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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 $Fg = m⋅g$ 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.

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 $daljša ročica pomeni večji navor - zato je lažje odpreti vrata daleč od tečajev$ .

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

Pogoji za statično ravnotežje

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

Ravnotežje sil: ∑F = 0, kar pomeni ∑Fx = 0 in ∑Fy = 0
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 $r = 0$ .

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

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.

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 $deluje na sredini, rn = 2,5 m$
md = 80 kg → Fgd = 784,8 N $deluje na rd = 2 m od A$
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.

Primer 2: Nosilec na dveh podporah - del 2

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

(FB ⋅ L) - $Fgn ⋅ L/2$ - (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 $716,13 + 559,17 = 1275,3 N ✓$ .

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 $gor +, dol -$ 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.

Povzetek za ponavljanje

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

∑F = 0 $vsota sil je nič - ni translacije$
∑M = 0 $vsota navorov je nič - ni rotacije$

Osnovne definicije:

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

Postopek reševanja:

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

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

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Tehnika

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Updated Apr 1, 2026

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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.

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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.

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.

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Seštevanje sil in ravnotežje

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 $daljša ročica pomeni večji navor - zato je lažje odpreti vrata daleč od tečajev$ .

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

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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:

Ravnotežje sil: ∑F = 0, kar pomeni ∑Fx = 0 in ∑Fy = 0
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 $r = 0$ .

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

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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.

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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 $deluje na sredini, rn = 2,5 m$
md = 80 kg → Fgd = 784,8 N $deluje na rd = 2 m od A$
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.

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Primer 2: Nosilec na dveh podporah - del 2

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

(FB ⋅ L) - $Fgn ⋅ L/2$ - (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 $716,13 + 559,17 = 1275,3 N ✓$ .

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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 $gor +, dol -$ 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.

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Povzetek za ponavljanje

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

∑F = 0 $vsota sil je nič - ni translacije$
∑M = 0 $vsota navorov je nič - ni rotacije$

Osnovne definicije:

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

Postopek reševanja:

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

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

We thought you’d never ask...

What is the Knowunity AI companion?

Where can I download the Knowunity app?

You can download the app in the Google Play Store and in the Apple App Store.

Is Knowunity really free of charge?

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Not only did it help me find the answer but it also showed me alternative ways to solve it. I was horrible in math and science but now I have an a in both subjects. Thanks for the help🤍🤍

David K

iOS user

Sudenaz Ocak

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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.

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iOS user

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

Aubrey

iOS user

Best app if you're in Highschool or Junior high. I have been using this app for 2 school years and it's the best, it's good if you don't have anyone to help you with school work.😋🩷🎀

Marco B

iOS user

Elisha

iOS user

Paul T

iOS user

Sile in ravnotežje teles: Razumevanje osnovnih pojmov

Sile, navori in ravnotežje - osnove

Seštevanje sil in ravnotežje

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Primer 1: Gugalnica v ravnotežju

Primer 2: Nosilec na dveh podporah - del 1

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Potence in koreni

Algebrski ulomki

Kotne funkcije v pravokotnem trikotniku in na enotski krožnici

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Geografija; Slovenija

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Sile in ravnotežje teles: Razumevanje osnovnih pojmov

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Sile, navori in ravnotežje - osnove

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Seštevanje sil in ravnotežje

Sign up to see the contentIt's free!

Pogoji za statično ravnotežje

Sign up to see the contentIt's free!

Primer 1: Gugalnica v ravnotežju

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Primer 2: Nosilec na dveh podporah - del 1

Sign up to see the contentIt's free!

Primer 2: Nosilec na dveh podporah - del 2

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Nasveti za reševanje in pogoste napake

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Povzetek za ponavljanje

We thought you’d never ask...

What is the Knowunity AI companion?

Where can I download the Knowunity app?

Is Knowunity really free of charge?

Most popular content in Tehnika

Strojna oprema računalnika

Računalniška omrežja

Inovativnost in podjetništvo v tehniki

Most popular content

Potence in koreni

Časi (ponovitev in poglobljeno)

Potence in koreni

Algebrski ulomki

Kotne funkcije v pravokotnem trikotniku in na enotski krožnici

Kombinatorika

Geografija; Slovenija

Biologija; uvod v biologijo in genetika

Potence

Can't find what you're looking for? Explore other subjects.

Students love us — and so will you.

4.6/5

4.7/5