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Updated Apr 1, 2026

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Sile in ravnotežje teles: Razumevanje osnovnih pojmov

Razumevanje sil, navorov in ravnotežja je ključno za fiziko in... Show more

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# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Sile, navori in ravnotežje - osnove

Statična ravnotežja srečuješ povsod - od gugalnic do premostitvenih konstrukcij. Gre za analizo pogojev, ki poskrbijo, da telo miruje kljub delovanju sil.

Sila (F) je vektorska količina, ki povzroča ali poskuša povzročiti spremembo gibanja. Določena je z velikostjo, smerjo in prijemališčem. Enota je Newton [N], kjer 1N = 1 kg⋅m/s².

Težišče (T) je točka, kjer si predstavljamo, da je zbrana vsa teža telesa. Sila teže Fg=mgFg = m⋅g vedno deluje navpično navzdol skozi težišče. Pri simetričnih homogenih telesih je v geometrijskem središču.

Navor (M) predstavlja vrtilni učinek sile. Nastane, ko sila deluje izven osi vrtenja. Formula: M = F⊥⋅r ali M = F⋅r⋅sin(α). Enota je Newton meter [Nm].

Pomembno: Pravokotna razdalja od osi vrtenja do nosilke sile je ključna za pravilno računanje navora.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Seštevanje sil in ravnotežje

Ker so sile vektorji, jih ne moreš preprosto seštevati kot števila. Pri kollinearnih silah (na isti premici) jih seštevamo ali odštevamo glede na smer, pri nekollinearnih pa uporabljamo paralelogramsko pravilo.

Za ravnotežje sil mora biti vsota vseh sil enaka nič: ∑F = 0. To pomeni, da se vsi vplivi med seboj izničijo. Razdelimo na komponente:

  • ∑Fx = 0 (vsota sil v smeri x)
  • ∑Fy = 0 (vsota sil v smeri y)

Ko je vsota sil enaka nič, telo ne pospešuje. Če je mirovalo, ostane pri miru.

Navor je odvisen od dveh dejavnikov: velikosti sile (močneje kot potisneš, večji je navor) in ročice daljsˇarocˇicapomenivecˇjinavorzatojelazˇjeodpretivratadalecˇodtecˇajevdaljša ročica pomeni večji navor - zato je lažje odpreti vrata daleč od tečajev.

Nasvet: Za smer navora uporabljaj dogovor: vrtenje v nasprotni smeri urinega kazalca je pozitivno (+), v smeri urinega kazalca negativno (-).

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Pogoji za statično ravnotežje

Telo je v statičnem ravnotežju, če miruje in se ne vrti. Hkrati morata biti izpolnjena oba pogoja:

  1. Ravnotežje sil: ∑F = 0, kar pomeni ∑Fx = 0 in ∑Fy = 0
  2. Ravnotežje navorov: ∑M = 0 okoli poljubne točke

Za ravnotežje navorov mora biti vsota vseh navorov okoli katerekoli izbrane točke enaka nič. To pomeni, da se vsi vrtilni učinki med seboj izničijo in telo se ne začne vrteti.

Izbira vrtišča je pomembna strategija - pametna izbira (npr. tam, kjer deluje neznana sila) lahko močno poenostavi računanje, ker postane navor te sile enak nič r=0r = 0.

Ključna misel: Oba pogoja morata biti izpolnjena hkrati - ni dovolj, če je izpolnjen le eden od njih.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Primer 1: Gugalnica v ravnotežju

Dva otroka sedita na gugalnici dolžine 4 m. Prvi otrok (30 kg) sedi 1,5 m od sredine. Kjer mora sedeti drugi otrok (25 kg)?

Podatki: m₁ = 30 kg → F₁ = 294,3 N, r₁ = 1,5 m; m₂ = 25 kg → F₂ = 245,3 N, r₂ = ?

Reševanje: Uporabimo pogoj za ravnotežje navorov. Vrtišče je na sredini gugalnice.

∑M = 0 → M₂ = M₁ F₂ ⋅ r₂ = F₁ ⋅ r₁

Izrazimo r₂: r₂ = (F₁ ⋅ r₁)/F₂ = (294,3 N ⋅ 1,5 m)/(245,3 N) ≈ 1,8 m

Odgovor: Drugi otrok mora sedeti 1,8 m od sredine na nasprotni strani.

Preverimo logiko: Lažji otrok mora sedeti dlje od vrtišča, kar se ujema z našim rezultatom.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Primer 2: Nosilec na dveh podporah - del 1

Homogen nosilec (50 kg, 5 m) leži na podporah A in B na konceh. Na 2 m od podpore A stoji delavec (80 kg). Izračunaj reakcijski sili.

Podatki:

  • mn = 50 kg → Fgn = 490,5 N delujenasredini,rn=2,5mdeluje na sredini, rn = 2,5 m
  • md = 80 kg → Fgd = 784,8 N delujenard=2modAdeluje na rd = 2 m od A
  • Neznanki: FA, FB

Prvi korak - ravnotežje sil: ∑Fy = 0 → FA + FB - Fgn - Fgd = 0 FA + FB = 490,5 N + 784,8 N = 1275,3 N

Imamo eno enačbo in dve neznanki. Potrebujemo še ravnotežje navorov.

Strategija: Izberi vrtišče v podpori A, ker bo navor od FA enak nič in poenostavil računanje.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Primer 2: Nosilec na dveh podporah - del 2

Drugi korak - ravnotežje navorov okoli točke A: ∑MA = 0

(FB ⋅ L) - FgnL/2Fgn ⋅ L/2 - (Fgd ⋅ 2 m) = 0

Vstavimo vrednosti: FB ⋅ 5 = (490,5 ⋅ 2,5) + (784,8 ⋅ 2) FB ⋅ 5 = 1226,25 + 1569,6 = 2795,85 FB = 559,17 N

Tretji korak - izračun FA: FA = 1275,3 - 559,17 = 716,13 N

Odgovor: Sila v podpori A je 716,13 N, v podpori B pa 559,17 N. Logično je sila v A večja, ker delavec stoji bližje tej podpori.

Preverimo: Vsota obeh reakcijskih sil mora biti enaka skupni teži 716,13+559,17=1275,3N716,13 + 559,17 = 1275,3 N ✓.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Nasveti za reševanje in pogoste napake

Ključni nasveti za test:

  • Skica je vse: Vedno najprej nariši skico in vriši vse sile. To je 90% rešitve.
  • Predznaki: Dosledno uporabljaj predznake za smeri gor+,dolgor +, dol - in navore.
  • Pametna izbira vrtišča: Izberi točko, kjer deluje neznana sila - njen navor bo enak nič.
  • Enote: Pazi na enote! Sile v N, razdalje v m, navori v Nm.

Pogoste napake:

  • Seštevanje sil kot skalarjev → Sile so vektorji, seštevaj po komponentah
  • Pozabiti na težo nosilca → Če ima telo maso, vriši silo teže v težišče
  • Napačna ročica → Ročica je vedno pravokotna razdalja od vrtišča do nosilke sile

Ne zamenjuj sile in navora - sila povzroča premikanje, navor pa vrtenje.

Zlato pravilo: Skica + pravilni predznaki + pametna izbira vrtišča = uspešno rešena naloga.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Povzetek za ponavljanje

Statično ravnotežje pomeni, da telo miruje in se ne vrti. Potrebna sta dva pogoja:

  • ∑F = 0 vsotasiljenicˇnitranslacijevsota sil je nič - ni translacije
  • ∑M = 0 vsotanavorovjenicˇnirotacijevsota navorov je nič - ni rotacije

Osnovne definicije:

  • Sila (F): potisk ali vlek, enota Newton [N]
  • Navor (M): vrtilni učinek sile, M = F ⋅ r, enota Newton meter [Nm]

Postopek reševanja:

  1. Nariši skico in prostoležni diagram (vse sile)
  2. Postavi koordinatni sistem
  3. Zapiši enačbe za ravnotežje sil Fx=0,Fy=0∑Fx = 0, ∑Fy = 0
  4. Izberi vrtišče in zapiši enačbo za ravnotežje navorov M=0∑M = 0
  5. Reši sistem enačb

Za test: Obvladaj osnovne pojme, znaj narediti skico in sistematično pristopi k reševanju. Vaja dela mojstra!



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The app is very easy to use and well designed. I have found everything I was looking for so far and have been able to learn a lot from the presentations! I will definitely use the app for a class assignment! And of course it also helps a lot as an inspiration.

Stefan S

iOS user

This app is really great. There are so many study notes and help [...]. My problem subject is French, for example, and the app has so many options for help. Thanks to this app, I have improved my French. I would recommend it to anyone.

Samantha Klich

Android user

Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

Anna

iOS user

Best app on earth! no words because it’s too good

Thomas R

iOS user

Just amazing. Let's me revise 10x better, this app is a quick 10/10. I highly recommend it to anyone. I can watch and search for notes. I can save them in the subject folder. I can revise it any time when I come back. If you haven't tried this app, you're really missing out.

Basil

Android user

This app has made me feel so much more confident in my exam prep, not only through boosting my own self confidence through the features that allow you to connect with others and feel less alone, but also through the way the app itself is centred around making you feel better. It is easy to navigate, fun to use, and helpful to anyone struggling in absolutely any way.

David K

iOS user

The app's just great! All I have to do is enter the topic in the search bar and I get the response real fast. I don't have to watch 10 YouTube videos to understand something, so I'm saving my time. Highly recommended!

Sudenaz Ocak

Android user

In school I was really bad at maths but thanks to the app, I am doing better now. I am so grateful that you made the app.

Greenlight Bonnie

Android user

very reliable app to help and grow your ideas of Maths, English and other related topics in your works. please use this app if your struggling in areas, this app is key for that. wish I'd of done a review before. and it's also free so don't worry about that.

Rohan U

Android user

I know a lot of apps use fake accounts to boost their reviews but this app deserves it all. Originally I was getting 4 in my English exams and this time I got a grade 7. I didn’t even know about this app three days until the exam and it has helped A LOT. Please actually trust me and use it as I’m sure you too will see developments.

Xander S

iOS user

THE QUIZES AND FLASHCARDS ARE SO USEFUL AND I LOVE Knowunity AI. IT ALSO IS LITREALLY LIKE CHATGPT BUT SMARTER!! HELPED ME WITH MY MASCARA PROBLEMS TOO!! AS WELL AS MY REAL SUBJECTS ! DUHHH 😍😁😲🤑💗✨🎀😮

Elisha

iOS user

This apps acc the goat. I find revision so boring but this app makes it so easy to organize it all and then you can ask the freeeee ai to test yourself so good and you can easily upload your own stuff. highly recommend as someone taking mocks now

Paul T

iOS user

 

Tehnika

17

Updated Apr 1, 2026

8 pages

Sile in ravnotežje teles: Razumevanje osnovnih pojmov

Razumevanje sil, navorov in ravnotežja je ključno za fiziko in vsakodnevno življenje - od gradnje mostov do odpiranja vrat. Ta snov ti razloži, zakaj se stvari ne premikajo ali ne vrtijo, ko so obremenjene.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

Sign up to see the contentIt's free!

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Sile, navori in ravnotežje - osnove

Statična ravnotežja srečuješ povsod - od gugalnic do premostitvenih konstrukcij. Gre za analizo pogojev, ki poskrbijo, da telo miruje kljub delovanju sil.

Sila (F) je vektorska količina, ki povzroča ali poskuša povzročiti spremembo gibanja. Določena je z velikostjo, smerjo in prijemališčem. Enota je Newton [N], kjer 1N = 1 kg⋅m/s².

Težišče (T) je točka, kjer si predstavljamo, da je zbrana vsa teža telesa. Sila teže Fg=mgFg = m⋅g vedno deluje navpično navzdol skozi težišče. Pri simetričnih homogenih telesih je v geometrijskem središču.

Navor (M) predstavlja vrtilni učinek sile. Nastane, ko sila deluje izven osi vrtenja. Formula: M = F⊥⋅r ali M = F⋅r⋅sin(α). Enota je Newton meter [Nm].

Pomembno: Pravokotna razdalja od osi vrtenja do nosilke sile je ključna za pravilno računanje navora.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

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Seštevanje sil in ravnotežje

Ker so sile vektorji, jih ne moreš preprosto seštevati kot števila. Pri kollinearnih silah (na isti premici) jih seštevamo ali odštevamo glede na smer, pri nekollinearnih pa uporabljamo paralelogramsko pravilo.

Za ravnotežje sil mora biti vsota vseh sil enaka nič: ∑F = 0. To pomeni, da se vsi vplivi med seboj izničijo. Razdelimo na komponente:

  • ∑Fx = 0 (vsota sil v smeri x)
  • ∑Fy = 0 (vsota sil v smeri y)

Ko je vsota sil enaka nič, telo ne pospešuje. Če je mirovalo, ostane pri miru.

Navor je odvisen od dveh dejavnikov: velikosti sile (močneje kot potisneš, večji je navor) in ročice daljsˇarocˇicapomenivecˇjinavorzatojelazˇjeodpretivratadalecˇodtecˇajevdaljša ročica pomeni večji navor - zato je lažje odpreti vrata daleč od tečajev.

Nasvet: Za smer navora uporabljaj dogovor: vrtenje v nasprotni smeri urinega kazalca je pozitivno (+), v smeri urinega kazalca negativno (-).

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

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Pogoji za statično ravnotežje

Telo je v statičnem ravnotežju, če miruje in se ne vrti. Hkrati morata biti izpolnjena oba pogoja:

  1. Ravnotežje sil: ∑F = 0, kar pomeni ∑Fx = 0 in ∑Fy = 0
  2. Ravnotežje navorov: ∑M = 0 okoli poljubne točke

Za ravnotežje navorov mora biti vsota vseh navorov okoli katerekoli izbrane točke enaka nič. To pomeni, da se vsi vrtilni učinki med seboj izničijo in telo se ne začne vrteti.

Izbira vrtišča je pomembna strategija - pametna izbira (npr. tam, kjer deluje neznana sila) lahko močno poenostavi računanje, ker postane navor te sile enak nič r=0r = 0.

Ključna misel: Oba pogoja morata biti izpolnjena hkrati - ni dovolj, če je izpolnjen le eden od njih.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

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Primer 1: Gugalnica v ravnotežju

Dva otroka sedita na gugalnici dolžine 4 m. Prvi otrok (30 kg) sedi 1,5 m od sredine. Kjer mora sedeti drugi otrok (25 kg)?

Podatki: m₁ = 30 kg → F₁ = 294,3 N, r₁ = 1,5 m; m₂ = 25 kg → F₂ = 245,3 N, r₂ = ?

Reševanje: Uporabimo pogoj za ravnotežje navorov. Vrtišče je na sredini gugalnice.

∑M = 0 → M₂ = M₁ F₂ ⋅ r₂ = F₁ ⋅ r₁

Izrazimo r₂: r₂ = (F₁ ⋅ r₁)/F₂ = (294,3 N ⋅ 1,5 m)/(245,3 N) ≈ 1,8 m

Odgovor: Drugi otrok mora sedeti 1,8 m od sredine na nasprotni strani.

Preverimo logiko: Lažji otrok mora sedeti dlje od vrtišča, kar se ujema z našim rezultatom.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

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Primer 2: Nosilec na dveh podporah - del 1

Homogen nosilec (50 kg, 5 m) leži na podporah A in B na konceh. Na 2 m od podpore A stoji delavec (80 kg). Izračunaj reakcijski sili.

Podatki:

  • mn = 50 kg → Fgn = 490,5 N delujenasredini,rn=2,5mdeluje na sredini, rn = 2,5 m
  • md = 80 kg → Fgd = 784,8 N delujenard=2modAdeluje na rd = 2 m od A
  • Neznanki: FA, FB

Prvi korak - ravnotežje sil: ∑Fy = 0 → FA + FB - Fgn - Fgd = 0 FA + FB = 490,5 N + 784,8 N = 1275,3 N

Imamo eno enačbo in dve neznanki. Potrebujemo še ravnotežje navorov.

Strategija: Izberi vrtišče v podpori A, ker bo navor od FA enak nič in poenostavil računanje.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

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Primer 2: Nosilec na dveh podporah - del 2

Drugi korak - ravnotežje navorov okoli točke A: ∑MA = 0

(FB ⋅ L) - FgnL/2Fgn ⋅ L/2 - (Fgd ⋅ 2 m) = 0

Vstavimo vrednosti: FB ⋅ 5 = (490,5 ⋅ 2,5) + (784,8 ⋅ 2) FB ⋅ 5 = 1226,25 + 1569,6 = 2795,85 FB = 559,17 N

Tretji korak - izračun FA: FA = 1275,3 - 559,17 = 716,13 N

Odgovor: Sila v podpori A je 716,13 N, v podpori B pa 559,17 N. Logično je sila v A večja, ker delavec stoji bližje tej podpori.

Preverimo: Vsota obeh reakcijskih sil mora biti enaka skupni teži 716,13+559,17=1275,3N716,13 + 559,17 = 1275,3 N ✓.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

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Nasveti za reševanje in pogoste napake

Ključni nasveti za test:

  • Skica je vse: Vedno najprej nariši skico in vriši vse sile. To je 90% rešitve.
  • Predznaki: Dosledno uporabljaj predznake za smeri gor+,dolgor +, dol - in navore.
  • Pametna izbira vrtišča: Izberi točko, kjer deluje neznana sila - njen navor bo enak nič.
  • Enote: Pazi na enote! Sile v N, razdalje v m, navori v Nm.

Pogoste napake:

  • Seštevanje sil kot skalarjev → Sile so vektorji, seštevaj po komponentah
  • Pozabiti na težo nosilca → Če ima telo maso, vriši silo teže v težišče
  • Napačna ročica → Ročica je vedno pravokotna razdalja od vrtišča do nosilke sile

Ne zamenjuj sile in navora - sila povzroča premikanje, navor pa vrtenje.

Zlato pravilo: Skica + pravilni predznaki + pametna izbira vrtišča = uspešno rešena naloga.

# Sile, navori in ravnotežje

Uvod v sile, navore in ravnotežje

To je osnova mehanike. Brez razumevanja ravnotežja ne moremo graditi
mostov

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Povzetek za ponavljanje

Statično ravnotežje pomeni, da telo miruje in se ne vrti. Potrebna sta dva pogoja:

  • ∑F = 0 vsotasiljenicˇnitranslacijevsota sil je nič - ni translacije
  • ∑M = 0 vsotanavorovjenicˇnirotacijevsota navorov je nič - ni rotacije

Osnovne definicije:

  • Sila (F): potisk ali vlek, enota Newton [N]
  • Navor (M): vrtilni učinek sile, M = F ⋅ r, enota Newton meter [Nm]

Postopek reševanja:

  1. Nariši skico in prostoležni diagram (vse sile)
  2. Postavi koordinatni sistem
  3. Zapiši enačbe za ravnotežje sil Fx=0,Fy=0∑Fx = 0, ∑Fy = 0
  4. Izberi vrtišče in zapiši enačbo za ravnotežje navorov M=0∑M = 0
  5. Reši sistem enačb

Za test: Obvladaj osnovne pojme, znaj narediti skico in sistematično pristopi k reševanju. Vaja dela mojstra!

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