Drs. Pristiadi Utomo, M.Pd.
BAB 8 Drs. Pristiadi Utomo, M.Pd.
FLUIDA
STANDAR KOMPETENSI : Menerapkan konsep dan prinsip mekanika klasik sistem kontinu dalam menyelesaikan menyelesaikan masalah. KOMPETENSI DASAR Setelah mempelajari bab ini Kamu dapat menganalisis hukum-hukum yang berhubungan dengan fluida statik dan dinamik serta penerapannya dalam kehidupan sehari-hari
Telur dapat melayang di dalam zat cair. Menurut hukum Archimedes pada keadaan itu massa jenis telur sama dengan massa jenis zat cair. Zat cair sebagai fluida memiliki gaya tekan ke atas sebesar tekanan dikalikan dengan luas penampang telur. Menurut Pascal tekanan itu akan ditrus dit ruskan kan ke seg segala ala ara arah h deng dengan an sama bes besarny arnya. a. Per Pernah nahkah kah kamu mel meliha ihatt ala alatt hid hidrol rolik ik pengangkat mobil di tempat pencucian mobil? Mobil dapat dinaikkan di atas pengisap yang didorong oleh gaya hidrostatik dan gaya ini hasil kali dari tekanan dengan luas penampang pengisap yang dipakai landasan mobil. Dalam bab ini Kamu akan mempelajari fluida statik dan fluida dinamik beserta hukum-hukum yang terkait di dalamnya serta penerapannya dalam kehidupan sehari-hari.
Drs. Pristiadi Utomo, M.Pd.
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Drs. Pristiadi Utomo, M.Pd.
Gerbang
Kapal selam dapat tenggelam dalam air laut dan dapat muncul lagi ke atas dengan mengisi atau mengos mengosongk ongkan an rongga rongga udara udara di dalamny dalamnyaa dengan dengan flui fluida da.. Dala Dalam m fisi fisika ka pros proses es-p -pro rose sess gera gerak k flui fluida da memerlukan pembahasan khusus mengingat menging at sifat-sifat flui fluida da yang yang berbe berbeda da denga dengan n sifa sifatt-si sifa fatt zat zat pada padat. t. Meka Mekani nika ka flui fluida da memb membah ahas as zat zat dala dalam m kead keadaa aan n berwujud cair atau gas dengan segala fenomenanya.
Kapal Selam
Mekanika fluida membatasi pembahasan gerak fluida dengan mengganggap fluida tidak mengalami perubahan volume sewaktu diberi tekanan. Dalam keadaan itu fluida disebut tidak kompresibel. Pembahasan fluida meliputi fluida statik dan fluida dinamik.
A. Fluida Statik Fluida Fluida merupakan merupakan istilah istilah untuk zat alir. alir. Zat alir dibatasi dibatasi pada zat mengalirkan seluruh bagian-bagiannya ke tempat lain dalam waktu yang bersamaan. Zat alir mencakup zat yang dalam wujud cair dan gas. Fluida statik meninjau fluida yang tidak bergerak. Misalnya air di gelas, air di kolam renang, air dalam kolam, air danau, dan sebagainya. Penggolongan Penggolongan fluida menurut menurut sifat-sif sifat-sifatnya atnya dibedakan menjadi menjadi dua, yaitu : 1. Fluida ideal 2. Fluida sejati
Gambar: Air dalam gelas adalah fluida statik
1) Fluida ideal
Ciri-ciri Fluida ideal adalah: a. Fluida Fluida yang tidak tidak kompresibe kompresibell (volumenya (volumenya tidak tidak berubah berubah karena perubaha perubahan n tekanan) tekanan) b. Berpin Berpindah dah tanp tanpaa mengal mengalami ami gese gesekan kan
2) Fluida sejati
Ciri-ciri Fluida sejati adalah: a. Kompr ompres esiibel bel b. Berpin Berpindah dah denga dengan n mengala mengalami mi gesek gesekan an
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Drs. Pristiadi Utomo, M.Pd. Sedangkan gaya-gaya yang bekerja pada fluida ada tiga macam yaitu: -
Kohesi, Kohesi, yaitu yaitu : gaya tarik-mena tarik-menarik rik antara partikel-pa partikel-partike rtikell yang yang sejenis sejenis
-
Adhesi, Adhesi, yaitu yaitu : gaya gaya tarik-m tarik-menarik enarik antara partikel-pa partikel-partike rtikell yang yang tidak tidak sejenis sejenis
-
tegan teganga gan n perm permuka ukaan an , yait yaitu u gaya gaya pada pada permuk permukaa aan n flui fluida, da, anggap anggapla lah h bahw bahwaa sete setete tess air air seolah-olah ada pembungkus.
1. Kohesi dan Adhesi Setetes air yang jatuh di kaca meja akan berbeda bentuknya bila dijatuhkan di sehelai daun talas. Mengapa demikian ? Antara molekul-molekul air terjadi gaya tarik-menarik yang disebut dengan gaya kohesi molekul air. Gaya kohesi diartikan sebagai gaya tarik-menarik antara partikel partikel zat yang sejenis. Pada saat air bersentuhan dengan benda lain maka molekul-molekul bagian luarnya tarik-menarik dengan molekul-molekul luar benda lain tersebut. Gaya tarik-menarik antara partikel zat yang tidak sejenis disebut gaya adhesi. Gaya adhesi antara molekul air dengan molekul kaca berbeda dibandingkan gaya adhesi antara molekul air dengan molekul daun talas. Demikian pula gaya kohesi antar molekul air lebih kecil daripada gaya adhesi antara molekul air dengan molekul kaca. Itulah sebabnya air membasahi kaca berbentuk melebar. Namun air tidak membas membasahi ahi daun daun talas talas melain melainkan kan tetes tetes air berbent berbentuk uk bulatbulat-bul bulat at menggel menggelind inding ing di permuk permukaan aan karena gaya kohesi antar molekul air lebih besar daripada gaya adhesi antara molekul air dan molekul daun talas. Gaya Gaya kohes kohesii maup maupun un gaya gaya adhes adhesii memp mempeng engar aruhi uhi bentu bentuk k perm permuka ukaan an zat cair cair dalam dalam wadahnya. Misalkan ke dalam dua buah tabung reaksi masing-masing diisikan air dan air raksa. Apa yang terjadi ? Permukaan air dalam tabung reaksi berbentuk cekung disebut meniskus cekung sedang sedangkan kan permuk permukaan aan air raksa raksa dalam dalam tabung tabung reaksi reaksi berben berbentuk tuk cembung cembung disebut disebut menisk meniskus us cembung. Hal itu dapat dijelaskan bahwa gaya adhesi molekul air dengan molekul kaca lebih besar daripada daripada gaya kohesi antar molekul air, air, sedangkan sedangkan gaya adhesi molekul air raksa dengan molekul kaca lebih kecil daripada gaya kohesi antara molekul air raksa.
Meniskus cekung dan meniskus cembung
Meniskus cembung maupun meniskus cekung menyebabkan sudut kontak antara bidang wadah (tabung) dengan permukaan zat cair berbeda besarnya. Meniskus cembung menimbulkan sudut kontak tumpul (> 90 90°°), sedangkan meniskus cekung menimbulkan sudut kontak lancip (< 90 °).
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Drs. Pristiadi Utomo, M.Pd. Gaya kohesi dan gaya adhesi juga berpengaruh pada gejala kapilaritas. Sebuah pipa kapiler kaca bila dicelupkan pada tabung berisi air akan dijumpai air dapat naik ke dalam pembuluh kaca pipa kapiler, sebaliknya bila pembuluh pipa kapiler dicelupkan pada tabung berisi air raksa akan dijumpa dijumpaii bahwa bahwa air raksa raksa di dalam dalam pembul pembuluh uh kaca kaca pipa pipa kapile kapilerr lebih lebih rendah rendah permuka permukaanny annyaa dibandingkan permukaan air raksa dalam tabung. Jadi kapilaritas sangat tergantung pada kohesi dan adhesi. Air naik dalam
Gambar: Pipa kapiler
pembuluh pipa kapiler dikarenakan adhesi sedangkan air raksa turun dalam pembuluh pipa kapiler dikarenakan kohesi. Perhatikan gambar berikut ini. .
Air
Raksa
Gambar: Pipa kapiler dalam tabung berisi air maupun air raksa
Pada air: Permukaanny Permukaannyaa cekung, pada pipa kapiler permukaannya permukaannya lebih tinggi, tinggi, karena adhesinya lebih kuat dari kohesinya sendiri. Pada raksa: Permukaanny Permukaannyaa cembung, sedangkan pada pipa kapiler permukaannya permukaannya lebih rendah, rendah, karena kohesi air raksa lebih besar dari adhesi antara air raksa dengan kaca.
2. Tegangan Permukaan Dalam zat cair ada partikel-partikel yang dikelilingi semacam bola dimana partikel itu sebagai pusatnya. Dalam bola itu adalah suatu medan. Perhatikan tiga buah partikel fluida A, B, C .
F2 F1
C
B A Gambar : Tiga buah partikel fluida yang terletak di te mpat-tempat berbeda memiliki keadaan gaya yang berbeda
Partikel Partikel A
=
dalam keadaan setimbang, setimbang, bekerja bekerja gaya-gaya gaya-gaya yang sama besar dari semua arah
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Drs. Pristiadi Utomo, M.Pd. Partikel B
=
karena F1 > F2 , gaya yang arahnya ke bawah lebih besar daripada gaya yang arahnya ke atas
Part artikel kel C
=
hany hanyaa ada ada gay gaya ke bawa bawah, h, hal hal inila nilah h yang ang dapa dapatt meny enyebab ebabka kan n tegan eganga gan n permukaan
Karena adanya kohesi, partikel-partikel pada permukaan air cenderung ditarik ke dalam. Sehingga zat cair membentuk permukaan yang sekecil-kecilnya. Dengan adanya adhesi, kohesi dan tegangan permukaan ketiganya dapat menentukan bentuk-bentuk permukaan zat cair. Bentuk permukaan itu misalnya cembung atau cekung.
a. Zat cair dalam bejana : Sudut kontak (θ (θ) yaitu sudut yang yang dibatasi oleh 2 bidang batas yaitu dinding tabung dan permukaan zat cair. Dengan pemahaman bahwa, •
dinding tabung : sebagai bidang batas antara zat cair dan tabung,
(θ = 1800) • permukaan zat cair : sebagai bidang batas antara zat cair dan uapnya (θ Menurut sudut kontaknya bentuk-bentuk permukaan zat cair dalam bejana : 1. Cekung
= air air dengan dinding gela elas, 0º < θ < 90º , zat cair membasahi dinding.
air Dinding kaca
2. Cemb Cembun ung g = air air raks raksaa den denga gan n din dindi ding ng gela gelas, s, 0º < θ < 90º, zat cair tidak membasahi dinding.
raksa Dinding kaca 3. Datar Datar = air dengan dengan dindin dinding g perak, perak, θ = 90º
air
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Drs. Pristiadi Utomo, M.Pd. Dinding perak
Tabel Sudut Kontak Zat Cair
Dinding
Air
Parafin Dinding perak Gelas pirex
Methylin Yodida
Gelas kali Timah hitam Pirex
Sudut Kontak 107° 107° 90° 90° 63° 63° 29° 29° 30° 30° 29° 29°
Efek pengurangan sudut kontak karena bahan pembasah kotoran atau campuran yang terdapat di dalam zat cair dapat merubah besarnya sudut kontak. Oleh pabrik banyak dibuat bahan-bahan kimia yang sangat tinggi potensinya sebagai zat pembasah. pe mbasah. Contoh : deterjen, rinso, dan lain-lain. Senyawa Senyawa-se -senya nyawa wa ini merubah merubah besarny besarnyaa sudut sudut kontak kontak yang yang semula semula besarny besarnyaa dari dari 90° 90° menjadi lebih kecil 90° 90°. Sehingga zat cair membasahi bahan. Sebaliknya zat yang membuat kain menjadi tahan air membuat sudut kontak air dengan kain menjadi lebih besar 90° 90°.
(a) Sebelum dicampur zat pembasah
(b) Setelah dicampur zat pembasah
b. Tegangan Tegangan Permukaan pada kawat yang dibengkokkan Gaya yang digunakan untuk menahan kawat
larutan sabun
supaya kawat dalam keadaan setimbang. kawat bengkok
F
kawat yang bisa digeser
= W1 + W2
γ =
W1
F 2l
Newton meter
Tegangan permukaan = gaya g aya per satuan W2
panjang
Kawat digeser sejauh sejauh s maka ada tambahan luas = l . s. Untuk menambah luas tersebut perlu dilakukan usaha dari luar W = F . s Usaha yang dilakukan per satuan luas adalah
F.s 2.l.s
=
F 2l
= γ
Usaha yang dilakukan per satuan luas adalah tegangan permukaan = besarnya energi per satuan luas Satuan tegangan permukaan = Newton meter =
Joule m
2
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Drs. Pristiadi Utomo, M.Pd.
Kerja Kelompok Neraca Torsi
Tujua ujuan n
: Men Menen enttukan ukan bes besarny arnyaa teg tegan anga gan n per perm mukaa ukaan n air air dan dan lar larut utan an sabun abun = massa jenis zat cair Alat-alat : y = tin tingg ggii per permu muka kaan an zat zat cai cair r - neraca torsi θ = sudut kontak - kawa kawatt yan yang g di dibeng bengko kokk kkan an = tegangan permukaan - kepi keping ng kaca kaca beja bejana na beri berisi si air air ata atau u laru laruta tan n sabun sabun r = jar jarii-jar -jarii pip pipaa kap kapiiler ler - anak timbangan g = perc percep epat atan an grav gravit itas asii
Dasar teori : a) Bila digunakan digunakan alat alat kawat kawat yang yang dibengkokk dibengkokkan an digunakan digunakan rumus γ = γ =
F 2l
l b) Bila yang digunakan digunakan keping keping kaca kaca maka maka menggunakan menggunakan rumus γ = γ =
F 2(l + d)
l Cara kerja: hitunglah gaya dari selisih jarum ketika akan lepas dan ketika lepas beban. Kerja Berpasangan Kerjakan soal-soal berikut bersama teman terdekatmu!
1. Sebuah kawat kawat berbentuk berbentuk segitiga sama sama sisi diletakkan diletakkan perlahan-la perlahan-lahan han di atas permukaan permukaan zat cair. cair. Tegangan permukaan zat cair 74 dyne/cm. Gaya oleh tegangan permukaan 1,776 dyne. Tentukan tinggi segitiga tersebut! 2. Sebu Sebuah ah pisau pisau sile silett uang uang beruk berukur uran an 3 cm x 11/2 cm, diletakkan di atas permukaan zat cair. Tegangan permukaan zat cair 72 dyne/cm. Tentukan berat minimum silet tersebut agar tidak tenggelam! 3. Untuk mengangkat mengangkat sebuah sebuah jarum yang yang panjangnya panjangnya 5 cm dari permukaan permukaan zat cair, cair, kecuali berat berat jarum itu sendiri, masih diperlukan gaya sebesar F Newton. Tegangan permukaan zat cair 63,1 dyne/cm. Tentukan F!
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Drs. Pristiadi Utomo, M.Pd.
Gaya yang menarik ke atas
:
2 π r γ r γ cos cos θ
Berat air
:
π r 2 ρ g y
2 π r γ r γ cos cos θ
= π r 2 ρ g y
γ =
ρgyr 2cosθ
Contoh:
1. Untuk Untuk mengangka mengangkatt sepoto sepotong ng kawat yang panjangn panjangnya ya 7,5 cm dari permuk permukaan aan air, air, kecuali kecuali gaya beratnya masih diperlukan gaya tambahan 1165 dyne. Berapakah besarnya tegangan permukaan air pada suhu tersebut ? Penyelesaian : l
= 7,5 cm
F
= 1165 dyne
γ
=
F 2l
=
1165 2.7,5
γ = γ = ... ?
=
1165 15
= 77,667
dyne cm
2. Etil alkohol alkohol naik naik 25 mm dari sebuah pipa pipa gelas yang yang berdiameter berdiameter 0,4 mm. Jika Jika massa massa jenis etil alkohol 0,79
gr Berapakah tegangan permukaan pada suhu tersebut. Sudut kontak cm3
antara etil alcohol dengan gelas : 30° 30° Penyelesaian : y
=
25 mm
d
=
0,4 mm
=
gr 0,79 3 cm
ρ θ
=
30° 30°
r
=
0,2 mm
γ = γ = ... ? γ = γ = =
=
ρgyr 2cosθ 0,02.0,79.980.2,5 2 cos 30° 38,71 1,73
= 22,35 dyne cm-1
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Drs. Pristiadi Utomo, M.Pd. Penyelesaian : γ = γ =
ρgyr 2cosθ
maka
y =
=
2γγcos ρgr
2.465. cos 128 0,2.13,6.980
y = 0, 0,215 cm
Kerja Mandiri Kerjakan soal-soal berikut ini! 1. Hitung Hitunglah lah tekanan tekanan (turun (turunnya nya tinggi) tinggi) pipa kapiler kapiler berdiame berdiameter ter 0,4 mm dan dileta diletakkan kkan vertika vertikall yang salah satu ujungnya dicelupkan dalam bak yang berisi air raksa. (massa jenis 13,6 g/cm3) dengan sudut kontak 1500, tegangan permukaan 450 dyne/cm. 2. Sebuah pipa pipa kapiler dimasukk dimasukkan an tegak lurus lurus ke dalam air raksa. raksa. Tegangan Tegangan permukaan permukaan air raksa raksa
0,5 N/m. Selisih tinggi air raksa didalam dan diluar pipa = ½
2 cm. Diameter kapiler =
1 6,28
cm ; massa jenis Hg = 13,6 g/cm3 ; g = 10 m/det m/det2. Tentukan besarnya sudut kontak antara air raksa dan dinding pipa.
3. Tekanan Hidrostatik. Tekanan adalah gaya per satuan luas yang yang bekerja dalam arah tegak lurus suatu permukaan. Tekanan disimbolkan dengan : p p =
F A
Tekanan hidrostatis adalah tekanan yang disebabkan oleh berat zat cair. cair. Tiap titik di dalam fluida tidak memiliki tekanan yang sama besar, tetapi berbeda-beda sesuai dengan ketinggian titik tersebut dari suatu titik acuan. Dasar bejana akan mendapat tekanan sebesar : P = tekanan udara luar + tekanan oleh gaya berat zat cair Po
(Tekanan (Tekanan Hidrostatik). p = po +
h +
Gaya berat fluida Luas penampang dasar bejana ρ. V . g
+
ρ . g . A . h
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Drs. Pristiadi Utomo, M.Pd.
Untuk konversi satuan tekanan adalah :1 atm = 76 cm Hg dan
1 atm = 105 N/m2 = 106
dyne/cm2 Untuk bidang miring dalam mencari h maka dicari lebih dahulu titik tengahnya (disebut : titik massa). Tiap titik yang memiliki kedalaman sama diukur dari permukaan zat cair akan memiliki tekanan hidrostatik sama
Gambar: Pada kedalaman yang sama tekanan hidrostatis bernilai sama asal zat cair sejenis p sejenis p1 = p2 = p3
Contoh:
1. Seekor ikan berada di dasar kolam air tawar sedalam h = 5 meter. meter. Hitunglah tekanan hidrostatis yang dialami ikan!
Penyelesaian
ph = . g . h ph = 1000 . 10 . 5 ph = 5 . 104 N/m2
Kerja Mandiri
1. Apab Apabil ilaa sebua sebuah h kapal kapal sela selam m meny menyel elam am seda sedala lam m 60 m, berapa berapa besar besar teka tekanan nan yang
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Drs. Pristiadi Utomo, M.Pd.
4. Hukum Pascal. Hukum Pascal berbunyisebagai berikut, tekanan yang bekerja pada fluida di dalam ruang tertutup akan diteruskan oleh fluida tersebut ke segala arah dengan sama besar. besar.
Gambar: Blaise Pascal
Contoh alat yang berdasarkan hukum Pascal adalah : pompa hidrolik, kempa hidrolik, alat pengangkat mobil. Perhatikan gambar bejana berhubungan di bawah ini. Permukaan fluida pada kedua kaki bejana berhubungan F1 A1
F2 A2
sama tinggi. Bila kaki I yang luas penampangnya penampang nya A1 mendapat gaya F1 dan kaki II yang luas penampangnya pe nampangnya A2 mendapat gaya F2 maka menurut Hukum Pascal harus berlaku : p1 = p2 F1 A1
F2 =
A2
atau
F1 : F2 = A1 : A2
Pada alat pengangkat mobil dengan gaya yang kecil dapat menghasilkan gaya angkat yang besar sehingga mampu mengangkat mobil
Gambar :
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Drs. Pristiadi Utomo, M.Pd. 1. Luas Luas penampang penampang penghis penghisap ap yang kecil dan yang besar besar dari suatu suatu pompa hidrolik hidrolik adalah adalah 6 cm2 dan 20 cm2. Jika pada penghisap yang kecil bekerja gaya 50 N, berapakah besar gaya timbul pada penghisap yang besar ? 2. Pompa hidrolik hidrolik mempunyai mempunyai penghisap penghisap dengan dengan luas luas penampang penampang 15 cm2 dan 3 dm2. Jika pada penghisap yang kecil diberi beban 400 N. Berapa besar gaya pada penghisap yang besar agar terjadi keseimbangan ? 3. Gaya besarnya besarnya 5 N pada penghisa penghisap p yang kecil dari dari suatu suatu pompa hidrolik hidrolik dapat dapat mengangkat mengangkat beban beratnya beratnya 600 N yang terdapat pada penghisap yang besar. Jika penghisap yang kecil berpenampang 400 cm2, berapakah luas penampang yang besar ? 4. Suatu kempa kempa hidrolik hidrolik dapat mengangkat mengangkat 1 ton ton mobil, jika jika diameter diameter penghisap penghisap besar besar 50 cm, diameter penghisap kecil 10 cm. Tentukan gaya yang harus dikerjakan pada penghisap kecil. 5. Sebu Sebuah ah kemp kempaa hidr hidrol olik ik memp mempun uny yai tora torak k yang yang berd berdia iame mete terr 20 cm dan dan 2 m untu untuk k mengangkat mobil. Pada torak kecil dilakukan gaya sebesar 100 N, sehingga torak besar naik setinggi 1 cm. Tentukan massa mobil dan berapa m turunnya torak kecil tersebut.
5. Hukum Utama Hidrostatik. Hukum utama hidrostatik berbunyi sebagai berikut, tekanan hidrostatis pada sembarang titik yang terletak pada bidang mendatar di dalam sejenis zat cair yang dalam keadaan seimbang adalah sama.
Gambar: Skema hukum utama hidrostatik
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Drs. Pristiadi Utomo, M.Pd. Besarnya gaya hidrostatika (Fh) yang bekerja pada bidang seluas A adalah : Fh = ph . A = Fh =
.g.h .A
.g.V
Dimana Fh = gaya gaya hidrostatika hidrostatika dalam SI (MKS) adalah Newton,
dalam CGS adalah dyne.
Kerja Berpasangan Kerjakan soal-soal berikut bersama teman sebangkumu!
1. Suatu bejana bejana berbentuk berbentuk pipa U mula-mula mula-mula diisi diisi dengan air raksa raksa yang massa massa jenisnya jenisnya 13,6 g/cm3, kemudian kaki kanan dituangkan 14 cm air lalu di atas air ini dituangkan minyak yang massa jenisnya 0,8 g/cm3, ternyata dalam keadaan setimbang selisih tinggi permukaan air raksa dalam kedua kaki 2 cm. Hitung berapa cm tinggi lajur minyak pada kaki kanan. 2. Dala Dalam m pipa pipa U terd terdapa apatt Hg (mass (massaa jeni jenisny snyaa 13,6 13,6 g/cm g/cm3). Pada kaki kiri dituangkan air setinggi 20 cm kemudian minyak (massa jenisnya 0,9 g/cm3) tingginya 8 cm. Pada kaki kanan ditambahkan alkohol (massa jenisnya 0,8 g/cm3) sehingga permukaan minyak dan permukaan alkohol sebidang. Berapa beda tinggi Hg pada kedua kaki pipa ? 3. Dala Dalam m pipa pipa U terd terdapa apatt Hg (mass (massaa jeni jenisny snyaa 13,6 13,6 g/cm g/cm3). Pada kaki kiri dituangkan air setinggi 30 cm. Berapa tinggi minyak pada kaki di sebelah kanan harus ditambahkan agar permukaan air dan permukaan minyak sebidang ? (massa jenis minyak 0,9 g/cm3). 4. Kaki kiri kiri dan kanan kanan sebuah pipa pipa U masing-mas masing-masing ing berdiamete berdiameterr 3 cm dan 11/2 cm, mula-mula diisi air raksa (
Hg
= 13,6 g/cm3). Kemudian kaki kiri diisi alkohol (massa jenis 0,8 g/cm3),
kaki kanan diisi bensin (massa jenisnya 0,7 g/cm3) setinggi 2 cm, sehingga tinggi air raksa di kaki kanan naik 1 cm. Hitunglah volume alkohol yang dituangkan. 5. Ke dala dalam m pipa pipa U yang berd berdia iame mete ter r
5 π
cm, mula-mula diisi air raksa (massa jenisnya 13,6
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Drs. Pristiadi Utomo, M.Pd.
Gambar: Archimedes
Ada tiga keadaan benda berada dalam zat cair antara lain sebagai berikut.
1) Benda tenggelam di dalam zat cair.
Berat zat cair yang dipindahkan = mc . g =
c
. Vc . g
Karena Volume Volume zat cair yang dipindahkan d ipindahkan = Volume Volume benda, maka : =
c
. V b . g
Gaya keatas yang dialami benda tersebut besarnya b esarnya : FA =
c
. V b . g
Dimana, Rapat massa benda FA = Gaya ke atas Rapat massa zat cair V b = Volume ume benda c W Berat benda di udara Vc = Volum olumee zat zat cair cair yang ang wc B e r a t s em u dipindahkan (berat benda di dalam zat cair). Benda tenggelam maka : FA < W b
FA
w
= = = =
c
. V b . g <
b
<
b
c
Selisih antara w dan FA disebut Berat Semu (wc)
. V b . g
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Drs. Pristiadi Utomo, M.Pd. 3) Benda terapung di dalam zat cair.
Misalkan sepotong gabus ditahan pada dasar bejana berisi zat cair, setelah dilepas, gabus tersebut akan naik ke permukaan zat cair (terapung) karena : FA > w c
. V b . g >
b
. V b . g c
>
b
Selisih antara w dan FA disebut gaya naik (Fn). Fn = FA - w Benda terapung tentunya dalam keadaan setimbang, sehingga berlaku : FA = W c
Vu
. Vc . g =
b
. V b . g
FA
= Gaya Gaya ke atas atas yang yang dial dialami ami oleh oleh bagian bagian benda benda yang yang
Vu Vc V b
tercelup di dalam zat cair. = Volume olume benda benda yang yang berada berada diper dipermuk mukaan aan zat cair cair.. = Volume olume benda benda yang yang terce tercelup lup di dala dalam m zat cair cair.. = Vu + Vc
V
FA =
c
. Vc . g
Benda terapung yang tepat diam diberlakukan diberlakukan keseimbangan keseimbangan benda yang mana resultan gaya pada benda sama dengan nol. Maka berlaku
FA = w
ρc.Vc.g. = ρ b.V b.g. Vc =
ρ b ρ c
V b
Karena V b
= Vu + Vc
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Drs. Pristiadi Utomo, M.Pd. Dengan Keterangan : FA = Gaya aya ke ke atas atas (N) F = Gaya naik (N) = Mass Massaa jeni jeniss gas gas peng pengis isii balon balon (kg (kg/m /m3) gas = Mass Massaa jeni jeniss uda udara ra = 1,3 1,3 kg/m kg/m3 ud w = Berat (N) V = Volume (m3) Kerja Kelompok Kerjakan soal-soal berikut bersama kelompokmu!
1. Sepotong Sepotong logam beratny beratnyaa di udara 4 N, tetapi tetapi beratnya beratnya tinggal tinggal 2,5 N bila dibenamka dibenamkan n dalam zat cair. Berapakah gaya tekan ke atas yang diderita benda? 2. Sebuah Sebuah silinde silinderr aluminiu aluminium m pejal mempuny mempunyai ai massa massa jenis jenis 2700 kg/m3, massanya 77 gram. Berat aluminium itu tinggal 450 N bila dibenamkan dalam minyak tanah Berapa massa jenis minyak tanah? 3. Sebuah benda benda terapung terapung di atas atas minyak minyak yang mempunyai mempunyai massa massa jenis jenis 0,9 g/cm g/cm3. Tinggi benda tersebut adalah 20 cm, sedangkan tinggi benda yang tidak tercelup adalah 2 cm. berapa massa jenis benda tersebut? 4. Sepotong Sepotong gabus terapung terapung di atas air air dengan ¼ bagian bagian terendam. terendam. Jika berat berat jenis air air adalah 1 grf/cm3, hitunglah berat jenis gabus! 5. Suatu Suatu benda dicelu dicelupkan pkan dalam dalam zat cair yang yang massa massa jenisnya jenisnya 1 gr/cm3, ternyata mendapat gaya ke atas sebesar 40 dyne. Bila massa jenis benda dua kali berat jenis air, berapakah volume benda itu? 6. Batang besi besi dalam air berat berat semunya semunya 372 N. Berapa berat berat semu besi tersebut tersebut dalam dalam cairan cairan yang densitasnya 0,75 g/cm3 jika berat besi 472 N? 7. Suatu Suatu gelas gelas beratny beratnyaa 25 N di udara, udara, 15 N di air, air, dan 7 N di dalam asam asam beleran belerang, g, hitung hitung massa jenis asam belerang!
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Drs. Pristiadi Utomo, M.Pd. 13. Sebuah kubus dari gabus dibebani dengan sepotong sepotong logam sehingga melayang dalam aseton. Jika massa logam 77 gram, massa jenis gabus 0,24 g/cm3, massa jenis logam 8,8 g/cm3, massa jenis aseton 0,8 g/cm3. Tentukan rusuk kubus! 14. Sebongkah es (massa jenisnya 0,9 g/cm3) terapung pada air laut (massa jenisnya 1,03 g/cm3). Jika es yang timbul di permukaan air laut 7,8 dm3. Hitunglah volume es! 15. Massa Massa jenis es 917 kg/m3. Berapa bagian es terletak di permukaan air? 16. Sebatang Sebatang kayu yang massa jenisnya jenisnya 0,6 g/cm3 terapung di dalam air. Jika bagian kayu yang ada di atas permukaan air 0,2 m3, tentukan volume kayu seluruhnya! 17. Sebuah Sebuah kubus kubus dari dari kayu kayu (massa (massa jenisny jenisnyaa 0,8 g/cm3), Mula-mula dibenamkan ke dalam bejana bejana kemudian kemudian dilepas dilepas sehingga naik ke permukaan permukaan gliserin (massa jenisnya jenisnya 1,25 g/cm3) dan ternyata 200 cm3 dari kayu tersebut berada di permukaan gliserin. Tentukan : a. gaya ke atas kayu kayu pada pada saat saat masih masih berada berada seluruhny seluruhnyaa dalam gliserin gliserin b. gaya aya na naik c. gaya gaya ke atas atas sete setelah lah benda benda setim setimban bang g d. Rusuk kub kubu us! 18. Sebuah Sebuah balon udara volumeny volumenyaa 400 m3, mengalami gaya naik 2200 N. Tentukan gaya ke atas dan berat total balon (g = 10 m/det2). 19. Sebu Sebuah ah balo balon n udar udaraa bervol bervolume ume 20 m3. Beri Berisi si H2 ( mass massaa
jeni jeniss 0,09 0,09 gr/ gr/dm dm3) bera beratt
perlengkapannya 100 N. Tentukan berat beban yang dapat diangkut!
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Drs. Pristiadi Utomo, M.Pd. Sewa Sewakt ktu u kele kelere reng ng dija dijatu tuhka hkan n ke dalam dalam beja bejana na kaca kaca yang yang beri berisi si cair cairan an yang yang hend hendak ak ditentukan koefisien viskositasnya, oleh gaya beratnya, kelereng akan semakin cepat jatuhnya. Tetapi etapi sesuai sesuai dengan dengan rumus rumus Stokes Stokes,, makin makin cepat cepat gerakan gerakanny nya, a, makin makin besar besar gaya gaya geseka gesekanny nnyaa sehingga akhirnya gaya berat itu tepat seimbang dengan gaya gesekan dan jatuhnya kelerengpun dengan kecepatan tetap sebesar v sehingga berlaku persamaan: w
=
Fs
m. g
=
6 π r η r η v
Akan tetapi sebenarnya pada kelereng juga bekerja gaya ke atas Archimedes sebesar berat cairan yang dipindahkan, yaitu sebesar: FA = ρc g V = ρc g
4 3
Fs
FA
3
π r
dengan V adalah volum kelereng dan ρc adalah massa jenis cairan. Dengan menuliskan: m = ρ b V = ρ b .
4 3
π r 3
w
dengan ρ b adalah massa jenis bahan pembuat kelereng, persamaan tersebut dapat ditulis menjadi: w = Fs + FA w − FA =Fs
4
π r 3 ρ b. g –
4
π r 3 ρ g = 6 π r η r η v
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Drs. Pristiadi Utomo, M.Pd. 2. Dian Diangg ggap ap ber bergera gerak k tanp tanpaa gese geseka kan n wala walaup upun un ada ada gera geraka kan n mate materi ri (tid (tidak ak memp mempun uny yai kekentalan) 3. Aliran Aliran fluida fluida adalah aliran aliran stasioner stasioner,, yaitu yaitu kecepatan kecepatan dan arah arah gerak gerak partikel partikel fluida fluida yang melalui suatu titik tertentu tetap. Jadi partikel yang datang kemudian di satu titik akan mengikuti jejak partikel-partikel lain yang lewat terdahulu.
1. Debit Fluida mengalir dengan kecepatan tertentu, misalnya v meter per detik. Penampang tabung alir berpenampang A, maka yang dimaksud dengan debit fluida adalah volume fluida yang mengalir persatuan waktu melalui suatu pipa dengan luas penampang A dan dengan kecepatan v. Q=
V t
Karena V = A.s maka persamaan debit menjadi : Q =
A.s t
dan
v=
s t
maka Q = A . v
2. Persamaan Kontinuitas Perhatikan fluida yang mengalir dalam sebuah pipa yang mempunyai ukuran penampang berbeda.
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Drs. Pristiadi Utomo, M.Pd.
Partikel – partikel yang semula di a, dalam waktu ∆t detik berpindah di b, demikian pula partikel yang semula di c berpindah di d. Apabila ∆t sangat kecil, maka jarak a-b sangat kecil, sehingga luas penampang di a dan b boleh dianggap sama, yaitu A1. Demikian pula jarak c-d sangat kecil, sehingga luas penampang di c dan di d dapat dianggap sama, yaitu A2. Banyaknya fluida yang masuk ke tabung alir dalam waktu ∆t detik adalah : ρ.A1.v1. ∆t dan dalam waktu ρ. waktu yang sama sejumlah sejumlah fluida meninggalka meninggalkan n tabung alir sebanyak sebanyak ρ A2.v2. ∆t. Jumlah ini tentulah sama dengan jumlah jumlah fluida yang masuk ke tabung alir sehingga : ρ.A1.v1. ∆t = ρ.A2.v2. ∆t
Jadi :
A1.v1 = A2.v2
Persamaan tersebut dinamakan persamaan Kontinuitas A.v yang yang merupak merupakan an debit debit fluida fluida sepanj sepanjang ang tabung tabung alir alir selalu selalu konstan konstan (tetap (tetap sama sama nilain nilainya) ya),, walaupu walaupun n A dan v masing masing-ma -masin sing g berbed berbedaa dai tempat tempat yang satu satu ke tempat tempat yang yang lain. lain. Maka Maka disimpulkan : Q = A1.v1 = A2.v2 = konstan Bila Bila A1v1.∆t.ρ t.ρ = m maka selama waktu ∆t massa sebanyak sebanyak m ini dianggap telah berpindah dari A1 ke A2. Kecepatannya berubah dari v1 menjadi v2. Bila ketinggiannya juga berubah dari h1
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Drs. Pristiadi Utomo, M.Pd. Hukum Bernoulli merupakan persamaan pokok fluida dinamik dengan arus streamline. Di sini berlaku hubungan antara tekanan, kecepatan alir dan tinggi tempat dalam satu garis lurus. Hubungan tersebut dapat dijelaskan sebagai berikut : Perhatikan gambar tabung alir a-c pada gambar pipa alir. Jika tekanan p1 ke kanan pada penampang A1, dari fluida di sebelah kirinya, maka gaya yang dilakukan terhadap penampang di a adalah p1A1, sedangkan sedangkan penampang di c mendapat gaya dari fluida dikanannya dikanannya sebesar sebesar p2A2, di mana p2 adalah tekanan terhadap penampang di c ke kiri. Dalam waktu ∆t detik dapat dianggap bahwa fluida di penampang a tergeser sejauh v1 ∆t dan fluida di penampang c tergeser sejauh v2 ∆t ke kanan. Jadi usaha yang dilakukan terhadap a adalah : p1A1v1 ∆t sedangkan usaha yang dilakukan pada c sebesar : - p2A2v2 ∆t Jadi usaha total yang dilakukan gaya-gaya tersebut besarnya : Wtot
=
W
=
W
=
W
=
(p1 A1 v1 - p2 A2 v2) ∆t p1A1 v1∆t – p2 A2 v2∆t p1 ρ p1 ρ
ρ A1 v1∆t −
m−
p 2 ρ
p 2 ρ
ρ A2 v2 ∆t
m
Dalam waktu ∆t detik fluida dalam tabung alir a-b bergeser ke c-d dan mendapat tambahan energi sebesar :
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Drs. Pristiadi Utomo, M.Pd. Contoh penggunaan Hukum Bernoulli : a) Semp emprotan b) Sayap Sayap pesa pesawa watt terb terban ang g c) Venturi enturi meter = alat yang yang digunakan digunakan untuk menentukan menentukan kecepatan kecepatan aliran aliran zat cair. cair. d) Pipa pi pitot e) Tower air
3. Viskositas (Kekentalan) Viskositas / kekentalan dapat dibayangkan sebagai gesekan antara satu bagian dengan bagian yang lain dalam fluida. F = ηA
V L
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Drs. Pristiadi Utomo, M.Pd.
Perhatikan skema semprotan berikut ini.
V
v1
Obat nyamuk cair mula-mula diam sehingga v1 = 0, sedangkan udara bergerak dengan kecepatan v2 karena didorong oleh pengisap. Tekanan p1 sama dengan p2 yaitu tekanan udara luar. Sehingga persamaan bernoulli menjadi: p1 + ½ ρ v12 + ρ g h1
=
p2 + ½ ρ v22 + ρ g h2
0 + ρ g h1
=
½ ρ v22 + ρ g h2
g h1
=
½ v22 + g h2
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Drs. Pristiadi Utomo, M.Pd.
Dari Dari persamaa persamaan n di atas dapat dapat dilihat dilihat bahwa bahwa v2 > v1 kita dapatkan p1 > p2 untuk luas penampang penampang sayap
F1 = p1 A dan F2 = p2 A dan kita dapatkan bahwa F1 > F2. Beda gaya pada
bagian bawah dan bagian atas (F1 – F2) menghasilkan gaya angkat pada pesawat terbang. Jadi, gaya angkat pesawat terbang dirumuskan sebagai : F1 – F2 = ½ ρ A(v22 – v12) Dengan ρ = massa jenis udara (kg/m3) 6. Venturimeter
Venturimeter adalah alat yang digunakan untuk menentukan kecepatan aliran zat cair. Dengan memasukkan venturimeter ke dalam aliran fluida kecepatan aliran fluida dapat dihitung menggunakan persamaan Bernoulli berdasarkan selisih ketinggian air atau selisih ketinggian raksa.
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Drs. Pristiadi Utomo, M.Pd. p1 + ½ ρ v12
=
p2 + ½ ρ v22
p1 − p2
=
½ ρ v22 − ½ ρ v12
∆
p
=
½ ρ (v22 − v12)
ρgh
=
½ ρ (v22 − v12)
gh
=
½ (v22 − v12)
Dengan menggunakan persamaan persamaan kontinuitas A1.v1 = A2.v2 untuk mendapatkan hubungan antara v2 dan v1, maka v1 dapat dihitung.
b. Ventu enturime rimeter ter dengan dengan Mano Manomete meterr
v1 P
v2 P
ρ
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Drs. Pristiadi Utomo, M.Pd. Kecepatan aliran fluida sebagai berikut.
v1 =
2ρ r gh ρ
7. Tower Air
Sebuah Sebuah bak penampu penampungan ngan air sebagi sebagi tower tower dengan dengan kran kran air yang dapat dapat memancarkan air melalui sebuah lubang baik di dasar maupun di ketinggian tertentu dapat di
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Drs. Pristiadi Utomo, M.Pd. Persamaan ini tidak lain adalah rumus gerak jatuh bebas. Sedangkan jarak jatuhnya fluida diukur dari titik proyeksi lubang air dihitung menggunakan persamaan gerak lurus beraturan. X = v2 . t
sedangkan waktu jatuh fluida
h = ½ g t2
Contoh: 1. Sebuah Sebuah tangki tangki terbuka terbuka berisi berisi air setinggi setinggi H. Pada jarak jarak h dari dari permukaan permukaan air dibuat dibuat suatu suatu lubang kecil, sehingga air memancar dari lubang itu. Berapa jauh air yang keluar dari tangki mengenai tanah ? Penyelesaian:
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Drs. Pristiadi Utomo, M.Pd.
A1 v1
h2
A2 v2 h
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Drs. Pristiadi Utomo, M.Pd.
Penyelesaian:
a. p1
= 8000 N/m2 + Bar
H
=1m
AA
= 20 cm2 = 0,002 m2
AB
= 10 cm2 = 0,001 m2
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Drs. Pristiadi Utomo, M.Pd. D
= 2 cm
h
= 300 cm
η
= 0,6 π cm2
v
=
2g h
=
2 980 300 = 76,68 cm/det
Q
= 76,68 0,6 π = 46,73 10-3π lt/det = 276 π 10-3 lt/mnt
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Drs. Pristiadi Utomo, M.Pd. v1 =
25,3 = 5,3
v1 = 5 cm/det
Rangkuman 1. Tekanan adalah gaya tiap tiap satuan satuan luas luas penampang, penampang, dirumuskan dirumuskan p =
F A
2. Tekanan hidrostat hidrostatis is yaitu yaitu tekanan pada kedalaman kedalaman tertentu tertentu zat cair yang tidak tidak mengalir, mengalir,
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Drs. Pristiadi Utomo, M.Pd.
Soal- soal Pilihan Ganda
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Drs. Pristiadi Utomo, M.Pd. d. 1,5 m/det e
0 75 m/det
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Drs. Pristiadi Utomo, M.Pd. d. 80 liter e
120 liter
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Drs. Pristiadi Utomo, M.Pd. 14.
Sebu Sebuah ah pipa pipa kapile kapilerr dimas dimasukk ukkan an trega tregak k lurus lurus dalam dalam alko alkohol hol (ma (mass ssaa jenis jenis = 0,8 g/cm g/cm3). Bila tegangan permukaan permukaan alkohol = 22 dyne/cm, berapa naiknya permukaan alkohol di
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Drs. Pristiadi Utomo, M.Pd. c. 827 dyne d
872 dyne
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