Ho sisinyeha ha sefofane se sekametseng ka mokhoa o batlang o le mothating o itseng ka matla a khohlano – ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha

1. Boima ba ntho = 2 kg, ho potlaka ka lebaka la matla a khoheli = 9.8 m/s ² , coefficient ea khohlano e sa fetoheng = 0.2, coefficient ea khohlano ea kinetic = 0.1. Na ntho e phomotse kapa e potlakile? Haeba ntho e potlakile, fumana (a) matla a net (b) boholo le tataiso ea ho potlaka ha lebokose!

tharollo

Tse tsejoang:

Boima (m) = 2 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s ²

Koefficient ea khohlano e sa fetoheng ( μs ₎ = 0.2

Koefficient ea kinetic friction ( μk ₎ = 0.1

Boima (w) = mg = (2)(9.8) = 19.6 Newtons

Karolo e rapameng ea boima (w _x ) = w sin 30 ^o = (19.6)(0.5) = 9.8 Newtons

Karolo e otlolohileng ea boima ba th (w _y ) = w cos 30 ^o = (19.6)(0.5√3) = 9.8√3 Li-Newton

Matla a tloaelehileng (N) = w _y = 9.8√3 Newtons

Matla a khohlano e sa fetoheng (fs ₎ = (0.2)(9.8√3) = 1.96√3 Newton = 3.39 Newton

Matla a khohlano ea kinetic (f _k ) = (0.1)(9.8√3) = 0.98√3 Newton = 1.69 Newton

Tharollo:

Ntho e phomotse haeba w _x < _fs , ntho e ntse e theohela tlase haeba w _x > _fs.

w _x = 9.8 Li-Newton le f _s = 3.39 Li-Newton.

(a) matla a letlooa

∑ F = w _x – f _k = 9.8 – 1.69 = 8.11 Li-Newton

(b) boholo le tataiso ea ho potlaka

∑ F = ma

8.11 = (2) a

= 4.05

Boholo ba ho potlaka = 4.05 m/s ² mme tataiso ya ho potlaka = ho theohela tlase.

2. Boima ba ntho = 4 kg, ho potlaka ka lebaka la matla a khoheli = 9,8 m/s ² . Koefficient ea kinetic friction = 0.2 le coefficient ea static friction = 0.4. Boholo ba matla F = 40 Newtons. Ntho e phomotse kapa e thella tlase? Haeba ntho e thella tlase, fumana (a) matla a net (b) boholo le tataiso ea potlakiso!

tharollo

Tse tsejoang:

Boima (m) = 4 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s ²

Koefficient ea khohlano e sa fetoheng ( μs ₎ = 0.4

Koefficient ea ho ferekana ha kinetic ( μk ₎ = 0.2

Boima (w) = mg = (4)(9.8) = 39.2 Li-Newton

Karolo e rapameng ea boima (w _x ) = w sin 30 ^o = (39.2)(0.5) = 19.6 Newtons

Karolo e otlolohileng ea boima (w _y ) = w cos 30 ^o = (392)(0..5√3) = 19.6√3 Li-Newton

Matla a tloaelehileng (N) = w _y = 19.6√3 Newtons = 33.95 Newtons

matla a khohlano a sa fetoheng (fs ₎ = μ _s N = (0,4)(33.95) = 13.58 Newtons

Matla a ho qabana ha kinetic (f _k ) = μ _k N = (0.2)(33.95) = 6.79 Newtons

F = 40 Newton

Tharollo:

Ntho e thella tlase haeba F < w _x + fs _. Ntho e thella holimo haeba F > w _x + _fs.

F = 40 Newtons, w _x = 19.6 Newtons le f _s = 13.58 Newtons.

F e kholo ho feta w _x + f _s kahoo ntho e thellela holimo.

(a) Matla a letlooa

∑ F = F – w _x – f _k = 40 – 19.6 – 6.79 = 13.61 Li-Newton

(b) Boholo le tataiso ea ho potlaka

∑ F = ma

6.4 = (4) a

= 1.6

Boholo ba ho potlaka ke 1.6 m/s ² mme tataiso ya ho potlaka e hodimo.

[ID ea sephutheloana sa wpdm='481′]

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