Units, Mass, Force, Weight & Basic Definitions PDF

1 T H E UNITS, MASS, F O R C E , WEIGHT & BASIC DEFINITIONS I H.I. Units : Before dealing w i t h the definitions o f such fundamental t e r m s as mass, I n...

1 T H E UNITS, MASS, F O R C E , WEIGHT & BASIC DEFINITIONS I H.I. Units : Before dealing w i t h the definitions o f such fundamental t e r m s as mass, I n n r , weight a n d the principles o f applied mechanisms i t is essential t o e x p l a i n i l i ' ' units used i n this book as they w i l l be the units o f the future n o t o n l y i n this unity b u t also i n o t h e r l e a d i n g c o u n t r i e s. They are k n o w n as S.I. u n i t s. | At present the u n i t s i n o u r c o u n t r y are o f M.K.S. system ; M stands f o r m e t r e 1111 ill of length), K for kilogramme (unit of mass), and S for second ( u n i t o f t i m e ). T h e l e t t e r s S.I. are a n a b b r e v i a t i o n f o r t h e f u l l n a m e Systems international d units ( I n t e r n a t i o n a l System o f U n i t s ). S.I. is the system o f units Introduced by a general conference o f l e a d i n g countries, i n c l u d i n g I n d i a , o n rlxhts a n d measures i n 1 9 6 0. I t is based u p o n m e t r i c u n i t s. A conspicuous 11 1111 re o f S.I. u n i t s is t h e a d o p t i o n o f a u n i t called N e w t o n as a u n i t o f w e i g h t " l li nee. Some o t h e r u n i t s i n S.I. w h i c h a t t r a c t a t t e n t i o n are : Joule ( u n i t o f w o r k or e n e r g y ). Pascal ( u n i t o f pressure) Watt (unit of power) : T h i s u n i t is used n o t o n l y for e l e c t r i c m o t o r s IMII also for i n t e r n a l c o m b u s t i o n engines w h i c h used horse-power as a u n i t so i n In S.I. u n i t s the u n i t h o r s e p o w e r hasno place. Six b a s i c quantities i n S.I. are : Quantity Unit Symbol Mass kilogramme kg Length meter m Time second s Temperature kelvin K Electric c u r r e n t ampere A Luminous intensity candela cd O t h e r units i n S.I. are d e r i v e d f r o m these basic u n i t s.  Elements of Mining Technology-3 /1.3 The units, mass, force, weight, & basic definitions/1.2 ii. The u n i t o f pressure, t h e Pascal (Pa), i.e. the pressure p r o d u c e d b y a A "metre per second" is k n o w n as a derived unit because i t measures force o f one N e w t o n a p p l i e d , u n i f o r m l y d i s t r i b u t e d , over a n area o f one q u a n t i t y , speed, i n terms of b o t h e r quantities, l e n g t h a n d t i m e. Units of one m ( 1 N / m ). 1 b a r = 100 k Pa. The value o f standard atmospheric 2 2 area a n d v o l u m e , are also derived u n i t s. pressure, i n t e r n a t i o n a l l y accepted, is 1 0 1 , 3 2 5 N / m. T h i s is v e r y 2 Mass. The u n i t o f mass, kilogramme, is the mass o f a p l a t i - n u m - close to t h e u n i t b a r (b) w h i c h is 1 0 N / m , i.e. 1 0 0 , 0 0 0 N / m. 5 2 2 i r i d i u m c y l i n d e r kept at the I n t e r n a t i o n a l Bureau o f Weights a n d Measures at Meteorologists generally use as t h e i r u n i t the bar, a n d quite often u n i t Sevres, near Paris. As the basic u n i t of mass is k i l o g r a m m e , the w o r d gramme millibar, w h i c h , as the n a m e i m p l i e s , is o n e - t h o u s a n d t h o f a bar. does n o t Fig. i n S.I. u n i t s ; e.g., density o f w a t e r is 100 k g / m. One should not 2 iii. The u n i t o f energy, J o u l e ( J ) , i.e. the w o r k done w h e n t h e p o i n t o f make the mistake o f w r i t i n g i t as 1 M g / m ; i t s h o u l d be w r i t t e n as 10 k g / m , 3 :i 3 a p p l i c a t i o n o f a force o f one N e w t o n is displaced t h r o u g h a distance o r 1000 k g / m. 3 o f one m e t r e i n the d i r e c t i o n o f the force. L e n g t h : The u n i t o f length, the metre, is distance between t w o lines iv. The u n i t o f p o w e r , t h e W a t t ( W ) , i.e. the rate of d o i n g w o r k at one o n a p l a t i n u m - i r i d i u m m bar w h i c h is kept at the I n t e r n a t i o n a l Bureau of j o u l e per second. Weights and Measures at Sevres. The measurment is made at 0°C and standard atmospheric pressure. A l l t h e d e r i v e d u n i t s i n mechanics can be expressed i n t e r m s o f t h e T i m e : The u n i t o f t i m e , the second, is based u p o n the mean sola basic u n i t o f l e n g t h , mass a n d t i m e. For example, day, this b e i n g the average t i m e between successive transits o f the sun d u r i n g ^. Mass Mass M L„-%\ the year. 1 second = 1/86400 o f a mean solar day. Density = — = — = — r = IML I Volume (Length) {hf 3 ' K This t i m e is i n d i c a t e d f a i r l y accurately b y a " q u a r t z " clock or w a t c T e m p e r a t u r e : The u n i t o f t h e r m o d y n a n i c t e m p e r a t u r e , k e l v i n , V e l o c i t y = -fe-' = ( L T " ) 1 the degree i n t e r v a l o n the t h e r m o d y a n m i c scale. The t e m p e r a t u r e o f the ic T freezing p o i n t o f w a t e r o n this scale is 273.15K, c o r r e s p o n d i n g to 0°C o n thj Change i n v e l o c i t y Celsius scale. Conversion f r o m °C to K e l v i n , for a l l p r a c t i c a l purposes : Acceleration = Time Temperature i n K e l v i n = 2 7 3 + t °C The t e m p e r a t u r e intervals o n the Klevin a n d Celsius scale are equ " ( T ) ( T ) - ( T f " 1 ' The sign o f degree (°) is n o t used for K e l v i n. Note: The i n t e r n a t i o n a l definitions o f the metre a n d the second a L, M a n d T denote the dimensions o f length, mass and t i m e respectively. m o r e c o m p l e x a n d are o f o n l y acdemical interest to students a n d readers The S.I. u n i t s are i n d e p e n d e n t o f g, the acceleration d u e to force o f books o n engineering. gravity, a n d are clear i n s t a t e m e n t , f o r example, The S.I. is a coherent system of units. A system of units is coherent the p r o d u c t o r q u o t i e n t o f any t w o u n i t quantities i n the system is the u n i t P ( N e w t o n s ) = M ( K i l o g r a m m e ) x f (m/s ) 2 the resultant quantity. P e r m i t t e d n o n S.I. units. The S.I. System has given a proper allowance A l l o t h e r units i n S.I. are derived f r o m the six basic units m e n t i o n for some n o n - S. I. u n i t s t h a t are i n use f o r a l o n g t i m e a n d have a b e a r i n g w i t h here a n d the c o m m o n l y used derived units are the f o l l o w i n g : MKS system. These u n i t s are p e r m i t t e d because they have c e r t a i n advantages i n p a r t i c u l a r fields, e.g. l i t r e , a t m o s p h e r e (for pressure), °C for t e m p e r a t u r e , i. T h e u n i t o f force, t h e N e w t o n ( N ) , i.e. t h a t force w h i c h w h are some o r t h e p e r m i t t e d n o n S.I. u n i t s. a p p l i e d to a mass o f one k i l o g r a m m e , w i l l give i t an acceleration 1 m/s. 2  The units, mass, force, weight, & basic definitions/1.4 _^ Elements of Mining Technology -3 /1.5 The prefixes attached to t h e u n i t to denote d e c i m a l m u l t i p l e or These p a r t i c u l a r prefixes are g r a d u a l l y f a l l i n g i n t o disuse. s u b m u l t i p l e o f the l a t t e r are as follows : I n t h i s b o o k the practice o f f o r m i n g groups o f t h r e e d i g i t s is n o t Multiplier rigidly adopted. S t a n d a r d for Prefix Abbreviation A t m a n y places i n this b o o k , for conversion of q u a n t i t i e s o f FPS o r One t h o u s a n d m i l l i o n. 10 9 giga G metric system o f units i n t o S.I. units use has been made o f u n i t y brackets ans One m i l l i o n S t r o u d s y s t e m o f c o n v e r s i o n. I n this system, devised by Professor S t r o u d , 10 6 mega M unity brackets are used. Since 6 0 m i n u t e s e q u a l one h o u r , i f w e w r i t e One thousand 10 3 kilo k One h u n d r e d 10 hecto* h 60 m i n 2 t h e n this is called a unity bracket, U n i t y brackets can be w r i t t e n w i t h Ten lh 10 1 deca* da One-teeth IO- deci* d lkN 1000 k N 1 1 min 60 s_ One-hundredth io- 2 ' centi* c either quantity as numerator, e.g. 60 s 1 min 1000 N lkN~ One-millionth io- 6 micro I n m o s t problems, unless the u n i t s of measurement are g i v e n as basic One-thousand m i l l i o n t h io- 9 nano n units, t h e y m u s t be changed i n t o basic u n i t s. Use o f S t r o u d system is c l a r i f i e d T h e 1 2 7 5 0 0 0 N / m = 1275 k N / m = 1.275 M N / m. 2 2 2 in t h e f o l l o w i n g e x a m p l e. I n u s i n g these m u l t i p l e s a n d s u b - m u l t i p l e s , o n l y one prefix s h o u l d Example: be used : thus 1 2 7 5 0 0 0 0 0 0 N / m 2 o r 1.275 G N / m , n o t 1.275 k M N / m. 2 2 Change 7 2 k m / h t o m/s. One t h o u s a n d N e w t o n s have to be w r i t t e n as 1 k N or as 1 0 N. 3 km 1000m 1 h 1 min For n o n - d e c i m a l figures the n u m e r i c a l values are w r i t t e n i n I n d i a as Ans. 7 2 k m / h = 72 1 km 60 m i n 60 s follows : 9 1 4 3 5 7 6 2 is w r i t t e n as 9,14,35,762 720 m since w e have such units as lakh (also w r i t t e n as lac) a n d crore; 1 36s lakh = 100000; 1 crore = 10000000. The m o d e r n practice i n foregin countries Basic terms. is to w r i t e a n u m e r i c a l value i n groups o f three d i g i t s s t a r t i n g f r o m the r i g h t h a n d side w i t h o u t the use o f c o m m a. Thus, the figure 9 1 4 3 5 7 6 2 is w r i t t e n as Mass and weight: 9 1 4 3 5 7 6 2. For w r i t i n g d e c i m a l figures the practice i n B r i t i s h a n d A m e r i c a n Mass is the measure o f q u a n t i t y o f m a t t e r i n a body a n d is a c o n s t a n t books is to f o r m groups of three digits c o m m e n c i n g f r o m the decimal p o i n t o n p r o p e r t y o f t h a t body. The u n i t o f mass i n MKS units as w e l l as i n S.I. u n i t s is the r i g h t h a n d side a n d also o n the left hand side. For example, kilogramme. There is a difference b e t w e e n mass a n d on weight. I f a person h o l d i n g 15692.43 as 15 692.43 a stone i n his h a n d releases his g r i p o n i t , is falls to the e a r t h. C l e a r l y a force 1 0. 5 4 7 3 7 as 1 0. 5 4 7 3 7 m u s t have p u l l e d t h e stone t o w a r d s the e a r t h. This force, w e c a l l a force o f g r a v i t y or g r a v i t a t i o n a l p u l l. We feel this w h e n w e h o l d the stone i n o u r Western countries have such h i g h units as, a m i l l i o n ( 1 0 0 0 0 0 0 ) , a hands. T h e g r a v i t a t i o n a l p u l l acts v e r t i c a l l y d o w n w a r d s a n d the w e i g h t o f a b i l l i o n (1 0 0 0 0 0 0 0 0 0 ) a n d a t r i l l i o n w h i c h is eqvalent to a m i l l i o n m i l l i o n i n b o d y indicates i t. A l l bodies are a t t r a c t t o w o r d s the centre o f the e a r t h b y t h i s America ( 1 0 0 0 0 0 0 0 0 0 0 0 0 ). g r a v i t a t i o n a l p u l l a n d f u r t h e r the b o d y f r o m t h e centre o f the e a r t h , less t h e  The units, mass, force, weight, & basic definitions/1.6 Elements of Mining Technology-3 /1.7 p u l l , less is its w e i g h t. A b o d y h a v i n g a definite mass w i l l have less w e i g h t at One N e w t o n is t h a t force w h i c h , w h e n a p p l i e d t o a b o d y h a v i n g a the t o p o f M o u n t Everest t h a n t h a t at the sea level. I f an a s t r o n a u t o n a flight mass o f 1 k i l o g r a m m e , gives i t a n a c c e l e r a t i o n o f 1 m per s e c o n d. 2 to the m o o n carries w i t h h i m a packet o f v i t a m i n pills w h i c h weighs one k i l o - 1 D y n e is t h a t force w h i c h , w h e n a p p l i e d to a b o d y h a v i n g a mass o f g r a m m e o n the e a r t h before the f l i g h t , the packet, assuming t h a t n o t h i n g was I g r a m m e , gives i t a n a c c e l e r a t i o n o f 1 cra/s ( 1 D y n e = 1 0 N ). I t is a very 2 - 5 removed or consumed o u t of it, w i l l have no weight w h e n the astronaut reaches I mall unit. a p o i n t d u r i n g the f l i g h t w h e r e the earth's gravitational p u l l is so negligible as R e a d e r s a c c u s t o m e d t o M K S u n i t s s o m e t i m e s use t h e w o r d. to be zero. I f at t h a t p o i n t i n the space the packet is w e i g h e d b y a s p r i n g " k i l o g r a m m e " w i t h o u t b e i n g specific, w h e t h e r i t refers t o mass o r w e i g h t. balance, the w e i g h t w i l l be zero t h o u g h the mass o f the packet is the same as When w e say "a b o d y o f 1 kg", the statement is n o t clear a n d causes confusion i t was o n t h e e a r t h. As the a s t r o n a u t approaches the m o o n , its g r a v i t a t i o n a l since the mass is expressed i n k i l o g r a m m e a n d the w e i g h t is also expressed i n p u l l w i l l case t h e s p r i n g balance t o record some w e i g h t o f the packet o f kilogramme, i n c o m m o n usage. M a n y of y o u m i g h t have observed the w e i g h i n g v i t a m i n pills, a n d o n the surface o f the m o o n the w e i g h t recorded w i l l be l / 6 t h machine at r a i l w a y stations or i n some d e p a r t m e n t a l stores w h i c h is operated o f the w e i g h t o n the earth, because the g r a v i t a t i o n a l p u l l o f the m o o n is 1/6"' by inserting a 10-paisa c o i n i n a slot o n the machine. You stand o n its p l a t f o r m , t h a t o f the earth. This should e x p l a i n the difference between the terms w e i g h t insert a 10-paisa c o i n a n d w i t h i n a second o r t w o , emerges a c a r d , t h e size o f and mass; The w e i g h t o f a b o d y is the force of gravity o the body. The i m p o r t a n t a r a i l w a y t i c k e t , o n w h i c h is t y p e d y o u r w e i g h t s t a t i n g , say, 60 k g. A c t u a l l y p o i n t t o n o t e is t h a t m a s s is c o n s t a n t b u t w e i g h t is v a r i a b l e. I he w o r d k g o n t h e card is w r o n g. I t s h o u l d be kgf. B u t t h i s indicates h o w The l a t t e r depends o n the place w h e r e the b o d y is s i t u a t e d. I f at a h e i g h t o f loosely the w o r d k g is o f t e n used. say, 4 0 0 k m above the earth's surface, a n astronaut weighs a n a r t i c l e o f mass M k g w i t h a s p r i n g balance a n d the g r a v i t a t i o n a l acceleration at the p o i n t is l k g f is t h e w e i g h t o f a b o d y h a v i n g a mass o f one k i l o g r a m m e. g m/s , t h e w e i g h t o f the article i n S.I. u n i t s , w i l l be M g N e w t o n s. 2 The g r a v i t a t i o n a l force o n a mass o f 1 k i l o g r a m m e , i n N e w t o n s , is the mass o f I kg m u l t i p l i e d b y t h e a c c e l e r a t i o n d u e to g r a v i t y ( i n m / s ). 2 As a l r e a d y stated, a s t a n d a r d piece o f p l a t i n u m - i r i d i u m is k e p t near Paris as a basis f o r the m e a s u r e m e n t o f mass a n d its mass is 1 k i l o g r a m m e. T h u s 1 k g f = 1 k g x 9. 8 1 m/s = 9. 8 1 k g m / s = 9. 8 1 N e w t o n s. 2 2 The masses o f o t h e r bodies are f o u n d f r o m comparison w i t h this standard and T h i s f o l l o w s f r o m t h e basic e q u a t i o n. t h e c o m p a r i s o n is m a d e b y w e i g h i n g. I f a b o d y has a mass d o u b l e t h a t o f the Force = Mass x acceleration standard, t h e force due to gravity a c t i n g o n i t is double the force acting o n the W h e n w e say t h a t a m i n e car weighs 1 tonne, i t means t h a t the w e i g h t s t a n d a r d. Mass is i n d e p e n d e n t o f g r a v i t a t i o n a l p u l l since any v a r i a t i o n i n the of the m i n e car is 1,000 k g f a n d expressed i n S.I. u n i t s t h e w e i g h t is 9 8 1 0 l a t t e r w i l l have a n e q u a l effect o n the s t a n d a r d mass. Newtons. T h i s represents t h e w e i g h t o f a mass o f 1,000 k i l o g r a m m e s. Since T h e f a m i l i a r b e a m balance compares t h e masses o f t w o bodies, a lhe w o r d k i l o g r a m m e - f o r c e (kgf) is used to express w e i g h t ( o r force), i t is k n o w n mass i n one p a n a n d the mass to be compared i n another p a n. A spring ippropriate t o use the w o r d "tonne-force" o r "tonnefe" to express the w e i g h t o f balance, o n the o t h e r h a n d , measures the force i n any d i r e c t i o n. a mass o f 1,000 k i l o g r a m m e s b u t text books r a r e l y use t h e w o r d t o n n e f e. The u n i t o f w e i g h t i n MKS system is k i l o g r a m m e - f o r c e (kgf) a n d i n Rigid Body: S.I. u n i t s i t is N e w t o n ( N ). A b o d y composed o f a very large n u m b e r o f particles whose positions If a m o v i n g body suddenly stops or changes its direction, one concludes lelative to one a n o t h e r d o n o t v a r y is considered to be a r i g i d body. A single t h a t s o m e t h i n g m u s t have caused i t ; i f a s t a t i o n a r y b o d y starts m o v i n g one particle is t a k e n to be a p o r t i o n o f m a t t e r whose d i m e n s i o n s are n e g l i g i b l e concludes that there must be something that caused the motion. This "something" and w h o s e p o s i t i o n m a y therefore be g i v e n as t h a t o f a m a t h e m a t i c a l p o i n t. i n the above t w o examples is Force w h i c h m a y be regarded as t h t w h i c h produces m o t i o n , destroys i t , o r changes the d i r e c t i o n a n d speed o f the dboy. Speed, Velocity and acceleration : A force has m a g n i t u d e , d i r e c t i o n a n d a p o i n t o f a p p l i c a t i o n. The u n i t o f force W h e n a b o d y is i n m o t i o n the rate at w h i c h i t is m o v i n g o n its p a r t h is the same as t h a t o f w e i g h t , viz. k g f in.MKS units a n d N e w t o n i n S.I. units. is called its speed o r velocity. A d i s t i n c t i o n is u s u a l l y m a d e b e t w e e n t h e terms O n the European c o n t i n e n t the s y m b o l k p or kgp (for k i l o p o n d ) is sometimes ipeed a n d velocity. W h i l e speed is d e f i n e d as the rate o n l y at w h i c h a b o d y is w r i t t e n i n place o f kgf. moving, the t e r m velocity implies d i r e c t i o n as w e l l as rate. Velocity is therefore  Elements of Mining Technology-3 /1.9 The units, mass, force, weight, & basic definitions/1.8 I f a b o d y moves i n a c i r c u l a r p a t h ( r o t a t i o n a l m o t i o n ) the a n g u l a r a vector q u a n t i t y b u t speed is a scalar quantity. A body moves w i t h u n i f o r m distance moves is measured i n radians a n d the angular v e l o c i t y is i n rad/sec. velocity w h e n t h r o u g h o u t the m o t i o n equal distances are covered i n equal Hie a n g u l a r acceleration is measured i n rad/sec. 2 times, however s m a l l or h o w e v e r large. ,. distance moved Velocity = There are 2n radians i n a circle. (Fig. 1.1.) R.P.M. x — = rad/sec. time taken Rest, motion and velocity are relative terms. Equations of linear and angular motion : M o t i o n o f a b o d y always means its m o t i o n w i t h reference to other For l i n e a r m o t i o n , i f body w h i c h m a y be at rest o r i n m o t i o n. Absolute m o t i o n is impossible to conceive. A l l o u r m o t i o n is w i t h reference to e a r t h w h i c h is supposed to be at u = v e l o c i t y o f a b o d y at start, rest for this concept. W h e n w e say, a t r a i n is m o v i n g at 4 0 k m p h , it means the v = v e l o c i t y o f a b o d y after t seconds, train's speed is 40 k m p h relative to the e a r t h , or the speed w i t h w h i c h it f = acceleration, appears to move to a n observer o n the e a r t h. s = distance covered w i t h i n t seconds, Consider a t r a i n m o v i n g at a u n i f o r m speed o f 30 m/sec. I f the engine puts i n m o r e p o w e r so t h a t the speed of the t r a i n changes, at a steady rate to and f o r a n g u l a r m o t i o n , i f 38 m/sec i n four seconds, the t r a i n has been accelerated by 2m/sec i n each cOj = a n g u l a r v e l o c i t y a t start, second. The acceleration is w r i t t e n as 2m/sec. Acceleration is defined as the 2 co = a n g u l a r v e l o c i t y after t s e c , 2 rate o f increase o f velocity. W h e n a body falls freely u n d e r the action o f earth's g r a v i t a t i o n a l p u l l , i t accelerates u n i f o r m l y ( c o n s i d e r i n g there is no air a = a n g u l a r acceleration resistance) a n d the value o f this acceleration is 9. 8 1 m/sec. The acceleration 2 9 = angle t u r n e d t h r o u g h i n t s e c , d u e t o g r a v i t y is w r i t t e n as g. Decceleration is the opposite of acceleration. A car m o v i n g at a speed of, say, 90 k m / h , i.e. 25 m/s deccelerates a n d reduces t h e n t h e equations are as f o l l o w s : its speed after a p p l i c a t i o n o f brakes a n d m a y e v e n t u a l l y stop. Decceleraion is for linear m o t i o n for a n g u l a r m o t i o n negative acceleration for purposes or calculations. v = u + ft co = co, + a t 2 The acceleration d u e to g r a v i t y varies f r o m place to place o n ourj e a r t h b u t i t varies so l i t t l e f r o m the sea level to M t. Everest, the highest p o i n t v = u + 2fs 2 2 co 22 = co, + 2 a 0 2 o n the e a r t h , t h a t t h e value is considered p r a c t i c a l l y constant and is t a k e n as 9. 8 1 m/s. cJl 1 2 = co, + - a t 2 1 s = ut + - f r 2 A b o d y m o v i n g i n a straight line has a l i n e a r v e l o c i t y a n d i f i t iSi 2 accelerated, i t is subjected to linear acceleration. A body m o v i n g i n a circular p a t h has an a n g u l a r v e l o c i t y a n d its acceleration is a n g u l a r acceleration. Consider a w h e e l r e v o l v i n g a t 1 8 0 r. p. m. = 3 r.p.s. 1 rev = 27i radians. Therefore a n g u l a r v e l o c i t y is 6n rad/sec. A n g u l a r v e l o c i t y o f every part o f t h e w h e e l is the same, b u t t h e l i n e a r velocities o f different parts differ. It is zero at the centre of the shaft a n d m a x i m u m at the periphery o f the wheel. In Fig. 1.2 the angular velocity o f p a r t i c l e P a n d Particle Q is the same b u t the linear v e l o c i t y o f Q is m o r e. The r e l a t i o n b e t w e e n l i n e a r v e l o c i t y V a n d a n g u l a r v e l o c i t y co is. v = tur, w h e r e r is t h e r a d i u s Fifi. 1.1. Arc AB = s = 9 radian x radius.  The units, mass, force, weight, & basic definitions /1.10 Elements of Mining Technology-3 /1.11 Example: A t r a i n m o v i n g at 36 k m / h is u n i f o r m l y accelerated so t h a t aft a n g u l a r acceleration = ^ ^ - 9 — — r a d / s 2 8 seconds its velocity is 54 m/s. Find the acceleration i n m/s and the distan 2 4x60 the t r a i n goes i n the i n t e r v a l. = 0.916 rad/s 2 Ans.: u = 36 k m / h. i.e. 10 m/s. 1 2 Since, w , = 0, t h e r e l a t i o n s h i p reduces to 0 = — «= t v = 54 k m / h , i.e. 15 m/s. A n g l e t u r n e d t h r o u g h i n the last t w o m i n u t e s can be calculated i f we To find acceleration : subtract f r o m the t o t a l angle t u r n e d the angular travel i n the first t w o minutes. Using v = u + ft Let t o t a l angle t u r n e d t h r o u g h i n 4 m i n u t e s be 0. Then 15 = 1 0 - 8 f 4 f = -m/s 2.-. 0 = ix0.916x(4x60) 2 v 8 _ 0.916x240x240 To find distance : 2 Using, distance s = average velocity x t i m e = 26380.8 radians 10 + 15 s = x 8 Let angle t u r n e d t h r o u g h i n first t w o m i n u t e s = 0 2 2 = 100 m.-. 9 2 = - x ( 0. 9 1 6 ) x (2 x 60f radians' 1 , A l t e r n a t i v e l y w e c o u l d have used either s = u t + — ft = 0. 9 1 6 x 120 x 60 radians = 6 5 9 5. 2 radians or v - u = 2fs 2 2 Example:.-. Angle t u r n e d t h r o u g h i n t h e last t w o o f the 4 m i n u t e s. A n electric m o t o r s t a r t i n g f r o m rest reaches a speed 2 1 0 0 r.p.m. i n J = 26380.8-6595.2 minutes. W h a t is the angular acceleration i n rad/s and h o w many revolutio: 2 = 1 9 7 8 5. 6 radians are t u r n e d i n the last t w o o f t h e four m i n u t e s. Ans.: 19785.6. 0 1, 0 = r e v o l u t i o n s = 3 1 4 8 rev. 271. Increase in angular velocity Angular acceleration = — = 3149 revolutions. Time taken Initial angular velocity = 0 Example: f i n a l angular velocity = 2 1 0 0 r.p.m. A l i f t is connected to a rope w h i c h w i n d s o n to a d r u m o f a r a d i u s 2 1 00 x 271 r a d / s 4 4 0 m m. The lift accelerates f r o m rest at the rate o f 0.33 m/s for 10s and t h e n 2 60 maintains constant velocity. Calculate the corresponding angular acceleration = 219.9 rad/s and a n g u l a r v e l o c i t y o f the d r u m i n u n i t s o f radians per second.  The units, mass, force, weight, & basic definitions /1.12 Ans.: Elements of Mining Technology-3 /1.13 D r u m radius = 4 4 0 m m = 0.44m T h e significance of t h i r d l a w w i l l be clear t o a n y one w h o has t r i e d t o A c c e l e r a t i o n o f the l i f t = 0.33 m / s 2 fire a g u n. As t h e g u n h e l d against t h e s h o u l d e r is fired t h e m o m e n t u m w i t h w h i c h t h e b u l l e t leaves t h e g u n results i n a r e c o i l against t h e shoulder. Let a n g u l a r acceleration o f the d r u m be a r a d / s 2 A n i n t e r e s t i n g a n d significant development based o n this l a w is t h e science of.'. pe w o u l d be ( 9. 8 N - 2 N ) because p a r t o f t h e g r a v i t i o n a l force ( 2 N ) w o u l d I f employed i n accelerating the mass, leaving the r e m a i n i n g 7.8 N counteracted I >y the rope. Fig. 1.2 Fig. 1.2 A. Banking of a curve. Still, or I Upwar- velocity Downward Equilibrium: constant acc" I acc" 2m/seci 2 2m/sec 2 A b o d y at rest is said t o be i n static e q u i l i b r i u m w h e n t h e forces acting o n i t keep i t steady a n d the system o f forces is said to be i n e q u i l i b r i u m. 9.8 N (9.8+2)N (9.8-2)N You m u s t have observed t h e game of t u g of w a r i n w h i c h t w o teams, each o f 9 - 1 1 players, p u l l a r o p e i n opposite d i r e c t i o n s. T h e t o t a l force of each t e a m acts o n the rope w h i c h is therefore subjected t o t w o forces i n opposite directions. W h e n the t w o teams exert equal force o n the rope, i t does n o t move | kg 1kg T b u t r e m a i n s steady. A t t h a t m o m e n t the rope is said t o be i n e q u i l i b r i u m a n d Floor reaction R the t w o forces a c t i n g o n t h e r o p e are s a i d t o be i n e q u i l i b r i u m. T h e forces a c t i n g o n t h e r o p e t r y i n g t o stretch i t , are tensile forces a n d t h e rope is i n Fig. 1.3 Fig. 1.3 (a) tension. A b o d y c a n n o t be i n e q u i l i b r i u m u n d e r the a c t i o n o f a single force, i \ : a n d i f t w o forces act o n a b o d y i t c a n o n l y be i n e q u i l i b r i u m i f t h e forces are of e q u a l m a g n i t u d e b u t act i n opposite d i r e c t i o n s a l o n g t h e same l i n e of A p e r s o n o f 7 0 k g stands i n a l i f t w h i c h moves w i t h a n acceleration of action. lm/s. W h a t w i l l be t h e pressure o n the f l o o r o f t h e l i f t (a) i f i t is a s c e n d i n g , 2 m -» B o w w h o i n t r o d u c e d i t i n 1 8 7 3. I n Graphic Statics, this n o t a t i o n is universally figures t h e r e s u l t a n t o f ab a n d be has t h e same m a g n i t u d e as vector ca b u t employed because i n c o m p l i c a t e d stress diagrams, i t becomes extremely easy the d i r e c t i o n is opposite t o t h a t o f the v e c t o r ca t o locate a p a r t i c u l a r force i n m a g n i t u d e , d i r e c t i o n a n d p o s i t i o n w h e n the — > — » — > n o t a t i o n is a d o p t e d. Bow's n o t a t i o n is i l l u s t r a t e d b y reference t o F i g. 1.15. ab + be = ac Force Polygon : I f m o r e t h a n t w o coplaner forces act at a p o i n t i t is easy t o o b t a i n their resultant b y a d d i n g the vectors as i n d i c a t e d above. I f m o r e t h a n 3 coplaner forces act at a p o i n t and are i n equilibrium they m a y be represented i n m a g n i t u d e a n d d i r e c t i o n b y the side o f a p o l y g o n t a k e n i n order. I f a system o f forces meets at a p o i n t a n d t h a t system is n o t i n e q u i l i b r i u m the p o l y g o n does n o t close, a n d the force n e e d e d t o b r i n g t h e system i n t o e q u i l i b r i u m ( c a l l e d e q u i l i b r a n t ) is represented b y the v e c t o r w h i c h j o i n t s the open ends o f t h e Fig. 1.15 incomplete polygon. Left - Space diagram or position diagram or configuration diagram. D r a w t o a suitable scale, t h e vectors ab, be to represent t h e forces Right - Vector diagram or force diagram F, a n d F i n m a g n i t u d e a n d d i r e c t i o n. The vector ac represents the resultant of 2 (i) With anticlockwise order starting with a F, a n d F (Fig. 1.16). 2 (ii) With clockwise order starting with a  The units, mass, force, weight, & basic definitions /1.34 Elements of Mining Technology-3 /1.35 If a set of coplaner force F F , F , F acting at a point O is i n equilibrium, 1} 2 3 4 -*->-» t h e i r r e s u l t a n t R m u s t be zero. Graphically, since t h e resultant is zero, i n t h e force or vector p o l y g o n , the last p o i n t e m u s t fall o n the first p o i n t a so that t h e ab + be = ac (resultant) vector ae is zero, i.e. t h e force p o l y g o n m u s t close. N o w d r a w the vector cd t o represent F i n m a g n i t u d e a n d d i r e c t i o n. 3 A n e x a m p l e o f a p p l i c a t i o n of space d i a g r a m a n d vector d i a g r a m t o T h e n a d = ac + c d = ab + be + c d , i.e. adrepresents t h e r e s u l t a n t of find o u t t h e stresses i n the b o t t o m of a crane a n d its s u p p o r t i n g members is —* i l l u s t r a t e d b y F i g. 1.17. The rope o f t h e crane raises a w e i g h t , W (say 1 k N ). F a n d F, a n d F. Similarly, d r a w the vector de t o represent F t o scale. J o i n 3 2 4 T h e r o p e coils r o u n d a w i n c h i n t h e crane i n d i c a t e d b y "effort". The b o o m is ae. T h e n t h e v e c t o r h i n g e d at i t s base t o the frame o f the crane a n d t h e u p p e r e n d is h e l d i n -»—»—» p o s i t i o n b y a m e m b e r a t t a c h e d t o t h e s t r u c t u r e o f t h e crane. T h e r e l a t i v e ae = a d + de positions o f ropes, b o o m a n d o t h e r s u p p o r t i n g members are as s h o w n i n tjie figure. ac = c d + d e To find o u t the forces i n different m e m b e r s o f t h e crane d r a w a space d i a g r a m as s h o w n i n t h e f i g u r e a n d m a r k o n i t t h e k n o w n forces W, ( 1 k N ) a n d effort ( 1 k N ) s h o w i n t t h e i r d i r e c t i o n s b y a r r o w s. D r a w a v e c t o r d i a g r a m s t a r t i n g w i t h p o i n t a, a n d d r a w ab, be. F r o m a a d r a w a l i n e a d p a r a l l e l t o force Q a n d f r o m c another l i n e c d p a r a l l e t o t h e force P. These parallelines m u s t m e e t at p o i n t t o give a closed figure as t h e system o f forces is i n e q u i l i b r i u m. T h e m a g n i t u d e of the forces Q a n d P can t h e n be measured. A l l the arrows o f the forces i n the vector/force d i a g r a m should f o l l o w a definite order (all clockwise or a l l anticlockwise). The force P i n the b o o m has, therefore to be i n t h e d i r e c t i o n c d a n d the force Q i n t h e t i e , i n t h e d i r e c t i o n da. Fig. 1.16 T h e vector ae represents therefore, i n m a g n i t u d e a n d d i r e c t i o n I r e s u l t a n t R o f t h e coplaner forces F , F , F , a n d F a c t i n g a t t h e p o i n t O. To 2 3 4 o b t a i n its p o s i t i o n o r l i n e o f a c t i o n , t h r o u g h t h e p o i n t O i n Fig. 1.16, d r a w a l i n e p a r a l l e l t o ae, w i t h letters A a n d E o n either side of i t. We have thus o b t a i n e d the resultant R complete i n m a g n i t u d e , d i r e c t i o n a n d p o s i t i o n. Fig. 1.17 I t s h o u l d be clear f r o m t h e foregoing that t o o b t a i n t h e r e s u l t a n t of a system o f coplaner forces acting at a p o i n t , i t is necessary t o d r a w the vectors T h e c o m p o n e n t o f a m a c h i n e o r s t r u c t u r e w h i c h is i n compression is ab, be, c d , etc. t o represent t h e forces t o a suitable scale, i n m a g n i t u d e and called a strut. I f a strut is broken whatever i t supports tends t o collapse i n w a r d s i n t o the s t r u t. The c o m p o n e n t of a m a c h i n e or s t r u c t u r e w h i c h is i n t e n s i o n is d i r e c t i o n. The l i n g j o i n i n g the first p o i n t a t o the last p o i n t e i n t h e vector called a tie. W h a t e v e r a tie supports tends t o f a l l a w a y f r o m i t i f the t i e is d i a g r a m represents the resultant i n m a g n i t u d e a n d d i r e c t i o n. This p o l y g o n of broken. vectors is k n o w n as t h e v e c t o r o r Force P o l y g o n.  The units, mass, force, weight, & basic definitions/1.36 Elements of Mining Technology-3 /1.37 Moment of a force : Total c l o c k w i s e m o m e n t Total c l o c k w i s e m o m e n t The moment of a force is defined as the product of the force multiplied by of couple about A o f couple a b o u t B the p e r p e n d i c u l a r distance of the force f r o m the p o i n t about w h i c h i t is acting. = P ( x ) + P ( d - x) = P ( d + y ) - Py I n Fig. 1.18 the m o m e n t o f the force a b o u t the p o i n t O is F x OA = Px + P d - Px = P d + Py - Py w h e r e OA is the perpendicular distance of the l i n e of a c t i o n of force F f r o m the = Pd = Pd p o i n t O. The m o m e n t o f a force m a y be clockwise or anticlockwise as s h o w n i n t h e f i g u r e a n d i t has a W h e n t u r n i n g a spanner, a couple is being exerted a n d the reaction of t u r n i n g effect a r o u n d t h e the n u t o n t h e spanner is equal a n d opposite t o the force a p p l i e d at the e n d o f p o i n t O. The t u r n i n g effect of the spanner. ( T h e r e m u s t always be equal a n d opposite forces a c t i n g ). I n the a force is u s u a l l y called the example i l l u s t r a t e d i n the Fig. 1.20 the l e n g t h of t h e spanner is considered t o turning moment. Other names be the a r m o f the couple a n d the m o m e n t of the couple is 8 0 x 0.200 = 16 N m. given t o t u r n i n g m o m e n t are I t is i m p o r t a n t t o r e m e m b e r t h a t t h e a r m l e n g t h is the p e r p e n d i c u l a r Clockwise Anticlockwise moment, torque and twisting distance. A force c a n have n o t r u n i n g m o m e n t a b o u t a p o i n t , i n its o w n l i n e moment moment moment. of a c t i o n. Fig. 1.18 Moment of force The p o i n t about w h i c h the force t u r n s is called a p i v o t or a f u l c u r m. F = 80N 1. A n y t w o couples o f equal I f a force acts t h r o u g h the p o i n t of r o t a t i o n i t has n o t u r n i n g effect or m o m e n t. m o m e n t i n the same plane are P r i n c i p l e o f m o m e n t states t h a t : I f a b o d y is at rest u n d e r t h e a c t i o n of e q u i v a l v e n t , ie. a couple c a n ) be replaced b y another couple several forces i n the same plane, t h e n the clockwise m o m e n t s about a n y p o i n t m u s t be equal t o t h e anticlockwise m o m e n t s a b o u t the same p o i n t ; o r the Reaction of the same moment, provided = 80N i t h a s be s a m e d i r e c t i o n , r e s u l t a n t m o m e n t o n the b o d y about a n y p o i n t is zero. 200 mm- clockwise, o r anticlockwise. Couple : Fig. 1.20 T w o p a r a l l e l forces, of equal m a g n i t u d e , b u t w h i c h act i n opposite directions, consitute a couple. A t y p i c a l couple is s h o w n i n Fig. 1.19. A person 2. A couple acting on a body a t t e m p t i n g t o open a screwed cap of a b o t t l e exerts a couple o n the cap. c a n o n l y be b a l a n c e d b y A couple is n o t equivalent t o a simple force a n d cannot be replaced b y a single a n o t h e r couple i n the same force. The perpendicular distance between the lines of action of the constituent p l a n e. N o s i n g l e force c a n p a r a l l e l force is called t h e a r m of the couple. I n Fig. 1.19 the forces P, P balance a couple. c o n s t i t u t e a couple a n d d is the a r m o f t h e couple. T h e m o m e n t of a couple, 3. A n y n u m b e r of couples i n also called t u r n i n g effect of the couple, about a n y p o i n t i n its plane is constant t h e same p l a n e acting u p o n a a n d equal t o P x. d. b o d y is equivalent t o a single c o u p l e w h o s e m o m e n t is Let us consider the t u r n i n g equal t o t h e algebric s u m o f effect o f the couple about the moments of the constituent a n y 2 points, A a n d B, as couples. Fig. 1.21 Increasing torque of a spanner by use of s h o w n i n Fig. 1.19 ( i i ). a pipe on the spanner. 4. I f t w o couples a c t i n g n one plane u p o n a b o d y the equal a n d opposite Fig. 1.19 Moment of couple the b o d y m u s t be i n e q u i l i b r i u m.  The units, mass, force, weight, & basic definitions/1.38 Elements of Mining Technology-3 /1.39 W h e n a couple acts u p o n a b o d y there c a n n o t be a n y m o t i o n i n a Example: A b e l l c r a n k lever has forces o f 6 0 N a n d 2 0 N a c t i n g as s h o w n i n straight l i n e because the couple has n o resultant force w h i c h w i l l b r i n g about Fig. 1.23. F i n d t h e r e s u l t a n t t u r n i n g m o m e n t o n t h e b e l l c r a n k lever. a m o t i o n i n a s t r a i g h t l i n e. U n d e r the i n f l u e n c e o f a couple a b o d y c a n o n l y Ans,: rotate, i f free, about an axis, t h r o u g h its centre o f g r a v i t y p e r p e n d i c u l a r t o the p l a n e o f t h e couple. Example: A r u l e 2 m l o n g is p i v o t e d at its centre; o n t h e r u l e are 3 forces o f 2 0 N , 5 0 N a n d 3 0 N a c t i n g as s h o w n i n Fig. 1.22. F i n d t h e v a l u e o f force W t o keep t h e r u l e i n e q u i l i b r i u m. Ans.: Let us consider t h e m o m e n t s o f forces a b o u t the p o i n t 0. For the r u l e t o be i n a state w 30 N 20 N 50 N of e q u i l i b r i u m , s u m of O clockwise moments = Fig. 1.23 A bell crank lever. 1—400-^—500—1 s u m o f t h e anticlockwise 1000 -4- 90() - Before a n y m o m e n t c a n be c a l c u l a t e d t h e r i g h t a n g l e d distances f r o m moments. the lines o f a c t i o n o f t h e forces t o t h e p i v o t m u s t b e f o u n d f o r each force. This Fig. 1.22 Distances in mm c a n b e d o n e b y c a l c u l a t i o n as f o l l o w s : IFOB = X S u m of clockwise moments = ( 2 0 x 500) N - m m + ( 5 0 x 900) N-mm... ' y = 55000 N-mm From traingle OAB, = cos 30° 400 Sum of anticlockwise moments = (30 x 400) N - m m + W x 1 0 0 0 N - m m... X = 400xcos30° X = 346.4 m m L o a d W s h o u l d p r o v i d e a n a n t i c l o c k w i s e m o m e n t e q u a l ttoo Y (55,000 - 1 2 0 0 0 ) N - m m = 4 3 0 0 0 N - m m. F r o m t r i a n g l e O C D , — — = cos 30° 5 ' 300 l m x W = 4 3 0 0 0 N - m m = 43 N m ( i n an anticlockwise direction) Y = 3 0 0 x cos 30° Y = 259.8 m m.and W = 43 N N o w c o n s i d e r i n g t h e m o m e n t s a b o u t O, c l o c k w i s e m o m e n t is 2 0 N x X = 2 0 N x 3 4 6. 4 m m = 6 9 2 8 N. m m.  Elements of Mining Technology-3 /1.41 The units, mass, force, weight, & basic definitions /1.40 Clockwise m o m e n t = Anticlockwise moments Anticlockwise moment, 60 N x Y = 6 0 N x 259.8 m m = 1 5 5 8 8 N - m m. Resultant m o m e n t o n b e l l c r a n k = 15588-6928 R,., x L s i n 60° = 9 8 1 N x - cos 60° 2 - = 8660 N - m m anticlockwise. Dividing by L Example: 981 N A u n i f o r m bar of mass 100 k g rests w i t h one e n d o n a r o u g h h o r i z o n t a l R „ , s i n 60° = ^ ^ - c o s 60° f l o o r a n d t h e o t h e r e n d against a s m o o t h v e r t i c a l w a l l. F r i c t i o n at t h e w a l l c a n be neglected. F i n d t h e m a g n i t u d e o f t h e r e a c t i o n at t h e w a l l , a n d t h e m a g n i t u d e a n d d i r e c t i o n o f the g r o u n d reaction w h e n the bar makes 60° t o the „ 981N. _ R x 0. 8 6 6 = — — x 0.5 horizontal. Ans.: I f w a l l is s m o o t h i offers n o f r i c t i o n a n d r e a c t i o n R o f w a l l is w = 283.2 N at r i g h t angles t o i t. Fig. 1.24 (Hi) F r o m F i g. 1.24 r e a c t i o n at f l o o r /? = /981 +(283-2) F > 2 2 = ^962361 + 80202 = 1021N Example: A B C D E is h o r i z o n t a l b e a m o f l e n g t h 1 7 m. A B '= 5 m , BC = 2 m , CD = 7 m a n d D E = 3 m. T h e b e a m is s i m p l y s u p p o r t e d at B a n d D , a n d t h e r e are d o w n w a r d loads o f 4 0 0 N , 1 0 0 0 N a n d 6 0 0 N at A , C, a n d E respectively. Fig. 1.24 Calculate t h e u p - w a r d reactions at B a n d D. ( T h e w e i g h t o f the b e a m c a n be neglected). R e f e r r i n g t o F i g. 1.24 ( i i ) , l e t t h e r e a c t i o n Rp at t h e f l o o r has a F i g. 1.25 is a d i a g r a m m a t i c r e p r e s e n t a t i o n o f t h e l o a d i n g. horizontal component H a n d a vertical component V; the horizontal component prevents t h e bar f r o s l i p p i n g W e i g h t of bar = 9 8 1 N a c t i n g d o w n w a r d at its Ans.; centre. I f w e t a k e m o m e n t s about B t h e s u p p o r t i n g force a t B, d e n o t e d b y ^ has n o t u r n i n g effect, a n d w e c a n f o r m a n e q u a t i o n w i t h one u n k n o w n. A p p l y i n g the general c o n d i t i o n s o f e q u i l i b r i u m : Taking m o m e n t s about B, a n d w o r k i n g t h r o u g h o u t i n units of N a n d m. U p w a r d forces = d o w n w a r d forces : 400N 1000N 600N V = 981N ---(1) B c D E Forces t o left = forces t o r i g h t H = R W ---(2) - — — 7 — 1 * 3 * Taking moments about any convenient point, e.g, the foot of the ladder, Fig. 1.25 A beam supported at two points since i t eliminates t w o t u r n i n g effects.  The units, mass, force, weight, & basic definitions /1.42 Elements of Mining Technology-3 / I A3 Clockwise m o m e n t s = a n t i c l o c k w i s e m o m e n t s (R A x 6) + ( 1 , 0 0 0 x 6 ) = ( 3 , 0 0 0 x 5) + ( 2 0 , 0 0 0 x 1.5) or R = 6,500 N ( 1 0 0 0 x 2) + ( 6 0 0 x 12) = ( 4 0 0 x ) + ( 1 ^ + 9 ) 2000 + 7200 = 2000 + 9 R Beam wt. D 3000N Distributed lead 1000N 4000N/m 300N/m 9 R = 7200 D 77777/fv/>///77 R D = 800 N RB| 1m -2m- -2.5m- -2.5m- T a k i n g m o m e n t s a b o u t D , R ^ h a v i n g n o t u r n i n g effect. -6m— Clockwise m o m e n t s — a n t i c l o c k w i s e m o m e n t s. -7m- -12m- ( 9 ) + ( 6 0 0 x 3) = ( 4 0 0 x 14) + ( 1 0 0 0 x 7) 9RB = 5 6 0 0 + 7 0 0 0 - 1 8 0 0 Fig. 1.26.-. R3=1200N. Check u p w a r d forces = + R A = 27,600 N Example: d o w n w a r d forces = 2 7 , 6 0 0 N Calculate t h e reactions R a n d A o f the beam as s h o w n i n Fig. 1.26. A n s w e r : R^ 2 1 , 1 0 0 N a n d R A = 6,500 N T h e l o a d o f t h e b e a m per u n i t l e n g t h is 3 0 0 N / m. A d i s t r i b u t e d l o a d o f 4 0 0 0 N / m is p l a c e d over the b e a m as s h o w n. T w o c o n c e n t r a t e d loads o f 3 0 0 0 N Example: a n d 1 0 0 0 N act o n t h e beam as s h o w n. F i g u r e 1.27 shows a shaft m o u n t e d i n bearings A a n d B. T h e w e i g h t of the p u l l e y s a n d shaft are g i v e n. D e t e r m i n e the l o a d o n each b e a r i n g. Ans.: T h e l o a d o f t h e b e a m c a n be r e p l a c e d b y a c o n c e n t r a t e d l o a d a c t i n g Ans.: t h r o u g h its centre o f gravity, o f m a g n i t u d e 3 6 0 0 N. t h e C.G. is 6 m from R. A Let t h e r e a c t i o n s , w h i c h w i l l be e q u a l t o t h e loads o n t h e bearings be S i m i l a r l y the d i s t r i b u t e d load can be replaced b y a concentrated l o a d o f 2 0 , 0 0 0 R andR _. A B N a c t i n g at a p o i n t 4.5 m f r o m R. A T a k i n g m o m e n t s a b o u t B, Taking moments about R A 1 x R A = 4 0 x 0.25 + 1 6 0 x 0.5 + 3 2 x 0.55 + 2 0 x 0.85.-. R A = 10 + 8 0 + 17-6 + 17 Clockwise m o m e n t s — a n t i c l o c k w i s e m o m e n t s.-. R = A 124.6 N. ( 3 , 0 0 0 x 1) + ( 2 0 , 0 0 0 x 4.5) + ( 3 , 6 0 0 x 6) + ( 1 , 0 0 0 x 12) —0.5 Shaft wt = R B x 6 4i 160 N 22L j or 3000 + 90,000 + 21,600 + 12,000 = R x6 2 i j1 or R,, = 2 1 , 1 0 0 N |l60N 20 N t 40 N T a k i n g m o m e n t about R B —0.15—4- 0.30- -0.30- i—0.25 Clockwise m o m e n t s = a n t i c l o c k w i s e m o m e n t s Fig. 1.27 Distances are in meters.  The units, mass, force, weight, & basic definitions /1.44 Elements of Mining Technology-3 /1.45 The v a l u e of can be calcualted i n a s i m i l a r w a y by t a k i n g m o m e n t s T h e r e f o r e b y the p r i n c i p l e of m o m e n t s , i n u n i t s of N m , a b o u t A a n d w e get R^ = 1 2 7. 4 N. The loads o n t h e bearings are a p p r o x i m a t e l y equal. W h e n d e s i g n i n g 56257t ( 2 4 0 x 0. 7 5 ) + ( 1 2 x 0. 2 ) = Pr essure x x 0.075 a system i n v o l v i n g l o a d o n b e a r i n g i t s h o u l d be kept i n m i n d that t h e l o a d on 4xl0 6 each b e a r i n g is a p p r o x i m a t e l y the same so that w e a r o n the bearings is nearly equal. 421.87571. (l80 + 2.4)Nm = x pressure N m Example: 4x10 A safety v a l v e o n a s t e a m boiler (Fig. 1.28) is j u s t o n the p o i n t of b l o w i n g o f f steam. W h a t s h o u l d be the steam pressure u n d e r t h e f o l l o w i n g So, Pr essure = 4 x 1 8 2 , 4 x 10 6 N /m 2 = 550.27 k N / m 2 circumstances ? 421.875*: (a) F is the f u l c r u m f o r t h e lever o r a r m o f t h e safety valve. Centre of gravity : (b) D i a m e t e r for t h e s t e a m passage at P is 7 5 m m. The centre of gravity of a b o d y is that p o i n t t h r o u g h w h i c h the resultant (c) T h e lever has a n effective l e n g t h L o f 7 5 0 m m f r o m t h e f u l c r u m and of t h e earth's p u l l u p o n t h e b o d y passes a n d a t w h i c h p o i n t t h e w e i g h t o f t h e w e i g h s 1 2 N. G is t h e p o i n t w h e r e its w e i g h t acts (Centre o f g r a v i t y ) tody c a n be c o n s i d e r e d as c o n c e n t r a t e d. The centre of g r a v i t y is occasionally a n d FG is 2 0 0 m m. referred t o as t h e centre of mass or centroid. For bodies of c o m m o n s y m m e t r i c a l (d) FV is 7 5 m m shapes t h e c e n t r e o f g r a v i t y is easily f o u n d o u t a n d is at f o l l o w i n g p o i n t s. W e i g h t W p l a c e d o n t h e l e v e r l is 2 4 0 N. A square or a rectangle : T h e i n t e r s e c t i o n o f d i a g o n a l s. Ans.