Mebele e 'meli e nang le boholo bo tšoanang ba ho potlaka - Tšebeliso ea mathata le litharollo tsa molao oa Newton oa motsamao

1. Boima bo habeli m₁ = 2 kg le m₂ = 5 kg di sebakeng se sekametseng mme di hokahantswe mmoho ka thapo jwalo ka ha ho bontshitswe setshwantshong. Koefficient ya kgatello ya kinetic pakeng tsa m₁ 'me ho sekamela ke 0.2 'me coefficient ea khohlano ea kinetic pakeng tsa m₂ 'me ho sekamela ke 0.1.

(a) Fumana hore na ba ho potlaka

(b) Fumana matla a khatello

Tse tsejoang:

Boholo 1 (m₁) = 2 kg

Boima ba 2 (m₂) = 4 kg

Khokahano ea kinetic pakeng tsa m₁ 'me sefofane se sekametseng (μ_k1) = 0.2

Khokahano ea kinetic pakeng tsa m₂ le sefofane se sekametseng (μ_k2) = 0.1

Ho potlakisa ka baka la matla a khoheli (g) = 9.8 m/s²

a) Boholo le tataiso ea ho potlaka

w₁ = boima ba 'mele 1 = m₁ g = (2 kg)(9.8 m/s²) = 19.6 Newton

w_1x = w₁ sebe 30^o = (19.6 N)(0.5) = 9.8 Newton

w_1y = w₁ Cos 30^o = (19.6 N)(0.87) = 17 Newton

N₁ = E matla a tloaelehileng ka m₁ = w_1y = 17 Newton

F_k1 = Matla a moferefere oa kinetic ho m₁ = μ_k1 N₁ = (0.2)(17 N) = 3.4 Newton

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w₂ = boima 2 = m₂ g = (4 kg)(9.8 m/s²) = 39.2 Newton

w_2x = w₂ sebe 60^o = (39.2 N)(0.87) = 34.1 Newton

w_2y = w₂ Cos 60^o = (39.2 N)(0.5) = 19.6 Newton

N₂ = Matla a tloaelehileng ho m₂ = w_2y = 19.6 Newton

F_k2 = Matla a moferefere oa kinetic ho m₂ = μ_k2 N₂ = (0.1)(19.6 N) = 1.96 Newton

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Boholo ba ho potlaka ha lebelo:

∑F_x = ma_x

w_2x > w_1x kahoo tataiso ea ho potlaka e tšoana le tataiso ea w_2x.

Matla a supang ho potlaka a matle mme matla a fapaneng le ho potlaka a mabe.

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

w_2x - F_k2 - w_1x - F_k1 = (m₁ +m₂ ) 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

Boholo ba ho potlaka = 3.16 m/s² . Tataiso ea ho potlaka = tataiso ea T₁ = tataiso ea w_2x

b) Boholo ba matla a khatello

Sebelisa molao oa bobeli oa Newton holim'a ntho ea 2:

w_2x - F_k2 - T₂ = m₂ a_x

34.1 N – 1.96 N – T₂ = (4 lik'hilograma)(3.16 m/s²)

32.14 N – T₂ = 12.64N

T₂ = 32.14 N – 12.64 N = 19.5 Newton

Matla a tsitsipano = T = T₁ = T₂ = 19.5 Newton

2. m₁ = 4 kg, m₂ = 2 kg. Fumana (a) boholo le tataiso ea ho potlaka (b) Boholo ba matla a khatello a hokahanyang m₁ le m₂ (c) boholo ba matla a kgatello a hokahanyang pulley le marulelo.

tharollo

w₁ = m₁ g = (4 kg)(9.8 m/s²) = 39.2 Newton

w₂ = m₂ g = (2 kg)(9.8 m/s²) = 19.6 Newton

a) Boholo le tataiso ea ho potlaka

∑F_y = ma_y

w₁ > w₂ kahoo tataiso ea ntho e tšoana le tataiso ea boima ba 1 (w₁)Matla a nang le tataiso e tšoanang le ea ho potlaka a matle 'me matla a nang le tataiso e fapaneng le ea ho potlaka a mabe.

w₁ - T₁ +T₂ - w₂ = (m₁ +m₂) a_y

w₁ - w₂ = (m₁ +m₂) a_y

39.2 N - 19.6 N = (4 lik'hilograma + 2 lik'hilograma) a_y

19.6 N = (6 kg) a_y

a_y = 19.6 N : 6 kg

a_y = 3.26 m/s²

Boholo ba ho potlaka = 3.26 m/s². Tataiso ea ho potlaka = tataiso ea w₁ .

b) Boholo ba matla a khatello a hokahanyang m₁ le m₂

Etsa kopo Molao oa bobeli oa Newton ka m₂ :

∑F_y = ma_y

w₁ - T₁ = m₁ a_y

39.2 N – T₁ = (4 kg)(3.26 m/s²)

39.2 N – T₁ = 13.04N

T₁ = 39.2 N – 13.04 N

T₁ = 26.16 Newton

Boholo ba matla a khatello a hokahanyang lintho = T = T₁ = T₂ = 26.16 Newton

c) Boholo ba matla a kgatello a hokahanyang pulley le marulelo.

Pulley e phomotse:

∑F_y = ma_y —— a_y = 0

∑F_y = 0

Matla a holimo a matle, matla a theohelang tlase a mabe:

T₃ - T₁ - T₂ = 0

T₃ = T₁ +T₂

T₁ le T₂ li na le boholo bo tšoanang, T₁ = T₂ = T = 26.16 N :

T₃ = 2T = 2(26.16 N) = 52.32 Li-Newton

3. Boloko ea 1 (m₁ = 10 kg) le boloko ba 2 (m₂ = 15 kg) e hokahantswe ka thapo hodima pulley e se nang khohlano. Khoefithara ya khohlano e sa fetoheng pakeng tsa boloko ba 2 ka ho sekamela = 0.6. Khoefithara ya khohlano ya kinetic pakeng tsa boloko ba 2 ka ho sekamela = 0.42. Fumana (a) Boholo ba matla a bonyane F a sebediswang hodima dintho e le hore dintho di potlakise ho ya hodimo (b) Fumana boholo ba matla a kgatello.

tharollo

w₁ = Boima ba boloko 1 = m₁ g = (10 kg)(9.8 m/s²) = 98 Newton

w₂ = Boima ba boloko 2 = m₂ g = (15 kg)(9.8 m/s²) = 147 Newton

w_2y = w₂ Cos 30^o = (147 N)(0.87) = 127.89 Newton

w_2x = w₂ sebe 30^o = (147 N)(0.5) = 73.5 Newton

N₂ = Matla a tloaelehileng holim'a boloko 2 = w_2y = 127.89 Newton

F_k2 = Matla a moferefere wa kinetic hodima boloko 2 = μ_k2 N₂ = (0.42)(127.89 N) = 53.7 Newton

F_s2 = Matla a khohlano e sa fetoheng bolokong 2 = μ_s2 N₂ = (0.6)(127.89 N) = 76.7 Newton

a) Boholo ba matla a fokolang a F a sebelisitsoeng linthong kahoo lintho li ile tsa potlaka ho ea holimo

∑F_x = ma_x —— a_x = 0

∑F_x = 0

Matla a eang holimo le a eang ka letsohong le letona a matle, matla a theohelang le a eang ka letsohong le letšehali a mabe.

F – F_k2 - w_2x - w₁ - T₂ +T₁ = 0

F – F_k2 - w_2x - w₁ = 0

F = F_k2 +w_2x +w₁

F = 53.7 N + 73.5 N + 98 N

F = 225.2 Newton

b) Boholo ba matla a khatello ea maikutlo

Sebelisa molao oa Newton oa ho sisinyeha holim'a boloko ba 1:

∑F_y = ma_y —— a_y = 0

∑F_y = 0

T₁ - w₁ = 0

T₁ = w₁ = 98 Newton

Sebelisa molao oa Newton oa ho sisinyeha holim'a boloko ba 2:

F – F_k2 - w_2x - T₂ = 0

T₂ = F – F_k2 - w_2x

T₂ = 225.2 N – 53.7 N – 73.5 N

T₂ = 98 Newton

Boholo ba matla a khatello = T₁ = T₂ = T = 98 Newton

4. Boloko ea 1 (m₁ = 16 kg) e lutse holim'a bokaholimo bo bataletseng 'me boloko ba 2 (m₂ = 12 kg) e robetse sebakeng se sekametseng hantle, e hokahantsoe ka thapo e fetang holim'a pulley e nyane, e se nang khohlano. Block 3 (m₃ = 5 kg) e lutse holim'a boloko 2. Koefficient ea khohlano ea kinetic pakeng tsa boloko 2 le bokaholimo bo otlolohileng ke 0,4. CoefKhohlano e sa fetoheng pakeng tsa boloko ba 2 le boloko ba 3 ke 0,3.

(A) Ha sistimi e lokollotsoe phomolong, boloko ea 3 le boloko ea 2 li ntse li thella hammoho?

(B) Haeba ho na le boloko ba 3, ho potlaka ha boloko ba 1 le boloko ba 2 ke eng?

Tharollo:

a) Ha sistimi e lokollotsoe phomolong, boloko ea 3 le boloko ea 2 li ntse li thella hammoho?

w₁ = E boima ba boloko 1 = m₁ g = (16 kg)(9.8 m/s²) = 156.8 Newton

w_1x = w₁ sebe 60^o = (156.8 N)(0.87) = 136.4 Newton

w_1y = w₁ Cos 60^o = (156.8 N)(0.5) = 78.4 Newton

N₁ = E matla a tloaelehileng a sebelisoang bolokong ba 1 ke sefofane se sekametseng = w_1y = 78.4 Newton

w₃ = E boima ba boloko 3 = m₃ g = (5 kg)(9.8 m/s²) = 49 Newton

N₂₃ = E matla a tloaelehileng a sebelisoang holim'a boloko ba 3 ke boloko ba 2 = w₃ = 49 Newton

N₃₂ = Nmatla a tloaelehileng a sebelisoang holim'a boloko ba 2 ke boloko ba 3 = N₂₃ = w₃ = 49 Newton

(N₂₃ 'me N₃₂ ke lipara tsa karabelo ea ketso)

F_s23 = E Matla a khohlano e sa fetoheng e sebediswang hodima boloko ba 3 ke boloko ba 2 = μ_s N₂₃ = (0.3)(49 N) = 14.7 Newton

F_s32 = E Matla a khohlano e sa fetoheng e sebediswang hodima boloko ba 2 ke boloko ba 3 =F_s23 = 14.7 Newton

(F_s23 'me F_s32 ke lipara tsa karabelo ea ketso)

w₂ = E boima ba boloko 2 = m₂ g = (12 kg)(9.8 m/s²) = 117.6 Newton

N₂ = E matla a tloaelehileng a sebelisoang nthong ea 2 ke bokaholimo bo otlolohileng = w₂ +N₃₂ = 117.6 Newtons + 49

Newton = 166.6 Newton

F_k2 = E Matla a khohlano ea kinetic holim'a block 2 = μ_k N₂ = (0.4)(166.6 N) = 66.64 Newton

Sebelisa molao oa Newton oa ho sisinyeha holim'a boloko ba 3:

∑F_x = ma_x

F_s23 =m₃ a_x

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

μ_s m₃ g = m₃ a_x

μ_s g = a_x

a_x = (0.3)(9.8 m/s²) = 2.94 m/s²

Lebelo le phahameng ka ho fetisisa la boloko ba 3 e le hore boloko ba 3 le boloko ba 2 li ntse li thella hammoho ke 2.94 m/s².

Jwale re bala boholo ba ho potlaka ha sistimi ka mora ho lokollwa phomolong.

Tsela eo boloko bo fallang ka yona = tsela eo bo potlakisang ka yona = tsela eo T e tsamayang ka yona₂ = tataiso ea w_1x.

∑F_x = ma_x

w_1x - T₁ +T₂ - F_k2 - F_s32 +F_s23 = (m₁ +m₂ +m₃) a_x

w_1x - F_k2 = (m₁ +m₂ +m₃ ) 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 e ntle, e bolela hore tataiso ea ho falla ha boloko kapa tataiso ea ho potlaka e tšoana le tataiso ea T₂ kapa tataiso ea w_1x.

Boholo ba ho potlaka ke 2.11 limithara / s² , lmatla ho feta 2.94 limithara / s² kahoo re ka etsa qeto ea hore boloko ba 3 le boloko ba 2 li ntse li thella hammoho ka mor'a ho lokolloa phomolong.

b) Boholo ba ho potlaka ha boloko 1 le boloko 2

∑F_x = ma_x

w_1x - F_k2 = (m₁ +m₂) a_x

—–> F_k2 = μ_k N₂ = μ_k w₂ = μ_k m₂ g = (0.4)(12 kg)(9.8 m/s²) = 47.04 Newton

136.4 N - 47.04 N = (16 lik'hilograma + 12 lik'hilograma) a_x

89.36 N = (28 kg) a_x

a_x = 89.36 N : 28 kg = 3.19 m/s²

[ID ea sephutheloana sa wpdm='493′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano
7. Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahantsoe ka lithapo le li-pulleys
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang