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 F_{x} ,*F _{y} *and F

_{z}by its edges.

Let *θ _{x}*= Angle made by force F with x-axis

*θ*

*= Angle made by force F with*

_{y}*y-axis*

*θ*

_{z}= Angle made by force F with z-axis

Now *F _{x} *=

*F*cos θ

_{x}*F*=

_{y }*F*cos θ

*… (1.26)*

_{y}*F*

*=*

_{z }*F*cos θ

_{z}

The cosines *of **θ _{x,} θ*

*and*

_{y }*θ*

_{z }are known as the direction cosines of the force

**F**and denoted by

*l = *cos θ_{x, }*m* = cos θ_{y} and *n* = cos θ_{z }…(1.27 A)

The three angles” are related by

cos^{2} θ_{x,}*+ *cos^{2} θ_{y} + cos^{2} θ_{z} = 1_{ }…(1.28)

or *l*^{2} + *m*^{2 + }*n*^{2 = }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 = F_{x }**i + **F_{y }**j + **F_{z }**k **…(1.29)

Magnitude of the vector **F** is given by

F = | F | =

But F* _{x}* = F cos θ

_{x}_{, }F

*cos θ*

_{y}*and F*

_{y}*cos θ*

_{z}*Substituting these values in equation (1.29), we get*

_{z.}F = (F cos θ* _{x}*)

**i**

_{ + }(F cos θ

*)*

_{y}**j +**and (F cos θ

*)*

_{z}**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 cos^{2} θ* _{x }*+ cos

^{2 }θ

*= 1. This is proved as,*

_{y}*θ*= 90 –

_{y }*θ*

_{x}*Hence cos θ*

_{y}= cos (90

– θ

*) = sin*

_{x}*θ*Or cos

_{x}^{2}θ

_{y}= sin

^{2}θ

*. Therefore, cos*

_{x}^{2 }

*θ*+ cos

_{x}^{2}θ

*= cos*

_{y }^{2}

*θ*+ sin

_{x}^{2}θ

*which is equal to 1.*

_{x}

**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(x _{1}, *

*y*

_{1},

*z*) and

_{1}*B(x*

_{2}, y_{2}, z_{2})Let *d *= Distance between *A *and *B *

θ* _{x}* = Angle made by force

**F**with

*x*-axis

θ* _{y} *= Angle made by force

**F**with

*y*-axis

θ* _{z}* = 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 *= *(x** _{2 }*– x

*),*

_{1}*dy*= (

*y*–

_{2}*y*)

_{1}and *dz *= *(z _{2} *–

*z*) … (1.32)

_{1}Also* dx *= *d *cos *θ _{x}, *

*dy*=

*d*cos θ

_{y}and * dz *= *d *cos θ_{2}

Also we know that

*F _{x} *=

*F*cos θ

*,*

_{x}*F*

*=*

_{y }*F*cos θ

*y*

and* F** _{z }*=

*F*cos θ

_{z}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 **= x**i** + y**j **+ z**k**

*(ii) *The position vector of a point *A *with co-ordinates *(x _{2}*,

*y*,

_{2}*z*

_{2}

*)*with respect to point

*B(x*,

_{1}*y*,

_{1}*z*

_{1}

*)*is given by (see Fig. 1.82

*a*)

**r** = *(x _{2 }*–

*x*

_{1})**i**+

*(y*–

_{2}*y*

_{1})**j**+

*(z*–

_{2}*z*)

_{1}**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 M _{z} = 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 = 4*i* + *j* + 4*k*

Co-ordinates of point B = (- 3, 4, -1)

Position vector of point B = – 3*i* + 4*j* – *k *

Moment about a-axis, M_{z} = 1900 Nm

Let magnitude of force = P

*As force is directed from A to B. Hence co-ordinates of A(4, 1, 4) means *x*_{1} = 4, *y*_{1} = 1, *z*_{1} = 4. The co-ordinates of B(- 3, 4, – 1) means *x*_{2} = – 3, *y*_{2} = 4 and *z*_{2} = – 1.

Therefore position vector is (*x*_{2} – *x*_{l}) **i** + (*y*_{2}, – *y*_{ l}) **j** + (*z*_{2} – *z*_{l}) **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 **F **

* 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: *

**F _{1}** == 80i,

**F**= 130k,

_{2 }**F _{3}** == -50j

**F**= 210k,

_{4}**F _{5}** = -110i + 190j – 50k.

**F _{6}** = 315j.

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

**Sol.** Given:

**F _{1}** == 80i,

**F**= 130k,

_{2 }**F _{3}** == -50j

**F**= 210k,

_{4}**F _{5}** = -110i + 190j – 50k.

**F _{6}** = 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 = F _{1} + F_{2} + F_{3} + F_{4} + F_{5} + F_{6} **

= 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 **r _{1}** = Position vector of force F

_{1}from point A

= 0 (as force F

_{1}is passing through A)

**r _{2}** = Position vector of force F

_{2}from point A = 0

**r**= Position vector of force F

_{3}_{3}from point A

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

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

**r _{4}** = Position vector of force F

_{4}with respect to point A

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

= (2i + 8k) – (16j) = (2i – l6j + 8k)

**r _{5}** = Position vector of force F

_{5}from point A

= 2i – l6j + 8k

**r _{6}** = Position vector of force F

_{6}from point A

= 2i – l6j + 8k.

(Force F_{5} and force F_{6} are acting at point B, hence their position vector will be same as of force F_{4}.)

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

M_{A} = r_{1} × F_{1} + r_{2} × F_{2} + r_{3} × F_{3} + r_{4} × F_{4} + r_{5} × F_{5} + r_{6} × F_{6}

= 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.

M_{A }= [- 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. **

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