1. Moment setrvačnosti

1. Moment setrvačnosti

FS-Pn-P019-Statika_TT MECHANIKA TUHHO TLESA 6. een loh 6.5 lovk na ebku 6.5 Zadn: Dobrovoln hasi Antonn . vystupuje po ebku openm o ze. Do jak vky me vystoupit, ne

mu ebk podklouzne? ebk povaujte za homogenn dlky = 4 m a hmotnosti m1 = 20 kg. Hmotnost hasie je m2 = 100 kg. ebk je open pod hlem = 60 od vodorovnho smru. Souinitel smykovho ten mezi ebkem a podlokou je stejn jako mezi ebkem a zd f = 0,3. 6.5 Vklad k een: Na soustavu hasie na ebku budeme nahlet jako na tuh tleso, kter mus bt v silov a momentov rovnovze.

Pesn podle definice rovnovn polohy tuhho tlesa rozepeme podmnky pro rovnost sil a pro momenty sil. 6.5 Vklad k een: 1. Podmnka rovnovhy ROVNOST SL: Svisl smr: F G1 FG2 FN A FTB

Vodorovn smr: FN B FTA 6.5 Vklad k een: 1. Podmnka rovnovhy ROVNOST SL: Prvn podmnku rovnovhy, kter je vyjdenm vektorovho soutu vech sil (viz obr.) psobcch na soustavu ebk + hasi

FG1 FG 2 FN A FN B FTA FTA 0 pepeme do sloky svisl a vodorovn. 6.5 Vklad k een: 2. Podmnka rovnovhy ROVNOST MOMENT SIL: Dal podmnku zskme z podmnky rovnosti moment sil.

Je jedno, ke ktermu bodu tyto momenty budeme potat, ale je dobr tento bod zvolit tak, aby zskan rovnice byla co nejjednodu. Zde je vhodn volit bod A nebo B, kde jsou psobit nkolika sil, kter se dky nulovmu ramenu v rovnici neprojev. 6.5 Vklad k een: 2. Podmnka rovnovhy ROVNOST MOMENT SIL: Pro uren momentu pouijeme bod A, co je vhodn pro uren

vzdlenosti x (dlka drhy hasie viz obrzek) M G1 M G2 M N B M N A M TA M TB 0 nulov rameno sly 6.5 Vklad k een: 2. Podmnka rovnovhy

ROVNOST MOMENT SIL: S ohledem na situaci na obrzku dosadme do vztah pro velikosti moment. Znamnka moment zvolme podle pravidla prav ruky: FG1 cos FG2 x cos FN B sin FTB cos 0 2 6.5 Vklad k een: Doplujc podmnky: Trojice zskanch rovnic m stle

jet velk mnostv neznmch. Z tohoto dvodu je nutn je jet doplnit o nsledujc podmnky (zkontrolujte platnost podle obrzku). Jde o vztahy mezi velikostmi sil FN a FT: FTA FN A f , FTB FN B f 6.5 Vklad k een: Vsledn soustava rovnic: FG1 FG 2 FN A FTB FN B FTA

FG1 cos FG2 x cos FN B sin FTB cos 0 2 FTA FN A f FTB FN B f 6.5 Zskan vsledky: eenm soustavy rovnic vyjdme hledanou neznmou x: m1 m2 f f 1 m1 x

tg 2 f 1 2m2 m2 Po selnm dosazen:x 2,6 m Zdroje a pouit literatura: [1] Mechanika tuhho tlesa. In: Wikipedia: the free encyclopedia [online]. San Francisco (CA): Wikimedia Foundation, 2001-2012 [cit. 2012-05-12]. Dostupn z:

http://cs.wikipedia.org/wiki/Mechanika_tuhho_tlesa. [2] VYBRAL, Bohumil. Statika tuhho tlesa: Studijn text pro eitele FO a ostatn zjemce o fyziku. [online]. [cit. 2012-05-13]. Dostupn z: http://fyzikalniolympiada.cz/texty/statika.pdf [3] HOFMANN, J. a M. URBANOV. Fyzika I. Praha: Vysok kola chemicko-technologick, 2005, s. 114-118.

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