A darasin kimiyyar lissafi da ya gabata, mun yi nazarin batutuwan da suka shafi yanayin barbashi da yanayin juyawa . A cikin yanayin yanayin barbashi, mun yi nazarin barbashi a cikin motsi na fassara ( motsi na tsaye , motsi na zagaye , motsi na parabolic ). A cikin yanayin yanayin juyawa, mun yi nazarin jikin masu tauri masu juyawa. A cikin wannan batu, mun yi nazarin abubuwa a cikin daidaito. Akwai nau'ikan daidaito guda biyu, wato daidaiton tsaye da daidaiton tsauri. A cewar dokar farko ta Newton, abu yana cikin daidaiton tsaye idan abu yana hutawa kuma abu yana cikin daidaiton motsi idan abu yana motsi a cikin saurin da ya dace. Wannan takarda ta fi mai da hankali kan tattaunawa kan daidaiton tsaye (abun da ke tsaye).
Ma'anar daidaiton jiki mai tauri wani muhimmin ilimi ne na asali kuma yana da amfani da yawa a rayuwar yau da kullun, musamman a fannin injiniyanci. Fahimtar da ƙididdige ƙarfin da ke aiki akan abu a cikin yanayin daidaito mara canzawa yana da matuƙar mahimmanci, musamman ga ƙwararrun fasaha (masu zane-zane ko injiniyoyi). Wajen tsara wani abu, ko gini, gada, abin hawa, da sauransu, masu zane-zane ko injiniyoyi suma suna ƙididdigewa a hankali ko tsarin gini, abin hawa, gada, da sauransu, yana iya jure ƙarfin da ke aiki a kansa don kada ya ruguje.
Misalin matsalolin
1. Ka yi la'akari da hoton da ke ƙasa! An ɗaure sandar 1,5 kg a tsaye a bango a gefe ɗaya, kuma an rataye fitila a wani nisa daga maƙallin. Jinkirin da ke cikin igiyar don kiyaye sandar a daidaito shine…
Tattaunawa
An sani:
Nauyin kaya (m 1 ) = 2 kg
Nauyin sandar (m2 ) = 1,5 kg
Saurin gudu saboda nauyi (g) = 10 m/s 2
Nauyin kaya (w)1) = (2)(10) = 20 Newtons
Nauyin sandar (w 2 ) = (1,5)(10) = 15 N
Tambaya: Girman tashin hankali a cikin igiyar don haka sandar ta kasance cikin daidaito
Amsa:
An ƙididdige tsawon igiyar ta amfani da dabarar Pythagorean:
Lissafa ƙarfin tashin hankali na igiya (T ):
Zaɓi axis na juyawa a wurin hinging. Yi la'akari da lokacin da ƙarfin ke aiki akan sandar.
Lokacin Ƙarfi na 1:
τ 1 = w 2 r 2 = (15 N) (0,4 m) = -6 nm
Lokacin ƙarfi 1 yana sa sandar ta juya a hannun agogo don haka lokacin ƙarfi 1 yana da alama mara kyau.
Lokacin Ƙarfi na 2:
τ 2 = w 1 r 1 = (20 N) (0,6 m) = -12 nm
Lokacin ƙarfi 2 yana sa sandar ta juya a hannun agogo don haka lokacin ƙarfi 1 yana da alama mara kyau.
Lokacin Ƙarfi na 3:
τ 3 = T r 3 zunubi θ = (T) (0,8) (0,6/1) = 0,48T
Lokacin ƙarfi na 3 yana sa sandar ta juya akasin agogo don haka lokacin ƙarfi na 3 yana da alama mai kyau.
Domin tsarin ya kasance a tsaye, lokacin ƙarfi da aka samu ( Στ) = 0
Στ = 0
τ 1 + τ 2 + τ 3 = 0
– 6 – 12 + 0,48T = 0
– 18 + 0,48T = 0
0,48T = 18
T = 18/ 0,48
T = 37,5 Newtons
Tushen tambaya:
Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a
