1. Baloog 2-kg ah ayaa ku yaal meel toosan oo u janjeerta xagal 37 o ah oo u jirta dhinaca toosan. Go'aami baaxadda xoogga dibadda ee lagu sameeyay baloogga, si balooggu uusan hoos ugu simbirixan diyaaradda. (sin 37 o = 0.6, cos 37 o = 0.8, g = 10 ms -2 , µ k = 0.2)
La yaqaan:
Cufnaanta (m) = 2 kg
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Miisaanka baloogga (w) = mg = (2)(10) = 20 Newtons
Dem 37 o = 0.6
Kos 37 o = 0.8
Isku-darka is -jiidjiidka dhaqdhaqaaqa (µ k ) = 0.2
Qaybta y ee miisaanka (w y ) = w cos 37 o = (20)(0.8) = 16 Newtons
Qaybta x ee miisaanka (w x ) = w sin θ = (20)(dembi 37) = (20)(0.6) = 12 Newtons
xoogga caadiga ah (N) = w y = 16 Newtons
La Doonayo : Xoogga dibadda (F)
Xalka :
wx = 12 Newton
Xoogga is-jiidjiidka dhaqdhaqaaqa (f k ) = µ k N = (0.1)(16) = 1.6 Newtons
Baaxadda xoogga dibadda F ee ku dhacay baloogga :
F + f k – w x = 0
F = w x – f k
F = 12 – 1.6
F = 10.4 Newton
Xoogga dibadda F wuxuu ka weyn yahay 10.4 Newtons.
2. Cufka baloogga = 2 kg, isku-darka is-jiidjiidka taagan µ s = 0.4 iyo θ = 45 o . Go'aami baaxadda xoogga F si baloogga uu u bilaabo inuu kor u simbirixdo.
La yaqaan:
Isugeynta is-jiidjiidka aan joogtada ahayn (µ s ) = 0.4
Xagasha (θ) = 45 o
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Cufka baloogga (m) = 2 kiilo garaam
Miisaanka baloogga (w) = mg = (2 kg)(10 m/s 2 ) = 20 kg m/s 2 = 20 Newtons
Qaybta x ee miisaanka (w x ) = w sin θ = (20)(dembi 45) = (20)(0.5√2) = 10√2 Newtons
Qaybta y ee miisaanka (w y ) = w cos θ = (20)(cos 45) = (20)(0.5√2) = 10√2 Newtons
La Doonayo : Baaxadda xoogga F
Xalka:
Balooggu wuxuu bilaabaa inuu kor u simbirixdo, haddii F ≥ wx + fs.
Qaybta x ee miisaanka:
w x = 10√2 Newton
qaybta y ee miisaanka :
w y = 10√2 Newton
Xoogga caadiga ah :
N = w y = 10√2 Newton
Xoogga is-jiidjiid la'aanta :
f s = µ s N = (0,4) (10√2) = 4√2
Baaxadda xoogga F si balooggu u bilaabo inuu kor u simbirixdo :
F ≥ w x + f s
F ≥ 10√2 + 4 √2
F ≥ 14√2 Newton
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