2 – x1) = (-1- 0) = -1, dy = (y2 – y1)= 2 – 0 = 2
dz = (z2 – z1) = (4 – 0)= 4
and d=√(〖dx〗^2+〖dy〗^2+〖dz〗^2 )=√((-1)^2+2^2+4^2 )=√21=4.5825
Now (F_A )_x=F_A/d dx=40/4.5825 (-1)=(-40)/4.5825=-8.728
(∴ FA = 40 N, d = 4.5825, dx = -1)
(F_A )_y=F_A/d dy=40/4.5825 (2)=80/4.5825=17.456
and (F_A )_z=F_A/d dz=40/4.5825 (4)=160/4.5825=34.912
Substituting these values in equation (i), we get
FA = – 8.728 i + 17.456} + 34.912 k …. (A)
Now calculate similarly force vectors FB and FC

Force Vector FB
FB = (FB)x i + (FB)y j + (FB)z k
The force FB = 10 N passes through point B(3, 0,-3) and orgin (0,0,0)
∴ dx = (3 – 0) = 3, dy =(0 – 0) = 0, dz = (-3 – 0) = -3,
and d=√(〖dx〗^2+〖dy〗^2+〖dz〗^2 )=√(3^2+0^2+〖(-3)〗^2 )=√18=4.242
Now (F_B )_x=F_B/d dx=10/4.242×(3)=30/4.242=7.072
(∵ FB = 10 N, dx = 3)
(F_B )_y=F_B/d dy=10/4.242 (0)=0
and (F_B )_z=F_B/d dz=10/4.242 (-3)=-7.072
Substituting these values in equation (ii), we get
FB = 7.072 i – 7.072 k

Force Vector FC
FC = (FC)x i + (FC)y j + (FC)z k …(iii)
The force FC = 30 N, passes through point C(2, – 2, 4) and origin (0, 0, 0)
∴ dx = 2 – 0 = 2, dy = – 2 – 0 = – 2, dz = 4 – 0 = 4
and d=√(〖dx〗^2+〖dy〗^2+〖dz〗^2 )=√(2^2+〖(-2)〗^2+4^2 )=√24=4.899
Now (F_C )_x=F_C/d dx=30/4.899×2=12.247
(F_C )_y=F_C/d dy=30/4.899 (-2)=(-60)/4.899=-12.247
and (F_C )_z=F_C/d dz=30/4.899 (4)=120/4.899=24.494
Substituting these values in equation (iii), we get
FC = (FC)x i + (FC)y j + (FC)z k
= 12.247 i – 12.247 ) + 24.494 k
The resultant force vector is obtained by adding force vectors FA, FB and FC [or by adding equations (i), (ii) and (iii)] ∴ Resultant force vector
= FA + FB + FC
= [(8.728 i + 17.456) + 34.912 k)] + [7.072 i -7.072 k] + [12.247 i – 12.247 j + 24.494 k] = [- 8.728 i + 7.072 i + 12.247 i] + [17.456j – 12.247}] + [34.912 k -7.072 k + 24.494 k] = 10.591 i + 5.209 j + 52.334 k. Ans.
∴ Magnitude of resultant force
=√(〖10.591〗^2+〖5.209〗^2+〖52.334〗^2 )=√(112.169+27.133+2738.847)
= 53.648 N. Ans.
Direction of the resultant force is obtained by finding unit vector in the direction of resultant force vector.
∴ Unit vector in the direction of resultant force vector 10.591 i + 5.209 j + 52.334 k is given by

=(10.591 i + 5.209 j + 52.0334 k)/√(〖10.591〗^2+〖5.209〗^2+〖52.334〗^2 )=(10.591 i + 5.209 j + 52.334 k)/53.648

=10.591/53.648 i+5.209/53.648 j+52.334/53.648 k=0.1974 i+0.0971 j+0.9755 k

∴ Direction of resultant force is 0.1974 i + 0.0971 j + 0.9755 k. Ans.

Problem 1.48. A force P = 911 N is directed from a point A (4, 1, 4) metres towards a point B(-3,4, 1) metres. Determine the force vector P.

Sol. Given:
Magnitude of P = 911 N
Co-ordinates of point A = (4,1,4)
∴ Position vector of point A = (4 i + 1 j + 4 k)
Co-ordinates of point B = (- 3,4, -1)
∴ Position vector of point B = (- 3 i + 4 j – k)
∴ Position* vector of force P
= Position vector of B
– Position vector of A
= (- 3 – 4) i + (4 – 1) j + (- 1- 4) k
= -7 i + 3 j – 5 k
Unit vector in the direction of force P
=(-7 i + 3 j – 5 k )/√(〖(-7)〗^2+3^2+〖(-5)〗^2 )
=(-7 i + 3 j – 5 k )/√(49+9+25)=(-7 i + 3 j – 5 k )/9.11
∴ Force vector P = Magnitude of P times unit vector in the direction of force
=9.11 ((-7 i + 3 j – 5 k))/9.11
=(911(-7))/9.11 i+(911×3)/9.11 j-(911×5)/9.11 k=-700 i+300 j-500k. Ans.

Problem 1.49. A force P is directed from a point A (4, 1, 4) metres towards a point B(-3,
4, – 1) metres. If it causes a moment Mz = 1900 Nm, determine the moment of P about x and
y-axis.

Sol. Given:

Co-ordinates of point A = (4, 1, 4)

Position vector of point A = 4i + j + 4k
Co-ordinates of point B = (- 3, 4, -1)
Position vector of point B = – 3i + 4jk

Moment about a-axis,         Mz = 1900 Nm

Let magnitude of force = P

*As force is directed from A to B. Hence co-ordinates of A(4, 1, 4) means x1 = 4, y1 = 1, z1 = 4. The co-ordinates of B(- 3, 4, – 1) means x2 = – 3, y2 = 4 and z2 = – 1.

Therefore position vector is (x2xl) i + (y2, – y l) j + (z2zl) k.

The force P is directed from A to B. Hence position vector of force P is given by

R = Position vector of point B – position vector of point A

= (3i + 4j – k) – (4i + j + 4k) = -7i + 3j – 5k

Now unit vector in the direction of force

Tho position vector of point A = (4i + j + 4k)

Moment of the force about origin O is given by cross product of position* vector of point A with the given force vector.

Let M = Moment about origin O

Then M = Position vector of point A × Force vector

* The moment of the given force about origin is also equal to the cross product of position vector of point B with given force vector. The result will be same.

**In the moment equation (i.e., M), the co-efficient of ‘k’ gives moment about z-axis.

Problem 1.50. Fig. 1.84(a) shows a rectangular parallel piped subjected to four forces in the directions shown. Reduce them to a resultant force applied at the origin and a moment

Sol. Given:

Force at C = 400 kN in the direction of CO

Force at F = 1000 kN in the direction of OF

Force at B = 800 kN in the direction of CB

Force at E = 707 kN in the direction ofAE.

Co-ordinates of point C = (5, 0, 0)

Position vector of point C = 5i + 0J + 0k = 5i

Fig. 1.84(a)

Co-ordinates of point B = (5, 3, 0)

Position vector of point B = (5i + 3j)
Co-ordinates of point F = (0, 3, 4)

Position vector of point F = (0i + 3j + 4k)

Co-ordinates of point E = (5, 0, 4)

Position vector of point E = (5i + 0J + 4k)

Co-ordinates of point A = (0, 3, 0)

Position vector of point A = (0i + 3j + 0k)
Position vector of force 400 kN

= Position vector of point O* – Position vector of point C
= (0i + 0j + 0k) – (5i) = – 5i

Position vector of force 1000 kN

= Position vector of point F – Position vector of O
= (0i + 3j + 4k) – (0i + 0j + 0k) = 3j + 4k

Position vector of force 800 kN

= Position vector of point B – Position vector of C

= (5i+3j) – (5i) = 3j

*Co-ordinates of point O is (0,0,0). Hence position vector of point O = (0i + 0j + 0k).

Position vector of force 707 kN

= Position vector of point E – Position vector of A
= (5i + 0J + 4k) – (0i + 3j + 0k) = (5i – 3j + 4k)

Now the unit vectors in the direction of forces can be obtained from the position vector or forces.

=(-5i)/√(〖(5)〗^2 )=(-5i)/5=-i
Unit vector in the direction of force 1000 kN
=(+3j+4k)/√(3^2+4^2 )=(3j+4k)/5
Unit vector in the direction of force 800 kN
=(+3j)/√(3^2 )=3j/3=j
Unit vector in the direction of force 707 kN
=(5i-3j+4k)/√(5^2+〖(-3)〗^2+4^2 )=(5i-3j+4k)/√(25+9+16)=((5i-3j+4k))/√50
Now different force vectors can be obtained by multiplying the force with the unit vector in the direction of the force.
∴ Force vector of 400 kN
= (400 kN) times the unit vector in the direction of 400 kN
= 400 times (- i) = – 400 I … (i)
Similarly, force vector of 1000 kN
= 1000 times unit vector in the direction of 1000 kN
=1000 ((3j+4k)/5)=200(3j+4k)=600j+800k …(ii)
Force vector of 800 kN
= 800 (unit vector in the direction of 800 kN)
= 800 (j) = 800j … (iii)
Force vector of 707 kN
= 707 (unit vector in the direction of 707 kN)
=707 ((5i-3j+4k))/√50=(707(5i+3j+4k))/7.07=100(5i-3j+4k)
=5000i – 3000j + 400k
∴ The resultant force vector (R) will be obtained by adding equations (i), (ii), (iii) and (iv)
∴ Resultant force vector R
=(-400i)+(600j+800k)+(800j)+(500i-300j+400k)
=(-400+500)i+(600+800-300)j+(800+400)k
=1000i+1100j+1200k. Ans.
The magnitude of resultant force R is given by
R=√(〖100〗^2+〖1100〗^2+〖1200〗^2 )+√(10000+1210000+1440000)=√2660000
Now let us find the moment of each force vector about O the origin.
We have to find resultant force applied at the origin and the moment.
The moment of the resultant force about origin will be zero, as resultant force is passing through the origin.
The moments of force 400 kN and 1000 kN about origin will also be zero as these forces passing through origin.
The position vector r of 800 kN
= Position vector of point C – Position vector of O = 5i
The position vector r of 707 kN with respect to origin
= Position vector of point A – Position vector of O = 3j
∴ Moment of force 800 kN about origin
= Position vector r of force 800 kN × Force vector 800 kN
= 5i × 800j (∴ Force vector of 800 kN = 800 j)
=5i×800j (∵Force vector of 800 kN=800j)
Moment of force 707 kN about origin
= Position vector r of force 707 kN × Force vector 707 kN
= 3j × (500i – 300j + 400k) = 1500j × i – 900j × j + 1200j × k
= 1500 (-k) – 900 (0) + 1200 (i) [∵ j x i = -k, j x j = 0, j x k = i] = -1500 k + 1200 i
= 1200 i – 5000 k
∴ Sum of moments of all forces at origin
= 4000 k + (1200 i – 1500 k) = 1200 i + 2500 k. Ans.
Problem 1.51. Three forces P, Q and R are applied along the diagonals of rectangular parallelepiped as shown in Fig. 1.85. If 2P = 2Q = R = 1000 N. Determine the equivalent force and moment at the origin.
Sol. Given:
2P = 2Q = R = 1000 N
∴P=1000/2=500 N,Q=500 N
R = 1000 N
Now position vector of force P
= Position vector of point O
– Position vector of point E
= (0i + 0j + 0k) – (4i + 0j + 3k)
= (- 4i – 3k)
Position vector of force Q
= Position vector of point D
– Position vector of point A
= (4i + 2j + 3k) – (4i + 0j + 0k)
= (2j + 3k)

Fig.1.85

Position vector of force R
= Position vector of point B – Position vector of point F
= (4i + 2j + 0k) – (0i + 0j + 3k) = (4i – 2j – 3k)
Unit vector in the direction of force P
=(-4i-3k)/√(〖(-4)〗^2+〖(-3)〗^2 )=(-4i-3k)/5
Unit vector in the direction of force Q
=(2j+3k)/√(2^2+3^2 )=(2j+3k)/√13=(2j+3k)/3.6
Unit vector in the direction of force R
=(4i+2j-3k)/√(4^2+2^2+〖(-3)〗^2 )=(4i+2j-3k)/√(16+4+9)=(4i+2j-3k)/√29=(4i+2j-3k)/5.385
Now force vector P
= Magnitude of P times the unit vector in the direction of P
=100(-4i-3k)=-400i-300k
Similarly, force vector Q
= Q times the unit vector in the direction of Q
=500((2k+3k)/3.6)=1000j/3.6+1500k/3.6=277.77j+416.66k
Also force vector R
= Magnitude of R times unit vector in the direction of R
=1000((4i+2j-3k)/5.385)=4000/5.385 i+2000/5.385 j-3000/5.385 k
=742.8i+371.4j-557.1k
∴ Resultant force vector
= Force vector P + Force vector Q + Force vector R
= (-400i – 300k) + (277.77j + 416.66k) + (742.8i + 371.4j – 557.1k)
= (-400 + 742.8)i + (277.77 + 371.4)j + (- 300 + 416.66 – 557.1)k
= 342.8i + 649.17j – 440.44k. Ans.
Magnitude of resultant force
=√(〖342.8〗^2+〖649.17〗^2+〖440.44〗^2 )=√(117511.84+421421.68+193987.39)
= 856.1 N. Ans.
Moment of all forces at O is given by
Mo = r1 × Force vector P + r2 × Force vector Q + r3 × Force vector R
Where r1 = Position vector of a point on force vector P = 0
(∵ Force P is passing through origin)
r2 = Position vector of a point on force vector Q = 4i or (4i + 2j + 3k)
r3 = Position vector of a point on force vector R = 3k or (4i + 2j + 0k)
∴ M0 = 0 × (- 400i – 300k) + 4i × (277.77) + 416.66k) + 3k × (742.8i + 371.4) – 557.1k)
= 0 + (4) (277.77)i × j + (4) (416.66)i × k + (3) (742.8)k× i + (3)(371.4)k× j + (3) (- 557.4)k×k=0
= (4)(277.77)k + (4) (416.66) (-j) + (3) (742.8)(j) + (3)(371.4) (-i) + 0
( ∵ i × j = k, i × k = -j, k × i =j. k × j =-i , k × k = 0)
= 1111.08k -1666.64i + 2228.4j -1114.2i
= – 1114.2i + (2228.4 – 1666.64)j + 1111.08k
= -1114.2i + 561.76j + 1111.08k. Ans.

Problem 1.52. A bar AB is acted upon by Six forces as shown in Fig. 1.86. The forces are:

F1 == 80i,                F2 = 130k,

F3 == -50j                F4 = 210k,

F5 = -110i + 190j – 50k.

F6 = 315j.

All forces are in Newtons. Determine the equivalent force and moment acting at A.

Sol. Given:

F1 == 80i,                F2 = 130k,

F3 == -50j                F4 = 210k,

F5 = -110i + 190j – 50k.

F6 = 315j.

Position vector of point A = 16j
Position vector of point B = (2i + 8k)
Position vector of point C = (i + 8j + 4k).
(i) To find the equivalent force
Equivalent force is the resultant force.
The resultant force R is given by

R = F1 + F2 + F3 + F4 + F5 + F6

= 80i + 130k + ( – 50j) + (- 210k) + (-100i + 190) – 50k) + 315)j
= (80-100)i + (-50 + 190 + 315)j + (130 – 210 – 50)k

= – 20i + 455j – 130k.   Ans.

Fig 1.86

(ii) To find the moment of all forces about point A

Let    r1 = Position vector of force F1 from point A
= 0 (as force F1 is passing through A)

r2 = Position vector of force F2 from point A = 0
r3 = Position vector of force F3 from point A

= Position vector of point C – position vector of point A

= (i + 8j + 4k) – (16j) = (i – 8j + 4k)

r4 = Position vector of force F4 with respect to point A

= Position vector of point B – position vector of point A
= (2i + 8k) – (16j) = (2i – l6j + 8k)

r5 = Position vector of force F5 from point A
= 2i – l6j + 8k

r6 = Position vector of force F6 from point A
= 2i – l6j + 8k.

(Force F5 and force F6 are acting at point B, hence their position vector will be same as of force F4.)

Moment* of all forces about point A is given by,

MA = r1 × F1 + r2 × F2 + r3 × F3 + r4 × F4 + r5 × F5 + r6 × F6
= 0 + 0 + (i – 8j + 4k) × (- 50j) + (2i – l6j + 8k) × (- 210k)

+ (2i – 16j + 8k) × (- 100i + 190j – 50k) + (2i- 16j + 8k) × (315j)

= (- 50i × j + 400j × j – 200k × j) + (- 420i × k + 3360j × k

– l680k × k) + (- 200i × i + 380i × j – 100i × k + 1600j × i

– 16 × 190 × j × j + 800j × k – 800k × i + 1520k × j

– 400k × k) + (630i × j – 16 × 315 × j × j + 2520 k × j)

But      i × j = k, j × j = 0, i × i = 0, k × k = 0, k × j = – i, i × k = – i.
j × k == i, j × i = – k, k × i = j, k × j = – i.

MA = [- 50k + 0 – 200 (- i)] + l- 420 (- j) + 3360i – 0]

+ [0 + 380k – 100(- j) + 1600 (- k) – 0 + 800i – 800j

+ 1520 (- i) – 0] + [630k – 0 + 2520 (- i)]

= – 50k + 200i + 420j + 3360i + 380k + 100j – 1600k + 800i – 800j – 1520i + 630k – 2520i
= (200 + 3360 + 800 – 1520 – 2520)i + (420 + 100 – 800)j + (- 50 + 380 – 1600 + 630)k

= 320i + 290j – 640k.  Ans.