Chương 2 - Thủy Tĩnh Học

THY TNH HC ThS L MINH LU

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CCHHNNGG 22 THY TNH HC

2.1 p sut thy tnh p lc.

Ly mt khi cht lng W ng cn bng (hnh 2 1). Nu chia ct khi bng mt mt phng tu ABCD v vt b phn trn, th mun gi phn di khi trng thi cn bng nh c ta phi thay th tc dng ca phn trn ln phn di bng mt h lc tng ng.

Trn mt phng ABCD, xung quanh mt im O

Hnh 2 1

tu ta ly mt din tch ; gi P l lc ca phn trn

tc dng ln , t s tbPP=

gi l p sut thy tnh

trung bnh. Nu din tch tin ti s 0, th t s P

tin ti gii hn p , gi l p sut thy tnh ti mt im, hoc ni gn l p sut thu tnh.

pP =

lim0 (2 1)

p sut thy tnh p l ng sut tc dng ln mt phn t din tch ly trong ni b mi trng cht lng ang xt.

Trong thu lc, lc P tc dng ln din tch gi l p lc thy tnh ln din tch y.

Ch : ngi ta thng gi tr s p ca p l p sut thy tnh v tr s P ca P

l p lc thy tnh. p sut c n v l 2mN hoc 2.sm

kg .

Trong k thut, p sut cn c o bng tmtphe (at) 1 at = 9,81.104 (N/m2) 1 at = 1(kG/cm2)

p lc c n v l Niutn (N) p sut cn c o bng chiu cao ct nc.

2.2 Hai tnh cht c bn ca p sut thy tnh.

Tnh cht 1: p sut thy tnh tc dng thng gc vi din tch chu lc v hng vo din tch y.

p sut thy tnh ti im O ly trn mt phn chia ABCD (hnh 2 2) l mt lc c th chia lm hai thnh phn: pn theo hng php tuyn ti im O ca mt


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ABCD v theo hng tip tuyn. Thnh phn c tc dng lm mt ABCD di chuyn, tc cht lng c th chuyn ng tng i, nhng nh gi thit ban u, cht lng ang xt trng thi tnh nn phi c = 0 v ch cn li thnh phn php tuyn pn. Thnh phn pn khng th hng ra ngoi c v cht lng khng chng li c sc ko m ch chu c sc nn. Vy p sut p ti im O ch c thnh phn php tuyn v hng vo trong.

t

Hnh 2 2 Hnh 2 3

Tnh cht 2: Tr s p sut thy tnh ti mt im bt k khng ph thuc hng t ca din tch chu lc ti im ny.

Ly mt phn t din tch ds c tm I v mt hnh tr v cng nh c tit din thng ds (hnh 2 3). y kia hnh tr c din tch dS' v tm I', y ny c hng bt k xc nh bi gc . Nhng kch thc v chiu di l nhng v cng nh.

Gi p v p' l nhng p sut, chng vung gc vi nhng mt tng ng. Theo nh ngha, ta c cc tr s p lc dF v dF' nh sau:

dF = p.dS dF' = p'.dS'

Hnh tr ny ng cn bng di tc dng ca nhng lc mt l v cng nh bc hai v ca nhng th tch l nhng v cng nh bc ba. Do ta c th b qua nhng lc th tch. Phng trnh ny chiu ln trc II', cho ta:

0cos' = dFdF (2 2) V nhng lc mt tc dng ln mt bn v vung gc vi II', trit tiu nhau.

Vy: pdS = p'.dS'cos; v dS = dS'cos nn ta rt ra: 'pp = (2 3)

Vy p sutt thy tnh ti im I l mt i lng v hng p, ch ph thuc v tr ca im I, ngha l trong h ta vung gc Oxyz th:

p = f(x, y, z) (2 4)

T hai tnh cht trn ca p sut thy tnh, ta thy r cc thnh phn tip tuyn u bng s khng v cc thnh phn php tuyn u bng nhau v bng p. V vy tens ng sut vit cho p sut thy tnh c dng

p 0 00 p 00 0 p


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2.3 Phng trnh vi phn c bn ca cht lng cn bng

Xt khi cht lng hnh hp v cng nh ABCDEFGH c cnh x, y, z (hnh 2 4) ng cn bng. iu kin cn bng l tng s hnh chiu trn cc trc ca cc lc mt v lc th tch tc dng ln khi bng khng.

Hnh 2 4.

Gi p l p sut ti trng tm M ca hnh hp, th p sut at5i trng tm mt

ADHE bng

2

. xxpp , ti trng tm mt BCGF bng

+2

. xxpp ; gi Fx l

thnh phn trn trc Ox ca lc th tch F tc dng ln ln mt n v khi lng cht lng, ta c th vit iu kin cn bng ca hnh hp theo phng x nh sau:

0.2

.2

. =+

+

zyxFzyxxppzyx

xpp x

rt gn ta c: 0=+

xFxp hoc 01 =

xpFx

Suy lun tng t i vi nhng hnh chiu cc lc trn cc trc Oy, Oz v vit ton b h thng phng trnh biu th s cn bng ca khi hnh hp, ta c:

01 =

xpFx

01 =

ypFy

(2 5)

01 =

zpFz

hoc 01 = gradpF

l h phng trnh vi phn c bn ca cht lng ng cn bng v cn gi l h phng trnh le (do le tm ra nm 1755). Phng trnh ny biu th quy lut chung v s ph thuc p sut thy tnh i vi to :

p = f(x, y, z)


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H (2 5) c th vit di dng vi phn ton phn ca p nh sau: nhn nhng phng trnh trong h (2 6) ring bit vi dx, dy, dz ri cng v i v, ta c:

( ) 01 =

+

+

++ dzzpdy

ypdx

xpdzFdyFdxF zyx

(2 6)

V p = f(x, y, z) ch l hm s ca to , nn ta co th vit c:

( ) 01 =++ dpdzFdyFdxF zyx

hoc dp = (Fxdx + Fydy + Fzdz) (2 7) Phng trnh (2 7) gi l phng trnh vi phn cn bng ca cht lng.

2.4 Mt ng p. Mt ng p l mt c p sut thy tnh ti mi im u bng nhau, tc l mt

c p = const, do dp = 0. Phng trnh vi phn ca mt ng p:

Fxdx + Fydy + Fzdz = 0 (2 8)

Tnh cht 1: Hai mt ng p khc nhau khng th ct nhau. Tnh cht 2: Lc th tch tc dng ln mt ng p thng gc vi mt ng p.

2.5 S cn bng ca cht lng trng lc Khi lc th tch tc dng vo cht lng ch l trng lc th cht lng c gi l

cht lng trng lc. Trong h ta vung gc m

trc Oz t theo phng thng ng hng ln trn, th i vi lc th tch F tc dng ln mt mt n v khi lng ca cht lng trng lc, ta c Fx = 0, Fy = 0 v Fz = -g (g l gia tc ri t do) (hnh 2 5)

Hnh 2 5.

1. Phng trnh c bn ca cht lng trng thi cn bng. T (1 7), thay Fx = 0, Fy = 0, Fz = -g ta c:

dp = -gdz (2 9)

Sau khi tch phn v chia cho g ta c:

constgpz =+

(vi = g) (2 10)


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T (2 10) xt ti hai im A v A0 ta c:

constp

zpz =+=+

00 (2 11)

hoc p = p0 + (z0 z) (2 11)'

Gi z0 l tung ca im trn mt t do v h l su ca im ang xt c tung z, ta c: h = z0 z.

Nn (2 11)' co th vit:

(2 12) p hp= 0 +

Phng trnh (2 11); (2 12) l phng trnh c bn ca thy tnh hc, biu th quy lut phn b p sut thy tnh trong cht lng ng cn bng. S hng

p

c th nguyn l di.

2. Mt ng p ca cht lng trng lc. Trong trng hp ang xt lc khi l lc trng trng, gia tc l gia tc ri t

do g, v vy trong h ta chn hnh chiu ca lc khi n v trn trc Ox, Oy, Oz s l: Fx = 0, Fy = 0, Fz = -g, cn phng trnh mt ng p c vit di dng:

- g.dz = 0; do g 0 nn z = const. Do vy mt ng p trong cht lng tnh ng nht s l cc mt nm ngang bt

k, trong c c mt thong, khng ph thuc vo hnh dng bng cha cht lng. Mt nm ngang cng s l mt phn cch ca hai loi cht lng cng cha trong mt bnh.

V d 1: Tm p sut ti mt im y b ng nc su 4m, trng lng n v ca nc = 9810 N/m3 ( = 1000kG/m3). p sut ti mt thong p0 = 98100N/m2 (p0 = 10.000kG/m2).

Gii: p dng cng thc (2 12) ta c: p = p0 + h = 98100 + 9810x4 = 137.340 N/m2 ( = 14.000kG/m2)

3. nh lut bnh thng nhau. Hai bnh thng nhau cha ng cht lng khc nhau v c p sut trn mt

thong bng nhau, cao ca cht lng mi bnh tnh t mt phn chia hai cht lng n mt thong s t l nghch vi trng lng n v ca cht lng, tc l:

1

2

2

1

=hh (2 13)

Trong h1, h2 l nhng cao ni trn ng vi nhng cht lng c trng lng n v 1, 2. Thc vy, p sut p1, p2 trn cng mt phn chia A B (hnh 2 6) bng nhau.


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Theo (2 12): p1 = p0 + 1.h1; p2 = p0 + 2.h2 nn 1.h1 = 2.h2 1

2

2

1

=hh

Nu cht lng cha hai bnh cng mt loi (1 = 2) th mt t do ca cht lng hai bnh cng mt cao; tc h1 = h2.

4. nh lut Patscan. Gi p0 l p sut ti mt ngoi ca mt th tch cht lng cho trc ng cn

bng (hnh 2 7a), p sut ti mt im A su h trong cht lng l: p = p0 + h.

Hnh 2 6. Hnh 2 7.

Nu tng p sut ngoi ln mt tr s p th p sut mi p' ti A s l:

p' = (p0 + p) + h; vy p sut mi ti A s tng ln mt lng bng p' p = p. Nh vy: " bin thin ca p sut thy tnh trn mt gii hn mt th tch

cht lng cho trc c truyn i nguyn vn n tt c cc im ca th tch cht lng ". Kt lun l nh lut Patscan.

My p thy lc lm vic theo nh lut Patscan: my gm hai xi lanh c din tch khc nhau, cha cng cht lng v c pttng di chuyn (hnh 2 8). Khi mt lc F nh tc dng ln n by th lc tc dng ln pittng nh s c tng ln

thnh P1, p sut xi lanh nh l 1

11

Pp = . Theo nh lut Patscan, p sut ti xi lanh

ln cng tng ln p1; vy tng p lc p2 tc dng ln pittng ln l:

1

12212 ..

P

pP ==

Hnh 2 8.

Nu coi P1, 1 khng i, mun tng P2phi tng 2.

Th d 1: P1 = 98,1N (10kG), d1 = 2cm d2 = 20cm, ta tnh c:

NP 98102

201,982

2 =

= (hoc 1000kG)

Thc t gia xilanh v pi1ttng c ma st


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nn: 1

212 .

PP =

trong hiu sut my p thy lc.

5. p sut tuyt i p sut d p sut chn khng. p sut tuyt i ptuyt hoc p sut ton phn l p sut p xc nh bi cng

thc c bn (2 12)

p = p0 + .h (2 14) Nu p sut tuyt i ptuyt ta bt i p sut kh quyn th hiu s l p sut

d pd hoc p sut tng i, tc l: pd = ptuyt pa (2 15)

Nu p sut mt thong l p sut kh quyn pa th:

pd = .h (2 16) p sut tuyt i bao gi cng l mt s dng, cn p sut d c th dng

hoc m: pd > 0 khi ptuyt > papd < 0 khi ptuyt < pa

Trong trng hp p sut d m th hiu s ca p sut khng kh v p sut tuyt i gi l p sut chn khng pck, hoc gi tt l chn khng.

pck = pa ptuyt (2 17) p sut chn khng l tr s p sut cn thiu lm cho p sut tuyt i bng

p sut kh quyn. Do c th gi p sut chn khng l p sut thiu. So snh (2 15) v (2 17) th thy p sut chn khng l tr s m ca p sut d, tc l:

pck pd (2 18) p sut ti mt im c th o bng chiu cao ct cht lng. Vy c th biu

th cc p sut nh sau:

ptuyt bng htuyt = tuyetp

pd bng hd = dup (2 19)

pck bng hck = ckp

Ta gi nhng cao htuyt, hd, hck l nhng cao dn xut ca nhng p sut ptuyt, pd, pck. Trong iu kin bnh thng, p sut kh quyn ti mt thong thng ly bng p sut ca ct thu ngn cao 760mm. Ngi ta quy c ly pa = 98100N/m2 (=1kG/cm2) v gi l tmtphe k thut. t mt phe k thut tng ng vi ct nc cao:

mp

h a 10981098100

===


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Hnh (2 9) cho bit v cch o p sut ti mt im bng chiu cao ct cht lng. Mun p sut tuyt i ti im A, ni bnh cha thng vi mt ng kn 1; ch ni t di mt thong ca cht lng trong bnh, c th t ngang, trn hoc t di im A (hnh 2 9 th ch ni t ngang im A). Trong ng kn phi ht ht khng kh p sut ti mt t do ca cht lng trong ng bng khng; khi khong cch thng ng htuyt t mt nc t do trong ng n ng nm ngang i qua im A biu th p sut tuyt i ti im A. Tr s p sut l: ptuyt = .htuyt.

Nu ng ni trn khng bt kn (hnh 2 9) m h ra khng kh (ng 2) th khong cch thng ng hd k t mt t do trong ng h n ng nm ngang i qua A biu th p sut d ti A; tr s l: pd = .hd.

Hnh 2 9.

Th d 2: Tm p sut tuyt i ptuyt v p sut d pd ti y ni hi su 1,2m, p sut ti mt thong l p0 = 196.200N/m2 (p0 = 21.200kG/m2), nc c = 9.810N/m3 ( = 1000kG/m3).

Gii: p sut tuyt i tnh theo (2 12):

ptuyt = p0 + h = 198.200 + 9.810x1,2 = 207.972N/m2 (=22.460kG/m2)

mp

h tuyettuyet 40,22810.9972.207

===

ct nc

pd = ptuyt pa = 207.972 - 98.100 = 109.872N/m2

mp

h dudu 20,11810.9872.109

===

ct nc


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Th d 3: Ti mt ct trc khi bm p sut chn khng l: pck = 68.670N/m2. Xc nh p sut tuyt i ti mt ct :

Gii: Theo (2 17): ptuyt = pa pck = 98.100 68.670 = 29.430N/m2.

6. ngha hnh hc v nng lng ca phng trnh c bn ca thy tnh hc.

ngha hnh hc: T phng trnh constpz =+

, c th ni rng tng s

cao hnh hc (z) i vi mt chun nm ngang v cao p sut (p ) l mt

hng s i vi ti bt k im no trong cht lng.

ngha nng lng: Ta thy rng khi cht lng ang xt mang mt th nng bng tng s v nng v p nng.

i vi mt n v trng lng, th nng bng: hz + hoc pz + v gi

l th nng n v; z gi l v nng n v; p gi l p nng n v.

7. phn b p sut thy tnh. p lc. Phng trnh c bn ca thy tnh hc (2 12) chng t rng i vi mt cht

lng trng lc nht nh, trong iu kin p sut mt t do p0 cho trc, p sut p l hm s bc nht ca su h; nh vy trong h to p, h phng trnh (2 12) c biu din bng mt ng thng. n gin ta gi thit p0 = pa khi pd = h.

S biu din bng th hm s (2 12) trong h to ni trn gi l phn b p sut thu tnh (hnh 2 10a).

..

45

Hnh 2 10


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Mun c phn b p sut tuyt i ch cn tnh tin ng OA' theo phng thng gc Oh mt on p0 v c O''A''. phn b p sut tuyt i l hnh thang vung gc OO''AA".

Nu ng thng ng trn xt s phn b p sut thy tnh khng bt u t mt t do , m bt u t mt su h' (im B hnh 2 10b), th p lc s l hnh thang vung BB'A'A (p sut d) hoc BB"A"A (p sut tuyt i).

Sau khi xt p lc trn nhng ng thng ng, ta c th v p lc trn ng thng nghing hoc ng gy khng kh khn g lm. Trong trng hp ny p lc cng l tam gic vung hoc hnh thang vung; (hnh 2 11) l th d v v p lc trn ng thng gy OAB:

Hnh 2 11.

Tam gic vung OAA' v hnh thang vung AA1'B'B l nhng p lc d tng ng vi on thng OA v AB. Mun v p lc tuyt i, ch cn tnh tin nhng cnh OA', A1'B' theo phng thng gc vi OA v AB i mt on

thng

0p v c c nhng hnh thang OO"A"A v AA1"B"B.

Cn v phn b p sut trn ng cong ta phi biu din bng th tr s p sut ti tng im theo phng trnh (2 12) ri ni li thnh ng cong ca phn b.

2.6 S cn bng ca cht lng trong bnh cha chuyn ng. Nghin cu s cn bng ca cht lng trong trng hp cc phn t cht lng

khng c chuyn ng tng i vi nhau nhng c chuyn ng i vi qu t: khi c khi cht lng chuyn ng nh mt vt rn, ta gi trng thi ny l trng thi tnh tng i ca cht lng, n xut hin khi bnh cha chuyn ng vi mt gia tc khng i, lc khi tc dng vo cht lng khng ch c trng lc m cn c c lc qun tnh. Ta nghin cu hai trng hp tnh tng i ca cht lng:

1. Khi bnh cha chuyn ng thng vi gia tc khng i.


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2. Khi bnh cha hnh tr trn quay u quanh trc thng ng ca bnh, h to gn cht vi bnh cha.

1. S cn bng ca cht lng ng trong bnh chuyn ng thng vi gia tc khng i.

Trng hp ny thng gp cc xe ch du, nc. Gi thit bnh cha ang chuyn ng thng vi gia tc khng i a. Mi phn t cht lng chu tc dng ca hai lc khi: trng lc G = mg v lc qun tnh R = ma, trong m l khi lng ca phn t cht lng. Vi h to nh hnh 2 12, hnh chiu Fx, Fy, Fz ca cc lc khi ln cc trc l:

Fx = a; Fy = 0; Fz = g

Hnh 2 12.

Mt ng p: Theo (2 8); phng trnh vi phn mt ng p vit thnh:

a.dx g.dz = 0 Tch phn ta co phng trnh mt ng p:

a.x + gz = const

Mt ng p nh vy l mt phng nghing, ta c mt h cc mt ng p song song lp thnh mt gc i vi mt nm ngang theo

gatg = .

S phn b p sut: Theo (2 7); c th vit:

dp = ( adx gdz) Sau khi tch phn, ta c

p = ax gz + C (C l hng s tch phn). Ti x = 0, z= H, c p = p0 (p0 l p sut ti mt thong), hng s tch phn s l:

C = p0 + gH Phng trnh xc nh p sut tnh tng i ti im vit c di dng:

p = ax gz + p0 + gH

Thay trong phng trnh ny = g v gh' = ax, ta c:

p = p0 + (H z) - h' Gi h l su ca im N tu trong cht lng k t mt thong nghing, ta

c: h = H (z + h')

Cui cng ta vit p = p0 + h. Nh vy tr v cng thc c bn tnh p sut thu tnh, ch cn ch rng h l su k t mt thong trong iu kin cn bng tng i.


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2. S cn bng ca cht lng ng trong bnh hnh tr trn quay u quanh trc thng ng qua tm bnh. Lc tc dng ln mi phn t cht lng bao gm: trng lc G = mg v lc qun

tnh ly tm F = m2r (trong l gia tc gc, r l khong cch t v tr phn t cht lng ta xt n trc quay). Theo h to nh trn hnh v (2 13), ly m = 1, hnh chiu Fx, Fy, Fz ca cc lc khi ln cc trc l:

Fx = 2x; Fy = 2y; Fz = g; trong x, y l hnh

Hnh 2 13.

chiu ca r ln trc x, y.

Mt ng p: Theo (2 8); c th vit:

2xdx + 2ydy g.dz = 0

Sau khi tch phn ta c:

Cgzyx =+ 22221

21

hoc: ( ) Cgzyx =+ 22221

hoc: Cgzr =2221

y l phng trnh ca mt parablt c trc quay Oz. vy mt ng p trong trng hp ny l mt h cc mt parablt (hnh 2 13), vi cc tr s C khc nhau.

Trn mt t do, khi x = y = 0 tc l r = 0, th z = z0, hng s tch phn bng:

C = gz0

Do phng trnh mt t do l:

)(21

022 zzgr r =

zr l tung ca im trn mt t do, cch trc quay l r.

Gi h' = zr z0 th phng trnh mt t do thnh: '21 22 ghr =

S phn b p sut: Theo (2 7); c th vit: dp = (2xdx + 2ydy gdz)

Sau khi tch phn ta c: 122

2

22Cgzyxp ++= ; trong C1 l hng

s tch phn. Trn mt t do: p = p0; khi r = 0, th z = z0; vy:

C1 = p0 + gz0


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v: 002

2 )2

( gzpgzrp ++=

hoc ( ) ( zzhgpzzgrpp ++=

+= 000

22

0 '2 )

gi h = h' + z0 z, th r rng h l su ca im ang xt tnh t mt t do cong trng thi tnh tng i, nh vy: p = p0 + h; ta tr v cng thc c bn tm p sut thy tnh.

2.7 p lc ca cht lng ln thnh phng c hnh dng bt k. Trong trng hp thnh rn l mt phng, th nhng p sut tc dng ln thnh

rn u song song vi nhau, do chng c mt p lc tng hp P duy nht. Ta nghin cu tr s v im t ca P.

1. Tr s p lc:

Hnh 2 14

Cn xc nh p lc P ca cht lng tc dng ln mt din tch phng c hnh dng bt k t nghing i vi mt thong mt gc (hnh 2 14).

p lc dP tc dng ln mt vi phn din tch d, m trng tm ca n t su h tnh bng:

dP = pd = (p0 + h)d

p lc P tc dng ln ton b din tch bng:

+==

dhppdP )( 0

Trn hnh phng ly h trc to Oxy nh hnh 2 14; ta c: h = zsin Vy:

+=+=+=

zdpdzdpdzpP sinsin)sin( 000 (2 20)

Tch phn , chnh bng m men tnh ca din tch i vi trc Oy. =

OySzd

Gi zc l tung ca trng tm C ca din tch , nh bit trong c hc l thuyt, c th vit: SOy = zc.

Gi hc l su ca C th: hc = zcsin

do :

sinc

Oyh

S = ; biu thc (2 20) vit thnh:

P = (P0 + hc) (2 21)

(ch rng biu thc (p0 + hc) l p sut tuyt i ti trng tm C ca din tch phng).


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Nh vy: p lc thu tnh ca cht lng tc dng ln din tch phng, ngp trong cht lng bng tch s ca p sut tuyt i ti trng tm din tch phng vi din tch y.

Nu p0 = pa, p lc d tc dng ln din tch phng ni trn bng:

P = hc (2 22)

Trong thc tin k thut, nhiu khi mt phng cn xt chu p lc thy tnh v mt pha, cn pha kia ca mt phng li chu p lc ca khng kh, trong trng hp mt phng ch chu tc dng ca p lc d m thi v p lc khng kh truyn t mt thong n mt phng cn bng vi p lc khng kh tc dng vo pha kh ca mt phng. V vy trong nhng trng hp tng t ch cn tnh p lc d theo (2 22).

p lc d thy tnh tc dng ln y phng ca bnh cha l trng hp ring ca p lc thy tnh tc dng ln mt phng. Nu din tch y v su h ca y gi khng i th p lc cht lng ln y thnh phng P = h khng ph thuc vo hnh dng bnh cha.

2. V tr tm p lc: im t ca p lc gi l tm p lc (tu theo p lc l p lc tuyt i hay

p lc d m tm p lc gi l tm p lc tuyt i hay tm p lc d) Ta gi D(z, y) l tm p lc d (hnh 2 14); cn xc nh cc to zD v yD

ca im D.

a) Xc dnh zD: Mmen ca p lc P i vi trc Oy bng:

M= PzD = hczD (2 23)

Tng s mmen i vi trc Oy ca p lc ln din tch vi phn bng (2 24); trong l

mmen qun tnh ca din tch i vi trc Oy. Cn bng (2 23) v (2 24) ta c:

====

yIdzhzdpzdM sinsin2 =

dzI y2

c

y

c

yD z

Ih

Iz

.sin

== (2 25)

Nh bit trong c hc, c th biu th mmen qun tnh ca din tch i vi trc Oy (Iy) bng mmen qun tnh ca din tch y i vi trc O'y' song song vi Oy v i qua trng tm C ca din tch (I0) nh sau:

20 zII y +=

Thay tr s Iy vo (2 25) ta c:

ccD z

Izz

0+= (2 26)


_ 23 _

Nh vy v tr ca tm p lc bao gi cng t su hn v tr ca trng tm.

ph lc 1 c cng thc tnh I0, zc v cho mt s hnh phng hay gp.

b) Xc dnh yD: Tng t nh lc xc nh zD, ta vit mmen cho trc Oz:

==

pydPyM D

Thay P theo (2 22) v ch rng hc = zcsin v p = h = zsin ta c:

=

zydyz Dc sinsin ; do c

D z

zydy

= (2 27)

2.8 p lc cht lng ln thnh phng hnh ch nht c y nm ngang.

Ta xt trng hp tng qut, khi thnh phng hnh ch nht t nghing vi mt nm ngang mt gc , c y rng b, chiu cao h; y trn hnh ch nht t su h1, y di t su h2 (hnh 2 15), p sut ti mt t do bng p sut khng kh p0 = pa.

V c th suy ra d dng p lc tuyt i khi bit p lc d, nn ta ch cn xt vic xc nh p lc d. Tr s tng p lc d trong trng hp ang nghin cu c th xc nh theo cng thc (2 22)

P = hc

Hnh 2 15

y = bh,

221 hhhc

+=

Vy:

bhhh

P2

21 += (2 28)

Tr s hhh2

21 + v phi ca phng trnh

(2 28) bng din tch ca p lc d AA'BB' (hnh 2 15)

hhh

221 +=

Vy cng thc (2 18), tr thnh:

P = b (2 29)

Ta c th ni rng: p lc d P tc dng ln hnh ch nht bng tch s din tch p lc vi b di y v trng lng ring ca cht lng.


_ 24 _

ng tc dng ca P tt nhin i qua trng tm th tch to bi p lc v hnh ch nht chu lc. Trn hnh (2 15) lc i qua trng tm ca p lc, v hnh chiu trng tm ca th tch ni trn ln p lc trng vi hnh chiu ca tm p lc.

Nu cnh trn hnh ch nht t ti mt t do (hnh 2 16a) th h1 = 0; p

Hnh 2 16.

lc thnh hnh tam gic vung gc c cnh khng bng nhau v c din tch bng: 22

1 hh= ; vy p lc P bng: hbhbP 22 ==

Trong trng hp ny, lc P i qua trng tm ca p (hnh 2 16a) tc l ti su 23

2 h . Nu hnh ch nht t thng ng th p lc trn thnh tam gic

vung cn (hnh 2 16b); do : 2

21 h= v 2

2bhbP == (2 30)

p lc d P i qua trng tm ca p lc, tc l su 232 h

Th d 4: Tnh p lc nc ln cnh cng ch nht c h = 3m, b = 2m, su nc thng lu H = 6m.

Gii: Ta ch cn tnh p lc d P. p dng (2 22), ta phi tnh hc. Theo (hnh 2 17):

Hnh 2 17

mhHhc 5,4236

2===

Theo (2 22):

P = hc = 9.810x4,5x3x2 = = 264.870N (P = 27000kG)

Tm p lc tnh theo (2 26) bng:

ccD h

Ihz

0+= ; 5,4

29

123.2

12

33

0 ====bhI

66,45,4.2.3

5,45,4 =+=Dz m


_ 25 _

Dng phng php gii:

p lc ln ca cng l hnh thang c din tch S1 (hnh 2 17). p lc d tnh theo cng thc:

( )[ ] ( )[ ] NbhHhHbSP 870.2642.2

363681.9.2

.1 =

+=

+==

ng tc dng ca lc P i qua trng tm ca p lc hnh thang. nh bit, trng tm hnh thang cch y ln mt on bng

3.

''2 a

BBBB

++ , trong B,

B', a ln lt l y ln, y b v chiu cao hnh thang. Vy trng tm hnh thang trng hp ang xt cch mt nc t do l:

mhHHhHHHzD 66,43

3.363.266

33.).(2 =

++

=++

=

2.9 p lc cht lng ln thnh cong.

Ni chung nu thnh cong c hnh dng bt k, th nhng p lc nguyn t khng hp li thnh mt p lc tng hp duy nht.

Trong mt s trng hp ring, nh mt cong l mt cu, mt tr trn xoay co ng sinh t nm ngang hoc thng ng, nhng p lc nguyn t u ng quy hoc u song song, do c mt p lc tng hp duy nht.

Ta ch nghin cu mt trng hp thng gp l p lc cht lng tc dng ln thnh cong hnh tr trn c ng sinh t nm ngang.

Ta c mt mt tr ABA'B' ng sinh nm ngang c di v cung AB

l mt cung trn. n gin vic tnh ton ta t h to sao cho trc nm ngang, th d trc Oy song song vi ng sinh, mt to nm ngang trng vi mt t do (hnh 2 18). p sut trn mt t do bng p sut khng kh: p0 = pa. Ta s xc nh thnh phn Px v Pz ca P (cn Py = 0), ri tm ra P theo:

22zx PPP +=

Hnh 2 19

Hnh 2 18


_ 26 _

Chng ta ch nghin cu p lc d, v phng php xc nh p lc tuyt i cng tng t.

Gi thit mt tr chu p lc cht lng t pha trn, pha di mt tr l kh.

Trn mt tr ly mt din tch nguyn t d, t su h (hnh 2 19); p lc nguyn t dP tc dng ln din tch bng: hddP =

Lc ny phn thnh hai thnh phn, thnh phn nm ngang dPx v thnh phn thng ng dPz (thnh phn dPy = 0).

xx hdhddPdP xdPxdP =

=

=

,, coscos

zz hdhddPdP zdPzdP =

=

=

,, coscos

Trong : dx = l hnh chiu ca din tch d ln mt ta

thng gc vi trc Ox v d

xdPd ,cos

z = l hnh chiu ca din tch d ln

mt ta nm ngang tc l ln mt t do.

zdPd ,cos

Thnh phn nm ngang Px ca p lc P xc nh bi: ==x x

xxx hddPP

Theo cng thc p lc d ln mt phng (2 22) th c th vit thnh:

xcx hP = (2 31)

Trong x l hnh chiu ca din tch ABA'B' ln mt zOy v hc l su ca trng tm ca x.

Thnh phn Px ny cn c th tnh ra d dng bng p lc, theo cng thc (2 29)

bP xx = (2 32)

trong x l din tch p lc. ng tc dng ca Px t su bng su ca trng tm din tch x.

Thnh phn thng ng Pz ca p lc P xc nh bi:

==z z

zzz hddPP

(2 33)

Ta nhn st rng tch phn bng th tch W ca hnh tr thng ng L,

gii hn b mt tr ABA'B', W c th tch bng:

x

xhd

bW z .= (2 34)

trong z l din tch hnh ABba.


_ 27 _

Vy thnh phn Pz chnh l trng lng G ca hnh lng tr L ni trn.

Pz = W = G (2 35)

ng tc dng ca thnh phn Pz i qua trng tm ca hnh lng tr L, hnh lng tr L gi l vt p lc. Vy thnh phn thng ng Pz bng trng lng ca vt p lc.

Nhng cng thc cho ta tm ba thnh phn Px, Py, Pz ca p lc P ln thnh cong l:

Px = hcx

Py = hcy (2 36)

Pz = W

Trong hc l su trng tm ca mt tr v ng thi cng l su trng tm nhng hnh chiu x, y

Kh 222 zyx PPPP ++= (2 37)

Ta nghin cu thm v vt p lc v phng ca Pz. Vt p lc l th tch gii hn bi thnh cong m ta xt, bn mt bn thng

ng, t ln cc mp ca thnh cong v ko di n khi mt ct t do hoc phn ko di ca mt t do ca cht lng. Trng lng ca vt p lc biu th thnh phn thng ng Pz ca p lc P. Trong trng hp mt cong l mt tr c ng sinh nm ngang, vt p lc thng biu th bi mt ct thng ng ca th tch ni trn v l din tch gii hn bi ng cong chu lc, hai ng thng ng i qua hai u ca ng cong v gp mt t do hoc phn ko di ca mt t do (th d mt ct ABab trn hnh 2 19).

Sau y l ba trng hp vt p lc: 1. Vt p lc c cht lng ngay trn mt cong (hnh 2 21a,b): c th cht

lng chim ton th vt p lc (hnh 2 21a) hoc ch c th chim mt phn ca vt p lc (hnh 21 b), trong c hai trng hp ny Pz u hng xung di. Ta quy c khi Pz hng xung di, vt p lc mang du (+)


_ 28 _

Hnh 2 21.

2. Vt p lc khng c cht lng ngay trn mt cong (hnh 2 21c,d): c th cht lng hon ton khng c trong vt p lc (hnh 2 21c) hoc c th ch chim mt phn vt p lc (hnh 2 21d), trong trng hp ny Pz u hng ln trn. ta quy c khi Pz hng ln trn, vt p lc mang du (-).

3. Mt cong c hnh dng hi phc tp, lm cho vt p lc c hnh dng phc tp: th d mt cong ACDB (hnh 2 22). Theo nh ngha v vt p lc ni trn, din tch ca mt thng ng ca vt p lc gm hai b phn: 1 v 2, 1 l din tch ca hnh BDE v 2 l din tch ca hnhACEb. xc nh hng v cc phn Pz1 ng vi 1 v Pz2 ng vi 2, ta c th phn ton b ng cong phc tp thnh nhiu on n gin ni trn. Th d on cong BDE phn thnh hai on BD v DE; vt p lc ng vi BD l hnh BDdb v theo nh quy c ni trn mang du (+), cn vt p lc ng vi DE l hnh DEbd mang du (-); tng s i s ca hai din tch BDdb v DEbd cho ta din tch BDE vi du (+). Ta cng chia ng cong ECA thnh hai on EC v CA ri cng tm vt p lc ng vi tng on, km theo du tng ng, sau cng cng i s nhng din tch ca vt p lc th tm c din tch ACEb vi du (-).

Nguyn tc dng vt p lc ni

Hnh 2 22

Hnh 2 23

trn tm phng hng cho thnh phn Pz p dng cho nhng trng hp m p sut d tc dng vo mt cong ln hn s khng: Pd > 0, tc khng c vn p sut chn khng (Pd < 0)

Th d 5: Tm tng p lc nc tc

dng ln mt ca cong AB di l = 3m, c din tch bng 1/4 din tch mt bn ca hnh tr trn m bn knh r = 1m (hnh 2 23). su nc h = 1m.

Gii: Tnh cc thnh phn Px v Pz ca tng

p lc P. p lc biu th thnh phn nm


_ 29 _

ngang ca tng p lc l tam gic vung cn ACD c din tch 221 hACD = ; vy

theo (2 32), ta c:

)1500(147152

3.1.981021 22 kGNlhlP ACDx =====

Thnh phn thng ng ca tng p lc biu din bi vt p lc ABO c din

tch ABO = 42r

ABO

= , do c th tch bng th tch 4

2lrlW ABO

==

Vy theo (2 35), ta c:

)2360(231034

3.1.14,3.98104

22

kGNlrWPz ====

V ngay pha trn mt cong khng c nc, nn Pz hng ln trn. Tng p lc tnh theo:

)2800(274702310314715 2222 kGNPPP zx =+=+=

ng tc dng ca tng p lc P i qua tm O lp vi ng nm ngang mt gc m:

56,1==z

x

PP

tg

Tc l = 57020'

2.10 nh lut csimt, s cn bng ca vt rn ngp hon ton v ni trn mt t do ca cht lng.

1. nh lut csimt:

Hnh 2 24

Ta xt lc thy tnh P tc dng vo mt vt rn c th tch W ngp hon ton trong cht lng (hnh 2 24). Mun vy ta xt thnh phn thng ng Pz' v thnh phn nm ngang Px' ca p lc P. Mun xc nh thnh phn thng ng Pz' ca P ta v mt tr thng ng m cc ng sinh ca mt tr u l nhng tip tuyn i vi mt ngoi ca vt rn; ng cong i qua tt c cc im tip xc gia mt tr v mt ngoi ca vt chia mt ngoi ca vt rn thnh hai phn khng kn: phn trn cde v phn di cfe. Lc Pz1' tc dng ln phn trn bng trng lng ca vt p lc abcde v hng thng ng; theo quy c v du ca vt p lc th P'z1 mang du (+).


_ 30 _

P'z1 = + Vabcde

Lc P'z2 tc dng ln phn di bng trng lng ca vt p lc abcfe v hng thng ng ln trn; P'z2 mang du (-):

P'z2 = - Vabcfe

Tng p lc thng ng P'z tc dng ln ton b mt kn ca cdfe bng:

P'z = P'z1 + P'z2 = (Vabcde Vabcfe) = Vcdef = - W (2 38)

v bao gi cng hng ln trn v bao gi cng c: '1'2 zz PP >

Mun xc nh thnh phn nm ngang Px ca P ta v mt tr nm ngang m cc ng sinh u tip xc vi mt ngoi ca vt rn; ng cong i qua tt c cc im tip xc gia mt tr v mt ngoi ca vt chia mt ngoi ca vt rn thnh hai thnh phn khng kn: phn tri kcm v phn phi kem.

V nhng hnh chiu thng ng k'c'm' v k'e'm' ca nhng mt kcm v kem bng nhau v trng tm ca nhng hnh chiu nhng su bng nhau, nn tng hp hai thnh phn tng p lc nm ngang bn tri v bn phi bng khng: P'x = 0; nh vy, ch cn li P = Pz.

Vy: Mt vt rn ngp hon ton trong cht lng chu tc dng ca mt p lc hng ln trn, c tr s bng trng lng khi cht lng b vt rn chon ch.

nh lut ny l nh lut csimt; p lc gi l lc csimt hoc lc y (cn gi l lc nng).

Phng ca lc csimt i qua trng tm D ca khi cht lng b vt rn chon ch, im D c gi l tm y (D khng phi l im t ca lc csimt)

nh lut csimt cng dng cho vt ni, tc l vt khng b chm hon ton trong cht lng v ni trn mt t do ca cht lng. Lc , p lc thy tnh tc dng ln phn b ngp trong nc bng trng lng khi cht lng b phn ngp ca vt rn chon ch.

2. S cn bng ca vt rn ngp hon ton trong cht lng: Trn c s nh lut csimt, nghin cu s cn bng ca mt vt rn ni

chung khng ng nht ngp hon ton trong cht lng, vt rn chu tc dng ca hai lc thng ng: trng lc G t ti trng tm C ca vt rn, hng xung di v lc y csimt t ti tm y D (tc l trng tm vt khi coi vt l ng cht), hng ln trn.

Mun vt ng cn bng tc l khng chm xung, khng ni ln, khng t quay th hai lc Pz v G phi bng nhau v t cng trn mt ng thng ng. V tr ca hai im G v D nh hng n tnh cht cn bng ca vt rn.

1. C thp hn D (h.2 25a) th s cn bng l n nh v nu y vt dch khi v tr cn bng th di tc dng ca ngu lc lp bi pz v G vt li tr v v tr c; 2. C cao hn D (h.2 25b) th s cn bng l khng n nh v nu y vt dch khi v tr cn bng th ngu lc lp bi Pz v G lm cho vt ln ngc i xa


_ 31 _

v tr c v chim v tr cn bng n nh; 3. C v D trng nhau (h.2 25c) ngha l trong trng hp vt ng cht th vt trng thi cn bng phim nh, ngha l vt ng cn bng vi bt k v tr ban u no. Vt rn khng trng thi cn bng nu Pz G, nu Pz < G th vt chm, nu Pz > G th vt ni ln.

Hnh 2 25

3. S cn bng ca vt rn ni trn mt t do ca cht lng: iu kin cn bng ca vt rn ni trn b mt t do ca cht lng khng ging

hn vi iu kin cn bng ca vt ngp hon ton trong cht lng. Ta xt mt vt ni trn mt nc, th d nh mt tm g, mt con tu..v Khi

vt trng thi ni th tt nhin iu kin Pz = G tho mn. Nu trng tm C ca vt ni thp hn tm y D th s cn bng ca vt ni

l n nh. Tuy nhin nu trng tm C cao hn tm y D, vt cha phi hon ton khng cn c th trng thi cn bng n nh.

Khi trng tm C cao hn tm y D m vt ni trng thi cn bng th c mt s khi nim

Hnh 2 26

nh sau: (hnh 2 26)

Mn nc l giao tuyn ca vt ni vi mt nc .

Mt ni l mt phng c chu vi l ng mn nc.

Trc ni l ng thng gc vi mt ni qua tm vt ni.

Chương 2 - Thủy Tĩnh Học

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