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Units and Dimensions

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 :

 (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/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.

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 (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°

Case II

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)

60=PQ

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.

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 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.

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 =

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.

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 = 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

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                                            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.