konkurentne komplanarne sile
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Konkurentne komplanarne sile
2. dio
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1. Zadavanje vektora sile
2. Projekcije sile na koordinatne osi
3. Rastavljanje sile na komponente
4. Rezultanta komplanarnih konkurentnih sila
5. Ravnoteža komplanarnih konkurentnih sila
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Vektor sile
• Hvatište sile
• Pravac (smjer) djelovanja• Intenzitet
1. Zadavanje vektora sile:
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Kvadranti (oktanti)
(N) k 2 j3i4F
:silaProstornab)
−+−=
(N) j3i4F
:silaaKomplanarna)
+−=
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2. Projekcije sile na koordinatne osi
y x F F F += jF iF F y x
⋅+⋅=
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2. Projekcije sile na koordinatne osi
F
F x
=α cos cos⋅= F F x
F
F y
=α sin sin⋅= F F y
x
y
F
F tg =α
y x F F F +=
jF iF F y x ⋅+⋅=
22
y x F F F +=
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2. Projekcije sile na koordinatne osi
jF iF F y x ⋅+⋅=
( ) ( )
( ) cossin
F sincosF
F sinF cosF
F F F y x
122
2222
222
222
=α+α
=α+α
=α⋅+α⋅
=+
Pravokutni trokut:
Osnovna jednadžba trigonometrije
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3. Rastavljanje sile na komponente
Grafiki prikaz [Mjerilo 1 cm :: n (kN)]
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Projekcije su “specijalni sluaj komponenata”
- usobno okomite
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Konkurentne sile
Pravci sila sijeku se u jednoj toci!
• Komplanarne konkurentne sile
• Prostorne konkurentne sile
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Konkurentne sile
Sustavi sila n>1
4. Rezultanta konkurentnih sila
5. Ravnoteža konkurentnih sila
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Sustav sila (n>1)
n=4
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4. Rezultanta konkurentnih silaa) Analitiko rješenje
b) Grafiko rješenje
b1) parcijalni paralelogramib2) poligon sila
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a) Analitiko rješenje n=2
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4.a) Rezultanta konkurentnih sila
x x Rx F F F 21 +=
y y Ry F F F 21 +=
22
Ry Rx R F F F +=
Rx
Ry
RF
F tg =α
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Rezultanta konkurentnog
komplanarnog sustava silaZa sustav od tri sile:
ili openito: 1.
2.
=
=
n
i
ix Rx F F 1
=
=
n
i
iy Ry F F 1
x x x Rx F F F F 321 ++=
y y y Ry F F F F 321 ++=
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Rezultanta konkurentnihkomplanarnih sila
Ponavljanje:
Projekcija rezultante u os x (y) jednaka je
sumi projekcija komponenata na os x (y).
=
=
=
=
n
1iiyRy
n
1iixRx
FF .2
FF .1
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4.b) Grafiko rješenje
• Parcijalni paralelogrami sila
• Poligon sila
Mjerilo na pr. 1 cm :: 5 kN
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b1) Grafiko rješenje• Parcijalni paralelogrami sila
Mjerilo 1 cm :: n kN
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b2) Grafiko rješenjen = 4
Poligon sila
Poetak rezultante u poetku prve sile šiljak rezultante u
šiljku zadnje sile.
Mjerilo 1 cm :: n kN
4321 F F F F F R +++=
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Sustav sila n = 5
Poligon
sila
Poligon sila
Mjerilo 1 cm :: n kN
54321 F F F F F F R ++++=
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5. Ravnoteža konkurentnih sila
a) Analitiko rješenje:
b) Grafiko rješenje
0F .2 0F .1n
1iiy
n
1iix ==
==
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5. a) Ravnoteža sila• Analitiko rješenje
na pr. za sustav od tri sile
0G10sinN20sinN .2
010cosN20cosN .1
0FFF .2
0FFF .1
BA
BA
y3y2y1
x3x2x1
=−⋅+⋅
=⋅−⋅
=++
=++
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5. a) Ravnoteža konkurentnihkomplanarnih sila
Ponavljanje:Suma projekcija svih sila u os x (y) jednaka
je nuli.
0F .2
0F .1
y
x
=
=
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5. b) Ravnoteža sila• Grafiko rješenje
Poligon sila mora biti zatvoren !
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Poligon sila je zatvoren Mjerilo 1 cm :: n kN
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Vježba – I. program
1. Projekcije sila na koordinatne osi
2. Rezultanta konkurentnih komplanarnih sila FR
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3. Rastavljanje sile F na komponente FA i FB
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4. Ravnotežna sila F4 = ?