A darasin kimiyyar lissafi da ya gabata, an yi nazari kan babban batun. yanayin ƙwayoyin cuta dan yanayin juyawaA cikin yanayin motsin barbashi muna nazarin barbashi a cikin motsin fassara (motsi kai tsaye, motsi na zagaye, motsin parabolic). A cikin yanayin juyawa, muna nazarin jikin da ke juyawa masu tauri. A cikin wannan batu, muna nazarin abubuwa a cikin daidaito. Akwai nau'ikan daidaito guda biyu: daidaiton tsaye da daidaiton motsi. A bisa ga dokar farko ta Newton, abu yana cikin daidaiton tsaye idan yana hutawa, kuma abu yana cikin daidaiton motsi idan yana motsi a cikin saurin da ya dace. Wannan takarda ta fi mai da hankali kan tattaunawa kan daidaiton tsaye (abubuwa a hutawa).
Konsep Daidaiton jiki mai tsauri Ilimi ne mai matuƙar muhimmanci kuma yana da amfani da yawa a rayuwar yau da kullun, musamman a fannin injiniyanci. Fahimtar da ƙididdige ƙarfin da ke aiki akan abubuwa a cikin yanayin daidaiton da ba ya canzawa yana da matuƙar muhimmanci, 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/s2
Nauyin kaya (w)1) = (2)(10) = 20 Newtons
Nauyin sandar (w)2) = (1,5)(10) = 15 N
An tambaya: Girman tashin hankali a cikin igiya 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:
τ1 = w2 r2 = (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:
τ2 = w1 r1 = (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:
τ3 = T r3 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 natse, lokacin ƙarfi (Στ) = 0
Στ = 0
τ1 + τ2 + τ3 = 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
