1. Boima bo habeli m 1 = 2 kg le m 2 = 5 kg di sekametse mme di hokahantswe mmoho ka thapo jwalo ka ha ho bontshitswe setshwantshong. Koefficient ya kgohlano ya kinetic pakeng tsa m 1 le ho sekamela ke 0.2 mme koefficient ya kgohlano ya kinetic pakeng tsa m 2 le ho sekamela ke 0.1.
(a) Fumana hore na li potlaka hakae
(b) Fumana matla a khatello
Tse tsejoang:
Boima ba 1 (m 1 ) = 2 kg
Boima ba 2 (m2 ) = 4 kg
Koefficient ea khohlano ea kinetic pakeng tsa m 1 le sefofane se sekametseng (μ k1 ) = 0.2
Koefficient ea khohlano ea kinetic pakeng tsa m2 le sefofane se sekametseng (μ k2 ) = 0.1
Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s 2
a) Boholo le tataiso ea ho potlaka
w 1 = boima 1 = m 1 g = (2 kg)(9.8 m/s 2 ) = 19.6 Li-Newton
w 1x = w 1 sin 30 o = (19.6 N)(0.5) = 9.8 Newton
w 1y = w 1 cos 30 o = (19.6 N)(0.87) = 17 Newtons
N 1 = Matla a tloaelehileng ho m 1 = w 1y = 17 Newtons
F k1 = Matla a khohlano ea kinetic ho m 1 = μ k1 N 1 = (0.2)(17 N) = 3.4 Li-Newton
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w 2 = boima 2 = m 2 g = (4 kg)(9.8 m/s 2 ) = 39.2 Li-Newton
w 2x = w 2 sebe 60 o = (39.2 N) (0.87) = 34.1 Newtons
w 2y = w 2 cos 60 o = (39.2 N)(0.5) = 19.6 Newtons
N 2 = Matla a tloaelehileng ho m 2 = w 2y = 19.6 Newtons
F k2 = Matla a khohlano ea kinetic ho m 2 = μ k2 N 2 = (0.1)(19.6 N) = 1.96 Newtons
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Boholo ba ho potlaka ha lebelo:
∑ F x = max
w 2x > w 1x kahoo tataiso ea ho potlakisa e tšoana le tataiso ea w 2x.
Matla a supang ho potlaka a matle mme matla a fapaneng le ho potlaka a mabe.
w2x - Fk2 - T2 +T1 - w1x - Fk1 = (m1 +m2) ax
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) ka x
a x = 18.94 N : 6 kg
a x = 3.16 m/s 2
Boholo ba ho potlaka = 3.16 m/s 2. Tataiso ea ho potlaka = tataiso ea T 1 = 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 – T 2 = m 2 a x
34.1 N – 1.96 N – T 2 = (4 lik’hilograma)(3.16 m/s 2 )
32.14 N – T 2 = 12.64 N
T 2 = 32.14 N – 12.64 N = 19.5 Newton
Matla a kgatello = T = T 1 = T 2 = 19.5 Newtons
2. m 1 = 4 kg, m 2 = 2 kg. Fumana (a) boholo le tataiso ea ho potlaka (b) Boholo ba matla a khatello a hokahanyang m 1 le m 2 (c) boholo ba matla a khatello a hokahanyang pulley le marulelo.
tharollo
w 1 = m 1 g = (4 kg)(9.8 m/s 2 ) = 39.2 Li-Newton
w 2 = m 2 g = (2 kg)(9.8 m/s 2 ) = 19.6 Li-Newton
a) Boholo le tataiso ea ho potlaka
∑ F y = may y
w 1 > w 2 kahoo tataiso ea ntho e tšoana le tataiso ea boima 1 ( w 1 ) . 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.
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 lik'hilograma + 2 lik'hilograma) a y
19.6 N = (6 kg) ka y
a y = 19.6 N : 6 kg
a y = 3.26 m/s 2
Boholo ba ho potlaka = 3.26 m/s 2. Tsela ea ho potlaka = tsela ea w 1.
b) Boholo ba matla a khatello a hokahanyang m 1 le m 2
Sebelisa molao oa bobeli oa Newton ho m2 :
∑ F y = may y
w 1 – T 1 = m 1 a y
39.2 N – T 1 = (4 lik’hilograma)( 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 Newton
Boholo ba matla a khatello a hokahanyang lintho = T = T 1 = T 2 = 26.16 Newtons
c) Boholo ba matla a kgatello a hokahanyang pulley le marulelo.
Pulley e phomotse:
∑ F y = may y —— a y = 0
∑ F y = 0
Matla a holimo a matle, matla a theohelang tlase a mabe:
T 3 – T 1 – T 2 = 0
T3 = T1 + T2
T 1 le T 2 li na le boholo bo tšoanang , T 1 = T 2 = T = 26.16 N:
T 3 = 2T = 2(26.16 N) = 52.32 Newton
3. Boloko ba 1 (m 1 = 10 kg) le boloko ba 2 (m 2 = 15 kg) tse hokahantsweng ka thapo hodima pulley e se nang khohlano. Khokahano ya khohlano e sa fetoheng pakeng tsa boloko ba 2 ka ho sekamela = 0.6. Khokahano 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
w 1 = Boima ba boloko 1 = m 1 g = (10 kg)(9.8 m/s 2 ) = 98 Newtons
w 2 = Boima ba boloko 2 = m 2 g = (15 kg)(9.8 m/s 2 ) = 147 Newtons
w 2y = w 2 cos 30 o = (147 N)(0.87) = 127.89 Li-Newton
w 2x = w 2 sebe 30 o = (147 N) (0.5) = 73.5 Newtons
N 2 = Matla a tloaelehileng holim'a boloko 2 = w 2y = 127.89 Newtons
F k2 = Matla a khohlano ea kinetic holim'a block 2 = μ k2 N 2 = (0.42)(127.89 N) = 53.7 Newtons
F s2 = Matla a khohlano e sa fetoheng bolokong 2 = μ s2 N 2 = (0.6)(127.89 N) = 76.7 Newtons
a) Boholo ba matla a fokolang a F a sebelisitsoeng linthong kahoo lintho li ile tsa potlaka ho ea holimo
∑ F x = max —— 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 – 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 Newton
b) Boholo ba matla a khatello ea maikutlo
Sebelisa molao oa Newton oa ho sisinyeha holim'a boloko ba 1:
∑ F y = may y —— a y = 0
∑ F y = 0
T 1 – w 1 = 0
T 1 = w 1 = 98 Newton
Sebelisa molao oa Newton oa ho sisinyeha holim'a boloko ba 2:
F – F k2 – w 2x – T 2 = 0
T 2 = F – F k2 – w 2x
T 2 = 225.2 N – 53.7 N – 73.5 N
T 2 = 98 Newton
Boholo ba matla a kgatello = T 1 = T 2 = T = 98 Newtons
4. Boloko ba 1 (m 1 = 16 kg) bo lutse holim'a bokaholimo bo bataletseng 'me boloko ba 2 (m 2 = 12 kg) bo lutse holim'a sekala se boreleli, se hokahaneng ka thapo e fetang holim'a pulley e nyane, e se nang khohlano. Boloko ba 3 (m 3 = 5 kg) bo lutse holim'a boloko ba 2. Koefficient ea khohlano ea kinetic pakeng tsa boloko ba 2 le bokaholimo bo bataletseng ke 0,4. Koefficient ea khohlano e sa fetoheng pakeng tsa boloko ba 2 le boloko ba 3 ke 0,3.
(a) Ha sistimi e lokollwa phomolong, boloko ba 3 le boloko ba 2 di ntse di thella mmoho?
(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 lokollwa phomolong, boloko ya 3 le boloko ya 2 di ntse di thella mmoho?
w 1 = Boima ba boloko 1 = m 1 g = (16 kg)(9.8 m/s 2 ) = 156.8 Newtons
w 1x = w 1 sin 60 o = (156.8 N)(0.87) = 136.4 Newton
w 1y = w 1 cos 60 o = (156.8 N)(0.5) = 78.4 Li-Newton
N 1 = Matla a tloaelehileng a sebelisoang holim'a boloko ba 1 ke sefofane se sekametseng = w 1y = 78.4 Newtons
w 3 = Boima ba boloko 3 = m 3 g = (5 kg)(9.8 m/s 2 ) = 49 Newtons
N 23 = Matla a tloaelehileng a sebelisoang holim'a boloko ba 3 ke boloko ba 2 = w 3 = 49 Newtons
N 32 = Matla a tloaelehileng a sebelisoang holim'a boloko ba 2 ke boloko ba 3 = N 23 = w 3 = 49 Newtons
(N 23 le N 32 ke para ya karabelo ya ketso )
F s23 = Matla a khohlano e sa fetoheng e sebediswang hodima boloko ba 3 ke boloko ba 2 = μ s N 23 = (0.3)(49 N) = 14.7 Newtons
F s32 = Matla a khohlano e sa fetoheng e sebediswang hodima boloko ba 2 ke boloko ba 3 = F s 23 = 14.7 Newtons
(F s23 le F s32 ke para ya karabelo ya ketso )
w 2 = Boima ba boloko 2 = m 2 g = (12 kg)(9.8 m/s 2 ) = 117.6 Li-Newton
N 2 = Matla a tloaelehileng a sebelisoang nthong ea 2 ke bokaholimo bo otlolohileng = w 2 + N 32 = 117.6 Newtons + 49
Newton = 166.6 Newton
F k2 = Matla a kgohlano ya kinetic hodima boloko 2 = μ k N 2 = (0.4)(166.6 N) = 66.64 Newtons
Sebelisa molao oa Newton oa ho sisinyeha holim'a boloko ba 3:
∑ F x = max
F s23 = m 3 a x
—–> Fs23 = μs N23 = μs w3 = μs m3 g
μ s m 3 g = m 3 a x
μ s g = a x
a x = (0.3)(9.8 m/s 2 ) = 2.94 m/s 2
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/s 2.
Jwale re bala boholo ba ho potlaka ha sistimi ka mora ho lokollwa phomolong.
Tsela eo boloko bo fallang ka yona = tsela eo boloko bo potlakang ka yona = tsela eo T 2 e tsamayang ka yona = tsela eo w 1x e tsamayang ka yona.
∑ F x = max
w1x - T1 +T2 - Fk2 - Fs32 +Fs23 = (m1 +m2 +m3) ax
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) ka x
a x = 2.11 m/s 2
a x e ntle, ho bolelang hore tataiso ea ho falla ha block kapa tataiso ea ho potlaka e tšoana le tataiso ea T 2 kapa tataiso ea w 1x.
Boholo ba lebelo ke 2.11 m/s 2 , bo ka tlase ho 2.94 m/s 2 kahoo re ka etsa qeto ea hore boloko ba 3 le boloko ba 2 li ntse li thella hammoho ka mor'a hore li lokolloe phomolong.
b) Boholo ba ho potlaka ha boloko 1 le boloko 2
∑ F x = max
w 1x – F k2 = (m 1 + m 2 ) a x
—–> Fk2 = μk N2 = μk w2 = μk m2 g = (0.4)(12 kg)(9.8 m/s2) = 47.04 Newton
136.4 N – 47.04 N = (16 kg + 12 kg) a x
89.36 N = (28 kg) ka x
a x = 89.36 N : 28 kg = 3.19 m/s 2
[ID ea sephutheloana sa wpdm='493′]
- Boima le boima
- Matla a tloaelehileng
- Molao oa bobeli oa Newton oa ho sisinyeha
- Matla a khohlano
- Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
- Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano
- Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano
- Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
- Motsamao ka lifting
- Motsamao oa 'mele o hokahantsoe ka lithapo le li-pulleys
- 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
- Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
- Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
- Motsamao o tšoanang ka selikalikoe se otlolohileng
- Matla a bohareng a motsamao o chitja o tšoanang
