Sinnaanta jirka oo ay ku xiran yihiin fiilooyin iyo boolal - adeegsiga dhibaatooyinka sharciga ugu horreeya ee Newton iyo xalalka

1. Sanduuq culeyskiisu yahay 5 kg ayaa saaran diyaarad u janjeerta xagal 30 o ah . Sanduuqa waxaa taageera xadhig. Go'aami xoogga xiisadda (T) iyo xoogga caadiga ah (N)!

Sinnaanta jirka oo ay ku xiran yihiin fiilooyin iyo boolal - adeegsiga dhibaatooyinka sharciga ugu horreeya ee Newton iyo xalalka 1

Solution

Sinnaanta jirka oo ay ku xiran yihiin fiilooyin iyo boolal - adeegsiga dhibaatooyinka sharciga ugu horreeya ee Newton iyo xalalka 2ΣFx = 0

T – w sin 30 o = 0

T = w sin 30 o

T = (5 kg)(9.8 m/s 2 ) sin 30 o

T = (49)(0.5)

T = 24.5 Newtons

F y = 0

N – w cos 30 o = 0

N = w cos 30 o

N = (49)(0.87)

N = 43 Newton

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2. Laba shay oo cufnaantoodu tahay m 1 = m 2 = 2 kg, oo ay ku xiran yihiin xarig aan cufnayn oo dul saaran shaandho aan isqabsanayn. Soo hel xoogagga xiisadda T 1 iyo T 2.

Sinnaanta jirka oo ay ku xiran yihiin fiilooyin iyo boolal - adeegsiga dhibaatooyinka sharciga ugu horreeya ee Newton iyo xalalka 3

Solution

Sinnaanta jirka oo ay ku xiran yihiin fiilooyin iyo boolal - adeegsiga dhibaatooyinka sharciga ugu horreeya ee Newton iyo xalalka 4

(a) Jaantuska jirka ee xorta ah ee shayga 1 (b) Jaantuska jirka ee xorta ah ee shayga 2

Ku dabaq sharciga ugu horreeya ee Newton diidmada 1aad:

F y = 0

T 1 – w 1 = 0

T 1 = w 1 = m 1 g = (2 kg)(9.8 m/s 2 ) = 19.6 N

Ku dabaq sharciga ugu horreeya ee Newton diidmada 2:

F y = 0

T 2 – w 2 = 0

T 2 = w 2 = m 2 g = (2 kg)(9.8 m/s 2 ) = 19.6 N

T 1 = T 2 = 19.6 N.

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3. Shay miisaan leh w A = 30 N iyo shay miisaan leh w B = 40 N, waxaa ku xiran xarig khafiif ah oo ka gudba shaandho aan is jiidjiid lahayn oo ah cuf aan la taaban karin. Go'aami isku-dhafka is jiidjiidka ugu badan ee u dhexeeya w B iyo dusha sare ee janjeedha, haddii nidaamku nasto.

Sinnaanta jirka oo ay ku xiran yihiin fiilooyin iyo boolal - adeegsiga dhibaatooyinka sharciga ugu horreeya ee Newton iyo xalalka 5

Solution

Sinnaanta jirka oo ay ku xiran yihiin fiilooyin iyo boolal - adeegsiga dhibaatooyinka sharciga ugu horreeya ee Newton iyo xalalka 6

(a) Jaantuska jirka ee xorta ah ee walaxda w A (b) Jaantuska jirka ee xorta ah ee walaxda w B

Ku dabaq sharciga ugu horreeya ee Newton si aad u diiddo w A jihada toosan (y):

F y = 0 (ma jiro dardargelin jihada toosan)

T – w A = 0

T = w A = 30 Newton

Ku dabaq sharciga ugu horreeya ee Newton si aad u diiddo w B jihada toosan (y) :

F y = 0

N – w B cos 45 o = 0

N = w B cos 45 o = (40)(0.7) = 28 Newtons

Ku dabaq sharciga ugu horreeya ee Newton si aad u diiddo w B jihada toosan (x):

F x = 0

F k + w B sin 45 o – T = 0

μ s N + w B dembi 45 o – T = 0

μ s (28) + (40) (0.7) - 30 = 0

μ s (28) + 28 – 30 = 0

μ s (28) = 30 – 28

μ s (28) = 2

μ s = 2/28

μ s = 0.07

Isugeynta is-jiidjiidka ugu badan ee u dhexeeya w B iyo dusha sare ee janjeedha = 0.07.

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  1. Walxaha isku dheelitirka hal-cabbir ah
  2. Walxaha dheelitirka laba-geesoodka ah
  3. Sinnaanta jirka oo ay ku xiran yihiin xadhkaha iyo boolalku
  4. Sinnaanta jirka ee diyaaradda u janjeerta

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