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:
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 = (mA +mB +mC) 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 TAB
ΣF = ma
TAB = mA a = 1 (2) = 2 Newton
Sebelisa foromo ea khatello ea thapo ho bala TBC
ΣF = ma
TBC = (mA +mB) 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/s2, fumana hore na ntho e potlakile hakae le hore na e tsitsitse hakae lithapong tsa boloko ka 'ngoe.
Tsejoa:
Boima ba ntho A (mA) = 6 kg
Boima ba ntho B (mB) = 3 kg
Koefficient ea khohlano ea block A (µ)k= 0.3
Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
Boima ba boloko A (w)A) = mA g = (6)(10) = 60 N
Matla a tloaelehileng holim'a block A (N)A) = wA = 60N
Boima ba boloko B (wB) = mB 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:
Fk = µk NA = (0,3)(60) = 18 Newton
Bala ho potlaka ha sistimi (a):
ΣF = ma
wB – Fk = (mA +mB) a
30 – 18 = (6 + 3) a
12 = 9 a
a = 12 / 9 = 1,3 m/s 2
Bala kgatello e ka hara thapo hodima boloko A (T)A):
ΣF = ma
TA - Fk = (mA) a
T A – 18 = (6)(1,3)
T A – 18 = 7,8
TA = 7,8 + 18 = 25,8 Newton
Bala kgatello thapong e hodima lehlasedi B (T)B):
ΣF = ma
wB - TB = mB (A)
30 – TB = 3 (1,3)
30 – T B = 3,9
TB = 30 - 3,9
TB = 26,1 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:
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) = mA g = (5)(10) = 50 Newton
Boima ba B (m B ) = 3 kg
Boima ba boloko B (wB) = mB g = (3)(10) = 30 Newton
Matla P = 60 N
Oa Batla: Ho potlakisa tsamaiso ea libalaka A le B (a) le tsitsipano thapong ea lebala A (T)A) le lehlaseli B (TB)
tharollo:
Bala ho potlaka ha sistimi o sebedisa foromo ya Molao wa Bobedi wa Newton.
ΣF = ma
wA - wB = (mA +mB) 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 – TA = 5 (2,5)
50 – TA = 12,5
TA = 50 – 12,5 = 37,5 Newton
Kgatelelo thapong e bolokong B:
ΣF = ma
T B – w B = m B a
TB – 30 = 3 (2,5)
T B – 30 = 7,5
TB = 7,5 + 30 = 37,5 Newton
