Tekanyo ea tsitsipano ea thapo

Lipotso tse 3 mabapi le equation ea khatello ea thapo

1. Setšoantšo se ka tlase se bontša liboloko tse tharo, e leng A, B le C tse fumanehang sebakeng se bataletseng se bataletseng. Haeba boima ba A = 1 kg, boima ba B = 2 kg le boima ba C = 2 kg le F = 10 N, joale fumana karolelano ea khatello thapong pakeng tsa A le B le khatello thapong pakeng tsa B le C.

Tse tsejoang:Tekanyo ea khatello ea thapo 1

Boima ba A (m A ) = 1 kg

Boima ba B (m B ) = 2 kg

Boima ba C (mC ) = 2 kg

Matla a ho tsikinyeha (F) = 10 N

Ho batloa: T AB : T BC

tharollo:

Bala ho potlaka ha sistimi u sebelisa foromo ea Molao oa Bobeli oa Newton:

ΣF = ma

F = (m A + m B + m C ) a

10 = (1 + 2 + 2) a

10 = 5 a

a = 10/5

a = 2 m/s 2

Sebelisa foromo ea khatello ea thapo ho bala T AB

ΣF = ma

T AB = m A a = 1 (2) = 2 Newton

Sebelisa foromo ea khatello ea thapo ho bala T BC

ΣF = ma

T BC = (m A + m B ) a = (1 + 2) (2) = (3)(2) = 6 Newtons

2. Ntho A e nang le boima ba 6 kg le ntho B e nang le boima ba 3 kg di hokahantswe ka thapo jwalo ka ha ho bontshitswe. Haeba coefficient ya kgohlano e le 0.3 mme g = 10 m/s 2 , fumana lebelo la ntho le kgatello thapong ya boloko ka bong.

Tsejoa:Tekanyo ea khatello ea thapo 2

Boima ba ntho A (m A ) = 6 kg

Boima ba ntho B (m B ) = 3 kg

Koefficient ea khohlano ea boloko A (µ k ) = 0.3

Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2

Boima ba boloko A (w A ) = m A g = (6)(10) = 60 N

Matla a tloaelehileng holim'a block A (N A ) = w A = 60 N

Boima ba boloko B (w B ) = m B g = (3)(10) = 30 N

Ho batloa: Ho potlaka ha sistimi (a) le kgatello thapong (T)

tharollo:

Bala matla a ho hohlana a kinetic, ke hore, matla a ho hohlana ha block A e tsamaya:

F k = µ k N A = (0,3)(60) = 18 Newtons

Bala ho potlaka ha sistimi (a):

ΣF = ma

wB – F k = (m A + m B ) a

30 – 18 = (6 + 3) a

12 = 9 a

a = 12 / 9 = 1,3 m/s 2

Bala kgatello e ka hara thapo e bolokong A (T A ):

ΣF = ma

T A – F k = (m A ) a

T A – 18 = (6)(1,3)

T A – 18 = 7,8

TA = 7,8 + 18 = 25,8 Li-Newton

Bala kgatello thapong e hodima lehlasedi B (T B ):

ΣF = ma

w B – T B = m B (a)

30 – T B = 3 (1,3)

30 – T B = 3,9

T B = 30 – 3,9

T B = 26,1 Li-Newton

3. Lintho tse peli A le B tse nang le boima ba 5 kg le 3 kg li hokahantsoe ka pulley e se nang khohlano. Matla a P a sebelisoa pulley ka lehlakoreng le holimo. Haeba liboloko ka bobeli li phomotse fatše qalong, lebelo la block A ke lefe, haeba boholo ba P e le 60 N?

Fumana hape kgatello thapong hodima diboloko A le B.

Tsejoa:Tekanyo ea khatello ea thapo 3

Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2

Boima ba A (m A ) = 5 kg

Boima ba boloko A (w A ) = m A g = (5)(10) = 50 Newtons

Boima ba B (m B ) = 3 kg

Boima ba boloko B (w B ) = m B g = (3)(10) = 30 Newtons

Matla P = 60 N

Ho batloa: Ho potlakisa tsamaiso ea libalaka A le B (a) le tsitsipano thapong ea lehlaseli A (T A ) le lehlaseli B (T B )

tharollo:

Bala ho potlaka ha sistimi o sebedisa foromo ya Molao wa Bobedi wa Newton.

ΣF = ma

w A – w B = (m A + m B ) a

50 – 30 = (5 + 3) a

20 = 8 a

a = 20/8

a = 2,5 m/s 2

Sebelisa mokhoa oa matla a khatello ho bala khatello thapong

Kgatelelo thapong e bolokong A:

ΣF = ma

w A – T A = m A a

50 – T A = 5 (2,5)

50 – TA = 12,5

T A = 50 – 12,5 = 37,5 Li-Newton

Kgatelelo thapong e bolokong B:

ΣF = ma

T B – w B = m B a

T B – 30 = 3 (2,5)

T B – 30 = 7,5

T B = 7,5 + 30 = 37,5 Li-Newton