3 su'aalood oo ku saabsan isle'egta xoogga is-qabsiga
1. Baloog A 3 kg ayaa miiska la saaraa ka dibna waxaa lagu xiraa xarig ku xiran dhagax B = 2 kg iyada oo loo marayo barkin sida lagu muujiyey. Cufnaanta iyo is jiidjiidka barkinta waa la dayacay. Dardargelinta cufisjiidadka g = 10 m/s 2. Go'aami dardargelinta nidaamka iyo xiisadda xarigga haddii:
a) Miis siman
b) jaantus qallafsan oo leh isku-darka isku-dhafka dhaqdhaqaaqa ee 0.4
La yaqaan
Cufka baloogga A (m A ) = 3 kg
Cufka dhagaxa B (m B ) = 2 kg
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Miisaanka baloogga A (w A ) = mg = (3)(10) = 30 Newtons
Miisaanka dhagaxa B (w B ) = mg = (2)(10) = 20 Newtons
La Doonayo: Dardargelinta nidaamka (a) iyo xiisadda xarigga (T)
Solution:
a) Miis siman
Xisaabi dardargelinta nidaamka adoo isticmaalaya qaacidada sharciga labaad ee Newton:
ΣF = ma
w B = (m A + m B ) a
20 = (3 + 2) a
20 = 5 a
a = 20 / 5 = 4 m/s 2
Xisaabi xiisadda xarigga adoo isticmaalaya qaacidada xiisadda xarigga:
Xiisadda xarigga ee baloogga A:
ΣF = m A a
T = m A a = (3)(4) = 12 Newtons
Xiisadda xarigga ee baloogga B:
ΣF = m B a
w B – T = (2)(4)
20 – T = 8
T = 20 – 8 = 12 Newton
b) jaantus qallafsan oo leh isku-darka isku-dhafka dhaqdhaqaaqa ee 0.4
Xoogga is-jiidjiididda dhaqdhaqaaqa:
F k = µ k N = (0,4)(30) = 12 Newtons
Xisaabi dardargelinta nidaamka adoo isticmaalaya qaacidada sharciga labaad ee Newton:
ΣF = ma
w B – f k = (m A + m B ) a
20 – 12 = (3 + 2) a
8 = 5 a
a = 8 / 5 = 1,6 m/s 2
Xisaabi xiisadda xarigga adoo isticmaalaya qaacidada xiisadda xarigga:
Xiisadda xarigga ee baloogga A:
ΣF = m A a
T – fk = m A a
T – 12 = (3)(1,6)
T – 12 = 4,8
T = 4,8 + 12 = 16,8 Newtons
Xiisadda xarigga ee baloogga B:
ΣF = m B a
w B – T = (2)(1,6)
20 – T = 3,2
T = 20 – 3,2 = 16,8 Newton
2. Shay culeyskiisu yahay 10 kg ayaa ku jira meel toosan. Isugeynta is-jiidjiidku waa 0.4, isku-darka is-jiidjiidku waa 0.35. g = 10 m/s 2. Haddii shay la siiyo xoog siman oo joogto ah oo ah 25 N, baaxadda xoogga is-jiidjiidku waa…
La yaqaan
Cufka shayga (m) = 10 kg
Isugeynta is-khilaafka taagan (µ s ) = 0.4
Isugeynta is-xoqidda dhaqdhaqaaqa (µk) = 0.35
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Xoogga toosan (F) = 25 N
Cufisjiidadka shayga (w) = mg = (10)(10) = 100 Newtons
Xoogga caadiga ah (N) = w = 100 Newtons
La Rabay: Cadadka is-jiidjiid aan joogto ahayn (fs s ) iyo kinetic (f k )
Solution:
Xoogga is-jiidjiid la'aanta::
f s = µ s N = (0,4) (100) = 40 Newton
Xoogga Kinetic Friction:
f k = µ k N = (0,35)(100) = 35 Newtons
Xoogga jiifa waa 25 Newtons oo keliya sidaa darteed wali ma dhaqaajin karo walxaha.
3. Cufnaanta baloogyada A iyo B ee jaantuska waa 10 kg iyo 5 kg siday u kala horreeyaan. Isugeynta is jiidjiidka u dhexeeya baloogga A iyo diyaaradda waa 0.2. Si looga hortago in baloogga A uu dhaqaaqo, miisaanka ugu yar ee baloogga C ee loo baahan yahay waa…
La yaqaan
Cufka baloogga A (m A ) = 10 kg
Cufka baloogga B (m B ) = 5 kg
Isku-darka is-jiidjiidka joogtada ah ee baloogga A (µ s ) = 0,2
Dardargelinta cuf-isjiidadka (g) = 10 m/s 2
Miisaanka baloogga A (w A ) = m A g = (10)(10) = 100 Newtons
Miisaanka baloogga B (w B ) = m B g = (5)(10) = 50 Newtons
Is-xoqid aan joogto ahayn (fs s ) = µ s N = (0,2)(w A + w C ) = (0,2)(100 + w C ) = 20 + 0,2 w C
La weydiiyay: Cufnaanta baloogga C si nidaamka loogu nasto
Jawaab:
Nidaamku wuu nastay sidaa darteed qaacidada sharciga koowaad ee Newton ayaa la isticmaalay:
ΣF = 0
w B – fs s = 0
50 – (20 + 0,2 w C ) = 0
50 – 20 – 0,2 w C = 0
30 – 0,2 w C = 0
30 = 0,2 w C
w C = 30 / 0,2 = 300 / 2 = 150 Newtons
Cufka baloogga C = 150 / 10 = 15 Kg
