Tezlanish kattaligi bir xil bo'lgan ikkita jism – Nyuton harakat qonunining qo'llanilishi masalalari va yechimlari

1. Ikki massali m1 ₌ 2 kg va m2 ₌ 5 kg qiyalik tekislikda joylashgan va rasmda ko'rsatilgandek, ip bilan bir-biriga bog'langan. m1 _{va qiyalik orasidagi kinetik ishqalanish koeffitsienti 0.2 ga, m2} va qiyalik orasidagi kinetik ishqalanish koeffitsienti esa 0.1 ga teng.

(a) Ularning tezlanishini aniqlang

(b) Taranglik kuchini aniqlang

Ma'lum:

Massa 1 (m1 ₎ = 2 kg

Massa 2 (m2 ₎ = 4 kg

_m1 va qiya tekislik orasidagi kinetik ishqalanish koeffitsienti (μk1 ₎ = 0.2

_m2 va qiya tekislik orasidagi kinetik ishqalanish koeffitsienti (μ _k2 ) = 0.1

Tortishish kuchi tufayli tezlanish (g) = 9.8 m/ ^s2

a) Tezlanishning kattaligi va yo'nalishi

w ₁ = og'irlik 1 = m ₁ g = (2 kg)(9.8 m/s ² ) = 19.6 Nyuton

w _1x = w ₁ sin 30 ^o = (19.6 N)(0.5) = 9.8 Nyuton

w _1y = w ₁ cos 30 ^o = (19.6 N)(0.87) = 17 Nyuton

N ₁ = m ga ta'sir qiluvchi normal kuch ₁ = w _1y = 17 Nyuton

F _{k1 = m1} ga kinetik ishqalanish kuchi = μ _k1 N1 = (0.2)(17 N) = 3.4 _Nyuton

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w ₂ = og'irlik 2 = m ₂ g = (4 kg)(9.8 m/s ² ) = 39.2 Nyuton

w _2x = w ₂ sin 60 ^o = (39.2 N)(0.87) = 34.1 Nyuton

w _2y = w ₂ cos 60 ^o = (39.2 N)(0.5) = 19.6 Nyuton

N2 = _{m2 ga} ta'sir qiluvchi normal kuch = _w2y = 19.6 Nyuton

F _{k2 = m2} ga kinetik ishqalanish kuchi = μ _k2 N2 = (0.1)(19.6 N) = 1.96 _Nyuton

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Tezlanishning kattaligi:

∑ F _x = ma _x

w _2x > w _{1x , shuning uchun tezlanish yo'nalishi w 2x} yo'nalishi bilan bir xil.

Tezlanish bo'ylab harakatlanuvchi kuchlar musbat, tezlanishga qarama-qarshi yo'nalishda harakatlanuvchi kuchlar esa manfiydir.

w_2x - F_k2 - T₂ +T₁ - w_1x - F_k1 = (m₁ +m₂) a_x

w _2x – F _k2 – w _1x – F _k1 = (m ₁ + m ₂ ) a _x

34.1 N – 1.96 N – 9.8 N – 3.4 N = (2 kg + 4 kg) a _x

18.94 N = (6 kg) a _x

a _x = 18.94 N: 6 kg

a _x = 3.16 m/ ^s2

Tezlanish kattaligi = 3.16 m/s2 ^. Tezlanish yo'nalishi = T1 yo'nalishi _{= w2x} yo'nalishi

b) Tortishish kuchining kattaligi

Nyutonning ikkinchi qonunini 2-obyektga qo'llang:

w _2x – F _k2 – T ₂ = m ₂ a _x

34.1 N – 1.96 N – T ₂ = (4 kg) (3.16 m/s ² )

32.14 N – T ₂ = 12.64 N

T2 = 32.14 N – 12.64 N = 19.5 _Nyuton

Kuchlanish kuchi = T = T ₁ = T ₂ = 19.5 Nyuton

[irp]

2. m1 ₌ 4 kg, m2 ₌ 2 kg. (a) tezlanishning kattaligi va yo'nalishini (b) m1 _va m2 _{ni bog'laydigan kuchlanish kuchining kattaligini} (c) blok va tomni bog'laydigan kuchlanish kuchining kattaligini aniqlang.

qaror

w ₁ = m ₁ g = (4 kg)(9.8 m/s ² ) = 39.2 Nyuton

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

a) Tezlanishning kattaligi va yo'nalishi

∑ F _y = may _y

w ₁ > w ₂ , shuning uchun jismning yo'nalishi og'irlik 1 ning yo'nalishi bilan bir xil ( w ₁ ) . Tezlanish bilan bir xil yo'nalishga ega bo'lgan kuchlar musbat, tezlanish bilan teskari yo'nalishga ega bo'lgan kuchlar esa manfiydir.

w ₁ – T ₁ + T ₂ – w ₂ = (m ₁ + m ₂ ) a _y

w ₁ – w ₂ = (m ₁ + m ₂ ) a _y

39.2 N - 19.6 N = (4 kg + 2 kg) a _y

19.6 N = (6 kg) bir _yil

a _y = 19.6 N : 6 kg

y = 3.26 m _/ ^s2

Tezlanish kattaligi = 3.26 m/s2 ^{. Tezlanish yo'nalishi =} _w1 yo'nalishi.

_{b) m1} va m2 _ni bog'laydigan kuchlanish kuchining kattaligi

Nyutonning ikkinchi qonunini _m2 ga qo'llang :

∑ F _y = may _y

w ₁ – T ₁ = m ₁ a _y

39.2 N – T ₁ = (4 kg)( 3.26 m/s ² )

39.2 N – T ₁ = 13.04 N

T ₁ = 39.2 N – 13.04 N

T ₁ = 26.16 Nyuton

Jismlarni bog'laydigan kuchlanish kuchining kattaligi = T = T ₁ = T ₂ = 26.16 Nyuton

c) Kasnak va tomni bog'laydigan kuchlanish kuchining kattaligi.

Kasnak dam olmoqda:

∑ F _y = may _y —— a _y = 0

∑ F _y = 0

Yuqoriga ko'taruvchi kuchlar ijobiy, pastga yo'naltiruvchi kuchlar esa salbiy:

T ₃ – T ₁ – T ₂ = 0

T3 = _T1 + _T2

T1 _va T2 bir xil kattalikka ega _, T1 ₌ T2 ₌ T = 26.16 N:

T3 = 2T = 2(26.16 N) = 52.32 _Nyuton

[irp]

3. Ishqalanishsiz shkiv orqali shnur bilan bog'langan 1-blok (m ₁ = 10 kg) va 2-blok (m ₂ = 15 kg). Nishabli 2-blok orasidagi statik ishqalanish koeffitsienti = 0.6. Nishabli 2-blok orasidagi kinetik ishqalanish koeffitsienti = 0.42. (a) Jismlarga ta'sir qilgan minimal kuch F ning kattaligini aniqlang, shunda jismlar yuqoriga qarab tezlashadi (b) Taranglik kuchining kattaligini aniqlang.

qaror

w ₁ = Blokning og'irligi 1 = m ₁ g = (10 kg)(9.8 m/s ² ) = 98 Nyuton

w ₂ = Blokning og'irligi 2 = m ₂ g = (15 kg)(9.8 m/s ² ) = 147 Nyuton

w _2y = w ₂ cos 30 ^o = (147 N)(0.87) = 127.89 Nyuton

w _2x = w ₂ sin 30 ^o = (147 N)(0.5) = 73.5 Nyuton

N ₂ = 2-blokdagi normal kuch = w _2y = 127.89 Nyuton

F _k2 = 2-blokdagi kinetik ishqalanish kuchi = μ _k2 N2 = (0.42)(127.89 N) = 53.7 _Nyuton

F _s2 = 2-blokdagi statik ishqalanish kuchi = μ _s2 N ₂ = (0.6)(127.89 N) = 76.7 Nyuton

a) Jismlar yuqoriga tezlanishi uchun ularga ta'sir qiladigan minimal kuch F ning kattaligi

∑ F _x = ma _x —— a _x = 0

∑ F _x = 0

Yuqoriga va o'ngga yo'naltirilgan kuchlar musbat, pastga va chapga yo'naltirilgan kuchlar esa manfiydir.

F – F _k2 – w _2x – w ₁ – T ₂ + T ₁ = 0

F – F _k2 – w _2x – w ₁ = 0

F = F _k2 + w _2x + w ₁

F = 53.7 N + 73.5 N + 98 N

F = 225.2 Nyuton

b) Kuchlanish kuchining kattaligi

Nyutonning harakat qonunini 1-blokda qo'llang:

∑ F _y = may _y —— a _y = 0

∑ F _y = 0

T ₁ – w ₁ = 0

T ₁ = w ₁ = 98 Nyuton

Nyutonning harakat qonunini 2-blokda qo'llang:

F – F _k2 – w _2x – T ₂ = 0

T ₂ = F – F _{k 2} – w _2x

T2 = 225.2 N – 53.7 N – 73.5 _N

T ₂ = 98 Nyuton

Kuchlanish kuchining kattaligi = T ₁ = T ₂ = T = 98 Nyuton

[irp]

4. 1-blok (m ₁ = 16 kg) gorizontal yuzada, 2-blok (m ₂ = 12 kg) esa kichik, ishqalanishsiz kasnak ustidan o'tuvchi shnur bilan bog'langan silliq qiyalik tekislikda yotadi. 3-blok (m ₃ = 5 kg) 2-blokda yotadi. 2-blok va gorizontal yuza orasidagi kinetik ishqalanish koeffitsienti 0,4 ga teng. 2-blok bilan 3-blok orasidagi statik ishqalanish koeffitsienti 0,3 ga teng.

(a) Tizim tinch holatdan chiqarilganda ham, 3-blok va 2-blok hali ham birga sirpanadi?

(b) Agar 3-blok mavjud bo'lsa, 1-blok va 2-blokning tezlanishi qanday?

yechim:

a) Tizim tinch holatdan chiqarilganda ham, 3-blok va 2-blok hali ham birga sirpanadimi?

w ₁ = Blokning og'irligi 1 = m ₁ g = (16 kg)(9.8 m/s ² ) = 156.8 Nyuton

w _1x = w ₁ sin 60 ^o = (156.8 N)(0.87) = 136.4 Nyuton

w _1y = w ₁ cos 60 ^o = (156.8 N)(0.5) = 78.4 Nyuton

N ₁ = Qiya tekislik tomonidan 1-blokka ta'sir qiluvchi normal kuch = w _1y = 78.4 Nyuton

w ₃ = Blokning og'irligi 3 = m ₃ g = (5 kg)(9.8 m/s ² ) = 49 Nyuton

N ₂₃ = 2-blok tomonidan 3-blokka ta'sir qiluvchi normal kuch = w ₃ = 49 Nyuton

N ₃₂ = 3-blok tomonidan 2-blokka ta'sir qiluvchi normal kuch = N ₂₃ = w ₃ = 49 Nyuton

(N ₂₃ va N ₃₂ harakat-reaktsiya juftligidir )

F _s23 = 2-blok tomonidan 3-blokka ta'sir qiluvchi statik ishqalanish kuchi = μ _s N ₂₃ = (0.3)(49 N) = 14.7 Nyuton

F _s32 = 3 _- blok tomonidan 2-blokka ta'sir qiluvchi statik ishqalanish kuchi = F _s23 = 14.7 Nyuton

(F _s23 va F _s32 harakat-reaktsiya juftligidir )

w ₂ = Blokning og'irligi 2 = m ₂ g = (12 kg)(9.8 m/s ² ) = 117.6 Nyuton

N ₂ = Gorizontal sirt tomonidan 2-jismga ta'sir qiluvchi normal kuch = w ₂ + N _{3 2} = 117.6 Nyuton + 49

Nyuton = 166.6 Nyuton

F _k2 = ₂ -blokdagi kinetik ishqalanish kuchi = μ _k N2 = (0.4)(166.6 N) = 66.64 Nyuton

Nyutonning harakat qonunini 3-blokda qo'llang:

∑ F _x = ma _x

F _s23 = m ₃ a _x

—–> F_s23 = μ_s N₂₃ = μ_s w₃ = μ_s m₃ g

μ _s m ₃ g = m ₃ a _x

μ _s g = a _x

a _x = (0.3)(9.8 m/s² ⁾ = 2.94 m/ ^s²

3-blok va 2-blok hali ham birga sirpanishi uchun 3-blokning maksimal tezlanishi 2.94 m/s2 ga ^teng.

Endi biz tizimning tinchlik holatidan chiqqandan keyingi tezlanishining kattaligini hisoblaymiz.

Blok siljishining yo'nalishi = blok tezlanishining yo'nalishi = T2 yo'nalishi _{= w1x} yo'nalishi.

∑ F _x = ma _x

w_1x - T₁ +T₂ - F_k2 - F_s32 + F._s23 = (m₁ +m₂ +m₃) a_x

w _1x – F _k2 = (m ₁ + m ₂ + m ₃ ) a _x

136.4 N – 66.64 N = (16 kg + 12 kg + 5 kg) a _x

69.76 N = (33 kg) a _x

a _x = 2.11 m/ ^s2

a _x musbat, ya'ni blok siljishining yo'nalishi yoki tezlanish yo'nalishi T2 yo'nalishi _{yoki w1x} yo'nalishi bilan bir xil.

Tezlanish kattaligi 2.11 m/s2 ^ga teng , bu 2.94 m/s2 ^dan past , shuning uchun biz 3-blok va 2-blok tinch holatdan chiqarilgandan keyin ham birga sirpanishini xulosa qilishimiz mumkin.

b) 1-blok va 2-blokning tezlanish kattaligi

∑ F _x = ma _x

w _1x – F _k2 = (m ₁ + m ₂ ) a _x

—–> F_k2 = μ_k N₂ = μ_k w₂ = μ_k m₂ g = (0.4)(12 kg)(9.8 m/s)²) = 47.04 Nyuton

136.4 N – 47.04 N = (16 kg + 12 kg) a _x

89.36 N = (28 kg) a _x

a _x = 89.36 N: 28 kg = 3.19 m/s ²

[wpdm_package id='493']

1. Massa va vazn
2. Oddiy kuch
3. Nyutonning ikkinchi harakat qonuni
4. Ishqalanish kuchi
5. Ishqalanish kuchisiz gorizontal sirt ustida harakat
6. Ishqalanish kuchi ta'sirida qo'pol gorizontal yuzada bir xil tezlanishga ega bo'lgan ikki jismning harakati
7. Ishqalanish kuchisiz qiyalik tekislikda harakat
8. Ishqalanish kuchi bilan qo'pol qiyalik tekislikda harakat
9. Liftda harakatlanish
10. Jismlarning harakati arqonlar va kasnaklar orqali bog'langan
11. Tezlanish kattaligi bir xil bo'lgan ikkita jism
12. Yassi egri chiziqni yaxlitlash – aylana harakatining dinamikasi
13. Egri chiziqni yaxlitlash – aylana harakatining dinamikasi
14. Gorizontal aylana bo'ylab bir tekis harakat
15. Bir tekis aylana harakatida markazdan qochiruvchi kuch