CHAPTER 7 TRANSPOSITION OF FORMULAE

Transcription

1 CHAPTER 7 TRANSPOSITION OF FORMULAE EXERCISE 30, Pge Mke d the subject of the formul: + b = c - d - e Since + b = c - d e then d = c e - b. Mke the subject of the formul: = 7 Dividing both sides of = 7 b 7 gives: = 7 3. Mke v the subject of the formul: pv = c Dividing both sides of pv = c b p gives: v = c p 4. Mke the subject of the formul: v = u + t Since v = u + t then v u = t nd dividing both sides b t gives: v u t = or = v u t 5. Mke the subject of the formul: + 3 = t Since + 3 = t then 3 = t nd dividing both sides b 3 gives: t or t 6. Mke r the subject of the formul: c = r 95

2 Dividing both sides of c = r b gives: c r or r = c 7. Mke the subject of the formul: = m + c Since = m + c then c = m c nd dividing both sides b m gives: m = or = c m 8. Mke T the subject of the formul: I = PRT Dividing both sides of I = PRT b PR gives: I T PR or T = I PR 9. Mke L the subject of the formul: XL f L Dividing both sides of XL XL f L b πf gives: f L or L = XL f 10. Mke R the subject of the formul: I = E R Multipling both sides of I = E R b R gives: I R = E nd dividing both sides b I gives: R = E I 11. Mke the subject of the formul: 3 Since 3 then 3 = Multipling both sides b gives: ( 3) = or = ( 3) 96

3 1. Mke C the subject of the formul: F = 9 5 C + 3 Rerrnging F = 9 C 3 gives: F 3 = 9 C 5 5 Multipling both sides b 5 9 gives: F3 C F 3 C 9 i.e. or C F

4 EXERCISE 31, Pge Mke r the subject of the formul: S = 1 r Multipling both sides of S = i.e. from which, 1 r b (1 r) gives: S(1 r) = S Sr = S = Sr nd dividing both sides b S gives: S S = r i.e. r = S S or r = 1 - S. Mke the subject of the formul: = ( d) d Multipling both sides of = d d b d gives: d = ( d) Dividing both sides b gives: d d nd d d = or d d Alterntivel, from the first step, d = ( d) i.e. nd d = - d d + d = from which, = d d d d i.e. = 3. Mke f the subject of the formul: A = 3(F f ) L 98

5 Multipling both sides of A = 3(F f ) L b L gives: AL = 3(F f) Dividing both sides b 3 gives: AL F f 3 nd f F AL or f = 3 3F AL 3 4. Mke D the subject of the formul: = AB 5CD Multipling both sides of = AB 5CD b D gives: D = AB 5C Dividing both sides b gives: D = AB 5C 5. Mke t the subject of the formul: R = R0(1 + t) Removing the brcket in R R 1 t R R R t 0 gives: 0 0 from which, R R0 R0 t R R0 nd R = t or t = R R0 R Mke R the subject of the formul: 1 R = 1 R R Rerrnging gives: R R R R R R R1 R i.e. R R R RR 1 1 RR1 Turning both sides upside down gives: R R R 1 99

6 7. Mke R the subject of the formul: I = E e R r Multipling both sides b (R + r) gives: i.e. nd I(R + r) = E e I R + I r = E e I R = E e I r nd dividing both sides b I gives: R = E e Ir I or R = E e r I 8. Mke b the subject of the formul: = 4b c Dividing both sides b 4 c gives: b b 4c or 4c Tking the squre root of both sides gives: b = 4c 9. Mke the subject of the formul: + b = 1 Rerrnging b 1 gives: b b 1 Turning both sides upside down gives: Multipling both sides b gives: b b b Tking the squre root of both sides gives: = i.e. = b b b b 100

7 10. Mke L the subject of the formul: t = L g Dividing both sides of t = L b gives: g t L g Squring both sides gives: t L g or L t g Multipling both sides b g gives: L = g t or L = gt Mke u the subject of the formul: v = u + s Since v = u + s then v - s = u or u = v - s Tking the squre root of ech side gives: u = v s 1. Mke the subject of the formul: N = Squring both sides of N = gives: N Multipling both sides b gives: N or + = N from which, = N The lift force, L, on n ircrft is given b: L = 1 v c where ρ is the densit, v is the velocit, is the re nd c is the lift coefficient. Trnspose the eqution to mke the velocit the subject. 101

8 Since L = 1 v c then L c v from which, velocit, v = L c 10

9 EXERCISE 3, Pge 6 1. Mke the subject of the formul: = mn mn Multipling both sides of = b gives: = m n nd fctorising gives: = m n Dividing both sides b (m n) gives: m n or m n Tking the squre root of both sides gives: m n. Mke R the subject of the formul: M = (R 4 - r 4 ) Dividing both sides of M = R 4 r 4 nd rerrnging gives: b gives: M R r M r R or R M r 4 4 Tking the fourth root of both sides gives: R = M 4 r 4 3. Mke r the subject of the formul: + = r 3 r Multipling both sides of + = Multipling the brckets gives: nd rerrnging gives: r 3 r b (3 + r) gives: ( + )(3 + r) = r 3 + r r = r r + r r = -3 3 Fctorising gives: r( + 1) = -3( + ) Dividing both sides b ( + 1) gives: r = 3( ) 1 103

10 Multipling numertor nd denomintor b -1 gives: r = 3( ) 1 4. Mke L the subject of the formul: m = L L rcr Multipling both sides of L m b (L + rcr) gives: m(l + rcr) = L L rcr Removing brckets gives: nd rerrnging gives: ml + mrcr = L mrcr = L - ml Fctorising gives: mrcr = L( - m) Dividing both sides b ( - m) gives: L = mrcr m 5. Mke b the subject of the formul: = b c b Multipling both sides b b gives: b b c nd rerrnging gives: Fctorising gives: c b b or b 1 c b b c Dividing both sides b 1 gives: b c 1 Tking the squre root of both sides gives: b = Hence, b = c c 1 1 c 1 6. Mke r the subject of the formul: = 1 r 1 r 104

11 Rerrnging b cross-multipling gives: 1 r 1 r Removing brckets gives: nd rerrnging gives: r r r r or r r Fctorising gives: r Dividing both sides b ( + ) gives: r Tking the squre root of both sides gives: r = 7. A formul for the focl length, f, of conve lens is: 1 f = 1 u + 1. Trnspose the formul to mke v v the subject nd evlute v when f = 5 nd u = 6 Rerrnging gives: f u v Turning ech side upside down gives: v = u f v f u uf uf u f When f = 5 nd u = 6, then v = uf (6)(5) 30 u f = The quntit of het, Q, is given b the formul Q = mc(t - t1). Mke t the subject of the formul nd evlute t when m = 10, t1 = 15, c = 4 nd Q = 1600 Removing the brckets in Q = mct t gives: Q = mct mct1 1 nd rerrnging gives: Q + mct1 mct or mct Q mvt1 Q mvt1 Q Q Dividing both sides b mc gives: t or t t1 or t t1 mc mc mc When m = 10, t 1 = 15, c = 4 nd Q = 1600, 105

12 Q t = t = 55 mc (10)(4) The velocit, v, of wter in pipe ppers in the formul h = 0.03Lv dg of the formul nd evlute v when h = 0.71, L = 150, d = 0.30 nd g = Epress v s the subject Multipling both sides of h = 0.03L v dg Dividing both sides b 0.03L gives: b dg gives: dgh = 0.03L v dgh v 0.03L or dgh v 0.03L Tking the squre root of ech side gives: v = dgh 0.03L When h = 0.71, L = 150, d = 0.30 nd g = 9.81, v = dgh (0.30)(9.81)(0.71) = L 0.03(150) 10. The sg S t the centre of wire is given b the formul: S = 3d(l d) 8 Mke l the subject of the formul nd evlute l when d = 1.75 nd S = 0.80 Squring both sides of S = 3d(ld) 8 gives: 3d l d S 8 Multipling both sides b 8 gives: 8S 3dl d Removing the brcket gives: Rerrnging gives: or 8S 3dl 3d 8S 3d 3dl 3dl 8S 3d Dividing both sides b 3d gives: l = 8S 3d 8S 3d 3d 3d 3d 106

13 i.e. l = 8S d 3d When d = 1.75 nd S = 0.80, l = 8S 8(0.80) d =.75 3d 3(1.75) 11. An pproimte reltionship between the number of teeth, T, on milling cutter, the dimeter of cutter, D, nd the depth of cut, d, is given b: 1.5D T D 4d Determine the vlue of D when T = 10 nd d = 4 mm. Multipling both sides of 1.5D T D 4d b D + 4d gives: T(D + 4d) = 1.5D Removing brckets gives: TD + 4dT = 1.5D Rerrnging gives: or Fctorising gives: 4dT = 1.5D TD 1.5D TD = 4dT D(1.5 T) = 4dT Dividing both sides b (1.5 T) gives: D = 4dT 1.5 T When T = 10 nd d = 4 mm, then D = 4dT 4(4)(10) T = 64 mm 1. A simpl supported bem of length L hs centrll pplied lod F nd uniforml distributed lod of w per metre length of bem. The rection t the bem support is given b: 1 F wl R = Rerrnge the eqution to mke w the subject. Hence determine the vlue of w when L = 4 m, F = 8 kn nd R = 10 kn Since R = 1 F wl then R = F + wl 107

14 nd R F = wl from which, w = R F L When L = 4 m, F = 8 kn nd R = 10 kn, w = (10) 8 1 = 3 kn/m The rte of het conduction through slb of mteril, Q, is given b the formul Q ka(t t ) d 1 where 1 t nd t re the tempertures of ech side of the mteril, A is the re of the slb, d is the thickness of the slb, nd k is the therml conductivit of the mteril. Rerrnge the formul to obtin n epression for t ka(t1 t ) Since Q then Qd = kat1 t d i.e. Qd ka = t1 t from which, t = t 1 - Qd ka r 14. The slip, s, of vehicle is given b: s = 1 100% where r is the tre rdius, ω is the v ngulr velocit nd v the velocit. Trnspose to mke r the subject of the formul. r Since s = 1 100% v nd then s 100 = 1 - r v r s = 1 - v 100 from which, r = v 1 s

15 15. The criticl lod, F newtons, of steel column m be determined from the formul L F EI n where L is the length, EI is the fleurl rigidit, nd n is positive integer. 1 Trnspose for F nd hence determine the vlue of F when n = 1, E = N / m, 6 4 I = m nd L = 1.1 m Since nd L F n EI then F n EI L F n EI L n i.e. F = EI L 1 When n = 1, E = N / m, I = m 6 4 nd L = 1.1 m, n 1 L 1.1 = N = MN 1 6 lod, F = EI The flow of slurr long pipe on col processing plnt is given b: Trnspose the eqution for r 4 pr V 8 Since nd 4 pr V then 8 8 Vpr 8V r 4 p 4 8 V from which, r = 4 p 109