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 / s_{2}

= 1000 gm x 100 cm / s_{2}

= 1000 x 100 x gm x cm / s_{2}

= 10^{5} 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= 10^{3 }and represented by ……. k

Mega= 10^{6} and represented by ……. M

Giga = 10^{9} and represented by ……. G

Tera = 10^{12} and represented by …….. T

Thus mega newton means 10^{6} newton and is represented by MN. Similarly, giga newton means 109 N and is represented by GN. The symbol TN stands for 10^{12} 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 /s^{2} = 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 10^{12} = 1 000 000 000 000

Gig a G 10^{9} = 1000 000 000

Mega M 10^{6} = 1 000 000

Kilo k 10^{3} = 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 :

** (A)Basic Units**

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/second^{2} m/s^{2}

Angualr velocity radian/second rad/s

Angular acceleration radianlsecond^{2} rad/s^{2}

Force Newton N =kg m/s^{2}

Momeq.t of force Newton metre Nm

Work, Energy Joule J = Nm =kg m^{2}/s^{2}

Torque Newton metre Nm

Power Watt W=J/s

Pressure Pascal Pa = N/m^{2}

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.

**Sol**. Given:

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 (60^{0} / 2) = 2p cos 30^{0}

= 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°

**Case II**

Let the magnitude of the two forces are P and Q.

Using equation (1.1) for case I.

136 = P^{2} + Q^{2}

Subtracting equation (ii) from equation (i), we get

196 – 136 = P^{2 }+ Q^{2} + PQ – (P^{2} + Q^{2})

60=PQ

Multiplying the above equation by two, we get 120 = 2PQ

Adding equation (iv) to equation (ii), we get 136 + 120 = P^{2} + Q^{2} + 2PQ

256 = P^{2} + Q^{2} + 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- Q^{2} or Q^{2} – 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.

**Sol.** Given:

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 30^{0} / 50 x 100 cos 30^{0})

= tan -1 0.366 = 20.10^{o}

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.

**Sol**. Given:

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 =

R^{2} = P^{2} + Q^{2} + 2PQ cos a

1500^{2} = p^{2} + (0.726 p)^{2} + 2p(0.726) x cos 90°

1500^{2} = p^{2} + 0.527 p^{2} + 0

= 1.527 p^{2}

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.

**Sol.** Given:

Sum of two concurrent forces = 270 N or P + Q = 270 N

Resultant, R = 180 N

Angle between force P and resultant R = 90^{0}

This means θ = 90^{0}

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 = 90^{0} Q sin a / P + Q cos a

But tan goo =90^{0} (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 – 90^{o}

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 R^{2} = P^{2} + Q^{2} + 2PQ cos a

= P^{2} + Q^{2} + 2P(-P)

( From equation (i), Q cos a = – P)

= p^{2} + Q^{2 }-2P^{2} = Q^{2} – p^{2} = ( 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 T_{1 }=Tension in chain No.1

T_{2} =Tension in chain No.2.

Applying Lame’s theorem at point C (Refer Fig. 1.10).

T_{1} / sin 150° = T_{2} / sin 120° = 1000 / sin 90°

T_{1} / sin 150° = T_{3} / sin 120° = 1000

T_{2} = 1000 sin 150° = 1000 x .5 = 500 N

T_{2} = 1000 sin 120° = 1000 x .866 = 866 N.

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**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

AC^{2} + BC^{2 }= 42 + 32 = 16 + g = 25

AB^{2} =52 = 25

AB^{2} =AC^{2} + BC^{2}

. . Triangle ABC is a right-angled triangle in

Which

< 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 T_{1} = Tension in chain AC

T_{2} = 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

T_{1} / sin of < BCE = T_{2} / sin of < ACE = 900 / sin 90°

T_{2 }/ sin 143° 8 = T_{2} / sin 126° 52 = 900 / 1

T_{2 }= 900 x sin 126° 52 = 720 N.