The following units of different systems are mostly used .:
1. C.G.S. (i.e., Centimetre-Gram Second) system of units.
2. M.K.S. (i.e., Metre-Kilogram-Second) system of units.
3. S.I. (i.e. , International) system of units.
1.2.1. C.G.S. System of Units. In this system, length is expressed in centimetre, mass in gram and time in second. The unit of force in this system is dyne, which is defined as the force acting on a mass of one gram and producing an acceleration of one centimetre per second square.
1.2.2. M.K.S. System of Units. In this system, length is expressed in metre, mass in kilogram and time in second. The unit of force in this system is expressed as kilogram force and is represented as kgf.
1.2.3. S.I. System of Units. S.l. is abbreviation for ‘The System International Units’. It is also called the International System of Units. In this system length is expressed in metre mass in kilogram and time in second. The unit of force in this system is Newton and is represented N. Newton is the force acting on a mass of one kilogram and producing an acceleration. of one metre per second square. The relation between newton (N) and dyne is obtained as One Newton = One kilogram mass x one metre / s2
= 1000 gm x 100 cm / s2
= 1000 x 100 x gm x cm / s2
= 105 dyne
When the magnitude of forces is very large, then the unit of force like kilo-newton and mega-newton is used. Kilo-newton is represented by kN.
One kilo-newton = 103 newton
or 1 kN = 103 N
and One mega newton= 106 Newton
The large quantities are represented by kilo, mega, giga and tera. They stand for :
Kilo= 103 and represented by ……. k
Mega= 106 and represented by ……. M
Giga = 109 and represented by ……. G
Tera = 1012 and represented by …….. T
Thus mega newton means 106 newton and is represented by MN. Similarly, giga newton means 109 N and is represented by GN. The symbol TN stands for 1012 N.
The small quantities are represented by milli, micro, nano and pico. They are equal to
Milli = 10-3 and represented by ……. m
Micro = 10-6 and represented by ……. µ
Nano = 10-9 and represented by …….. n
Pico = 10-12 and represented by ……. p.
Thus milli newton means 10-3 newton and is represented by mN. Micro newton means
10-6 N and is represented by µN.
The relation between kilogram force (kgf) and newton (N) is given by One kgf = 9.81 N
Weight of a body is the force with which the body is attracted towards earth. If
W = weight of a body, m = mass in kg, then W = m x g Newtons
If mass, m of the body is 1 kg, then its weight will be,
W = 1 (Kg) x 9.81 m /s2 = 9.81 N.
Table 1.1 shows the multiples and sub-multiples of the S.I. units prefixes.
Table 1.1. S.I. Prefixes
Prefix Symbol Multiplying factor
Tera T 1012 = 1 000 000 000 000
Gig a G 109 = 1000 000 000
Mega M 106 = 1 000 000
Kilo k 103 = 1000
Milli m l0-3 = 0.001
Micro µ 10-6 = 0. 0.000 001
Nano n 10-9 = 0.000 000 001
Pi co p 10-12 = 0.000 000 000 001
The basic, supplementary and some derived units with dimensions are given as :
Physical quantity Notation or Unit Dimension or Symbol
Length metre m
Mass kilogram kg
Time Second s
Electric Current Ampere A
Temperature Kelvin K
Luminous Intensity Candella cd
(B) Supplementary Units
Plane angle Radian rad
Solid angle Steridian sr
(C) Derived Units
Acceleration metre/second2 m/s2
Angualr velocity radian/second rad/s
Angular acceleration radianlsecond2 rad/s2
Force Newton N =kg m/s2
Momeq.t of force Newton metre Nm
Work, Energy Joule J = Nm =kg m2/s2
Torque Newton metre Nm
Power Watt W=J/s
Pressure Pascal Pa = N/m2
Frequency Hertz Hz= s-1
Problem 1.1. Two forces of magnitude 10 N and 8 N are acting at a point. If the angle between the two forces is 60°, determine the magnitude of the resultant force.
Force P= 10 N
Force Q = B·N
Angle between the two forces, a = 60°
The magnitude of the resultant force (R) is given by equation (1.1)
Problem 1.2. Two equal forces are acting at a point with an angle of 60° between them. If the resultant force is equal to 20 x .J3 N, find magnitude of each force.
Sol. Given : Angle between the force, a = 60°
Resultant, R = 20 x √3
The forces are equal. Let P is the magnitude of each force.
Using equation (1.3), we have
R = 2p cos a / 2 20 x √3 = 2p x cos (600 / 2) = 2p cos 300
= 2p x √3 / 2 = p x √3
P = 20 x √3 / √3 = 20 N.
Magnitude of each force = 20 N. Ans.
Problem 1.3. The resultant of the two forces, when they act at an angle of60o is 14 N. If the same forces are acting at right angles, their resultant is .J 136 N . Determine the magnitude of the two forces.
Sol. Given: Case I
Resultant, R1 = 14 N
Angle, a = 60°
Let the magnitude of the two forces are P and Q.
Using equation (1.1) for case I.
136 = P2 + Q2
Subtracting equation (ii) from equation (i), we get
196 – 136 = P2 + Q2 + PQ – (P2 + Q2)
Multiplying the above equation by two, we get 120 = 2PQ
Adding equation (iv) to equation (ii), we get 136 + 120 = P2 + Q2 + 2PQ
256 = P2 + Q2 + 2PQ or (16)2 = (P + Q)2
16 =P + Q
p = (16- Q)
Substituting the value of P in equation (iii\ we get
60 = (16 – Q) x Q = 16Q- Q2 or Q2 – 16Q + 60 = 0
. . This is a quadratic equation.
= 16 + 4 / 2 and 16 – 4 / 2 = 10 and 6.
Substituting the value of Q in equation (v ), we get
P = (16- 10) or (16- 6) = 6 or 10.
Hence the two forces are 10 N and 6 N. Ans.
Problem 1.4. Two forces are acting at a point 0 as shown in Fig. 1. 7. Determine the resultant in magnitude and direction.
Force P =50 N, Force Q = 100 N
Angle between the two forces, a = 30°
The magnitude of the resultant R is given by equation (1.1) as
The resultant R is shown in Fig. 1.8.
The angle made by the resultant with the direction of P is given by equation (1.2) as
tan θ = Q sin a / P + Q sin a
θ = tan -1 (Q sin a / P + Q sin a) = tan -1 ( 100 x sin 300 / 50 x 100 cos 300)
= tan -1 0.366 = 20.10o
Angle made by resultant with x-axis = 9 + 15° = 20.10 + 15 = 35.10°. Ans.
Problem 1.5. The resultant of two concurrent forces is 1500 Nand the angle between the forces is 90°. The resultant makes an angle of 36° with one of the force. Find the magnitude of each force.
Resultant, R = 1500 N
Angle between the forces, a = 90°
Angle made by resultant with one force, 9 = 36°
Let P and Q are two forces.
Using equation (1.2), tan θ = Q sin a/ P + Q cos 90°
tan 36° = Q sin 90° / P + Q cos90° = Q x 1 / P + Q x 1 = Q /P or 0.726 = Q / P
= Q = 0.726 P
Using equation (1.1) , R =
R2 = P2 + Q2 + 2PQ cos a
15002 = p2 + (0.726 p)2 + 2p(0.726) x cos 90°
15002 = p2 + 0.527 p2 + 0
= 1.527 p2
Substituting the value of Pin equation (i), we get
Q = 0.726 x 1213.86 = 881.26 N. Ans.
Alternate Method. Refer to Fig. 1.8 (a). Consider triangle OAC.
Using sine rule, we get
sin 90° / R = sin 36°Q = sin 54° / P
sin 90° / R = sin 36°Q
Q = R sin 36° / sin 90° (where R = 1500 N)
= 1500 x 0.5877 / 1 = 881.67 N.
Also , we have sin 90° / R = sin 54° / p
P = R sin 54° / sin 90° = 1500 x 0.8090 / 1
= 1213.52 N.
Problem 1.6. The sum of two concurrent forces P and Q is 270 N and their resultant is 180 N. The angle between the force P and resultant R is 90°. Find the magnitude of each force and angle between them.
Sum of two concurrent forces = 270 N or P + Q = 270 N
Resultant, R = 180 N
Angle between force P and resultant R = 900
This means θ = 900
Find: (i) Magnitude of P and Q
(ii) Angle between P and Q (i.e., angle a)
Using equation (1.2), tan θ = Q sin a / P + Q cos a
tan = 900 Q sin a / P + Q cos a
But tan goo =900 (i.e., infinity). This is only possible when P + Q cos a= 0
P= -Q cos a
The above result can also be obtained by using alternate method.
Alternate Method. Refer to Fig. 1.8 (b). Consider triangle OAC in which 8 = 90°, LOAC
= 180 – a, LACO = a- θ = a – 90o
Using sine rule, we get sin 90° / Q = sin (180 –a) / R = sin (a – 90) / P
From first and last terms, we get sin 90° / Q = sin ( a – 90) / P
or 1 / Q = – cos a / P [ sin (a – 90) = sin [-(90 –a)] (90 –a)] = – sin (90 –a) = – cos a]
p = – Q cos a
This is the same result as given by equation (i) above.
Using equation (1.1), Squaring to both sides, we get R2 = P2 + Q2 + 2PQ cos a
= P2 + Q2 + 2P(-P)
( From equation (i), Q cos a = – P)
= p2 + Q2 -2P2 = Q2 – p2 = ( Q + P)( Q _ P)
or 1802 = 270 (Q – P) ( R = 180,. Q + P = 270)
or 32400 = 270(Q- P)
Q / P = 32400 / 270 = 120
P + Q 270 (given)
Adding the above two equations, we get 2Q = 270 + 120 = 3go
Q = 195 N. Ans.
And P = 270-Q = 270- 195·= 75 N. Ans.
Value of angle a
Substituting the values of P and Q in equation (i),
P = – Q cos a or 75 = – 195 a cos a = -75 / 195 = -0.3846
a = cos -1 ( -0.3846) = 112.618°
Problem 1.7. A weight of 1000 N is supported by two chains as shown in Fig. 1.9.
Determine the tension in each chain.
Sol. Given: Weight at C = 1000 N
< CAB = 30°
< CBA = 60°
< ACB =goo
In right-angled triangle ADC,
< ACD = goo – 30° = 60°
In right-angled triangle BDC,
< BCD = goo – 60° = 30°
< ACE = 180° – 60° = 120°
<BCE =180° – 30° = 150°
Let T1 =Tension in chain No.1
T2 =Tension in chain No.2.
Applying Lame’s theorem at point C (Refer Fig. 1.10).
T1 / sin 150° = T2 / sin 120° = 1000 / sin 90°
T1 / sin 150° = T3 / sin 120° = 1000
T2 = 1000 sin 150° = 1000 x .5 = 500 N
T2 = 1000 sin 120° = 1000 x .866 = 866 N.
Problem 1.8. A weight of 900 N is supported by two chains of lengths 4 m and 3 m as shown in Fig. 1.11. Determine the tension in each chain.
Sol. Given: Weight at C = 900 N
Length, AC = 4 m
Length, BC = 3 m
Length, AB = 5 m
In triangle ABC
AC2 + BC2 = 42 + 32 = 16 + g = 25
AB2 =52 = 25
AB2 =AC2 + BC2
. . Triangle ABC is a right-angled triangle in
< ACB = 90°
sin a = BC / AB = 3/ 5 = 0.6
a = 36° 52 and a + b =90°
b = 90° – a = 90° – (36° 52 ) = 53° 8
let T1 = Tension in chain AC
T2 = Tension in chain BC
In right-angled triangle BDC,
θ l = 90° – p = 90°- 53° 8′ = 36° 52′
< ACE = 180° – 91 = 180° – 53° 8′ = 126° 52′
< BCE = 180°-92 = 180°-36° 52′= 143° 8′ and LACE= 90°.
Applying Lame’s theorem at C
T1 / sin of < BCE = T2 / sin of < ACE = 900 / sin 90°
T2 / sin 143° 8 = T2 / sin 126° 52 = 900 / 1
T2 = 900 x sin 126° 52 = 720 N.