10 Dynamics Ntho e hokahaneng ka thapo hodima pulley Mochini wa Atwood – Mathata le Ditharollo
1. Sekotwana sa A se nang le boima ba 5 kg, se behilweng hodima sefofane se bataletseng se bataletseng. Sekotwana sa B se nang le boima ba 3 kg se leketlileng pheletsong e 'ngoe ea thapo e hokahantsoeng le sekotwana sa A hodima pulley. Ho potlaka ka lebaka la matla a khoheli ke 10 m/s 2 . Ho potlaka ha dikotwana ka bobedi ke eng ?
Tse tsejoang:
Boima ba boloko A (mA) = 5 kg
Boima ba boloko B (m B ) = 3 kg
Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
Boima ba boloko B (w B ) = m B g = (3)(10) = 30 Newtons
Ho batloa: Ho potlakisa diboloko ka bobedi (a)
Tharollo:
Sefofane se bataletseng se boreleli kahoo ha ho na matla a khohlano. Matla a potlakisang li-block ka bobeli ke boima ba block B.
ΣF = ma
w B = (m A + m B ) a
30 = (5 + 3) a
30 = 8 a
a = 30/8
a = 3.75 m/s 2
2.
Ho latela setšoantšo se ka holimo, 
(1) ho potlaka ha ntho = 0
(2) ntho e tsamaya ka lebelo le sa fetoheng
(3) ntho e phomotseng
(4) ntho e a tsamaya haeba boima ba ntho bo le bonyenyane ho feta matla a hulang ntho eo.
Tharollo:
(1) Ho potlaka ha ntho = 0.
Matla a marang-rang:
∑ F = ma –> ho potlaka (a) = 0
∑ F = 0
F 1 + F 2 – F 3 = 12 + 24 – 36 = 36 – 36 = 0 N
(2) Ntho e tsamaya ka lebelo le sa fetoheng
Ho se potlakise ho bolela ntho e phomotseng kapa e tsamayang ka lebelo le sa fetoheng.
(3) ntho e phomotseng
Ha ho matla a letlooa ho bolela ntho e phomotseng.
(4) ntho e a tsamaya haeba boima ba ntho bo le bonyenyane ho feta matla a hulang ntho eo.
Boima bo sebetsa ka lehlakoreng le otlolohileng, ha matla a ho hula a sebetsa ka lehlakoreng le otlolohileng.
Ntho e tsamaya ka tsela e otlolohileng kahoo ke matla a otlolohileng feela a sebetsang hodima ntho eo
3. Haeba coefficient ea kinetic friction pakeng tsa block A le bokaholimo ba tafole e le 0.1. Potlako e bakoang ke matla a khoheli ke 10 m/s2, joale matla a sebetsa holim'a block A e le hore sistimi e tsamaee ka letsohong le letšehali ka 2 m/s 2.
Tse tsejoang:
Boima ba boloko A (mA) = 30 kg
boima ba boloko A (w A ) = (30 kg)(10 m/s 2 ) = 300 kg m/s 2 kapa 300 Newtons
Boima ba boloko B (m B ) = 20 kg
boima ba boloko B (w B ) = (20 kg)(10 m/s 2 ) = 200 kg m/s 2 kapa 200 Newtons
Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
Koefficient ea kinetic friction ( μk ) = 0.1
Ho potlaka ha sistimi (a) = 2 m/s 2 (ho ya ka letsohong le letshehadi)
Matla a khohlano ea kinetic (f k ) = μ k N = μ k w A = (0.1)(300) = 30 Newtons
Ho batloa: Boholo ba matla F
Tharollo:
Molao oa bobeli oa Newton:
ΣF = ma
Ntho A e ya ka letsohong le letshehadi:
F – f k – w B = (m A + m B ) a
F – 30 – 200 = (30 + 20)(2)
F – 230 = (50)(2)
F – 230 = 100
F = 230 + 100
F = 330 Newton
4. Lintho tse peli, A = 2 kg le B = 6 kg, li khomaretsoe pheletsong e 'ngoe ea thapo holim'a pulley, joalo ka ha ho bontšitsoe setšoantšong se ka tlase. Haeba ho potlaka ka lebaka la matla a khoheli e le 10 ms -2 joale ho potlaka ha ntho B ke eng.
Tse tsejoang:
Boima ba ntho A (mA) = 2 kg, mB = 6 kg, g = 10 m/s2
boima ba ntho A (w A ) = (m A )(g) = (2)(10) = 20 N
boima ba ntho B (w B ) = (m B )(g) = (6)(10) = 60 N
Ho batloa: Ho potlakisa ntho b (ho potlakisa ha sistimi).
Tharollo:
w B > w A e le hore ntho B e theohele tlase, ntho A e theohele holimo
ΣF = ma
w B – w A = (m A + m B ) a
60 – 20 = (2 + 6) a
40 = (8) a
a = 5 m/s 2
5. Lintho tse peli tse hokahantsoeng ka thapo holim'a pulley e boreleli, joalo ka ha ho bontšitsoe setšoantšong se ka tlase. Haeba m 1 = 1 kg, m 2 = 2 kg, 'me ho potlaka ka lebaka la matla a khoheli ke 10 ms -2 , joale matla a khatello T ke afe.
Tse tsejoang:
Boima ba ntho 1 (m1) = 1 kg
Boima ba ntho 2 (m2 ) = 2 kg
Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
boima ba ntho 1 (w 1 ) = m 1 g = (1 kg)(10 m/s 2 ) = 10 kg m/s 2 kapa 10 Li-Newton
boima ba ntho 2 (w 2 ) = m 2 g = (2 kg)(10 m/s 2 ) = 20 kg m/s 2 kapa 20 Newtons
Ho batloa: Matla a tsitsipano (T)?
Tharollo:
w 2 > w 1 kahoo m 2 e theohela tlase , m 1 e nyolohela hodimo.
Molao oa bobeli oa Newton oa motsamao :
ΣF = ma
w 2 – w 1 = (m 1 + m 2 ) a
20 – 10 = (1 + 2) a
10 = (3) a
a = 3.3 m/s 2
Ho potlaka ha sistimi = 3.3 m/s 2.
m2 e theohela tlase:
w 2 – T 2 = m 2 a
20 – T 2 = (2)(3.33)
20 – T 2 = 6.66
T 2 = 20 – 6.66
T 2 = 13.3 Newton
m 1 e nyolohela hodimo:
T 1 – w 1 = m 1 a
T 1 – 10 = (1)(3.3)
T 1 – 10 = 3.33
T 1 = 10 + 3.33
T 1 = 13.3 Newton
Matla a kgatello (T) = 13.3 Newtons.
6. Boima ba m 1 = 6 kg le boima ba m 2 = 4 kg. Bokaholimo bo bataletseng bo boreleli. Ho potlaka ka lebaka la matla a khoheli ke 10 m/s 2 . Potlako ea sistimi ke efe?
Tse tsejoang:
Boima ba m1 = 6 kg
boima ba m2 = 4 kg
ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
boima ba w 1 = m 1 g = (6 kg)(10 m/s 2 ) = 60 kg m/s 2 kapa 60 Li-Newton
boima ba w 2 = m 2 g = (4 kg)(10 m/s 2 ) = 40 kg m/s 2 kapa 40 Li-Newton
Ho batloa: Ho potlaka ha sistimi (a)
Tharollo:
m 1 holim'a sefofane se boreleli se otlolohileng ntle le khohlano e le hore sistimi e potlakisoe ke boima ba boloko ba 2.
Sebelisa molao oa bobeli oa Newton:
∑ F = ma
w 2 = (m 1 + m 2 ) a
40 N = (6 kg + 4 kg) a
40 N = (10 kg) a
a = 40 N / 10 kg
a = 4 m/s 2
7. Liboloko tse peli, boloko ka 'ngoe e na le boima ba 2 kg, e hokahantsoe ka thapo holim'a pulley, joalo ka ha ho bontšitsoe setšoantšong se ka tlase. Sepakapaka se otlolohileng le pulley li boreleli. Haeba boloko B e huloa ke matla a rapameng a 40 Newtons, joale ho potlakisa boloko ke eng. Ho potlaka ka lebaka la matla a khoheli ke 10 m/s 2.
Tse tsejoang:
boima ba boloko A (mA) = boima ba boloko B (mB) = 2 kg
Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
Matla a F = 40 N
boima ba ntho A (w A ) = mg = (2)(10) = 20 N
Ho batloa: Ho potlaka ha sistimi (a)?
Tharollo:
Sebelisa molao oa bobeli oa Newton:
∑ F = ma
F – w A = (m A + m B ) a
40 – 20 = (2 + 2) a
20 = (4) a
a = 20/4
a = 5 m/s 2
8. Boima ba boloko A = 2 kg le boima ba boloko B = 1 kg. Boloko B qalong ha e phomotse, ebe e potlakiswa ho ya tlase ho fihlela e fihla fatshe. Potlako ka lebaka la matla a khoheli ke 10 m/s 2 . Boholo ba matla a kgatello ke bofe.
Tse tsejoang:
Boima ba boloko A (mA) = 2 kg
Boima ba boloko B (m B ) = 1 kg
Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
Boima ba boloko B (w B ) = m B g = (1)(10) = 10 Li-Newton
Ho batloa: Boholo ba matla a khatello ea maikutlo (T)
Tharollo:
Hlokomoloha matla a khohlano.
Ho potlaka ha sistimi (a)
∑ F = ma
w B = (m A + m B ) a
10 = (2 + 1) a
10 = 3 a
a = 10/3
Matla a khatello ea maikutlo (T)
Matla a tsitsipano bolokong A:
∑ F = ma
T = m A a = (2)(10/3) = 20/3 = 6.7 Li-Newton
Matla a kgatello hodima boloko B:
∑ F = ma
w B – T = m B a
10 – T = (1)(10/3)
10 – T = 3.3
T = 10 – 3.3 = 6.7 Newton
Matla a khatello ea maikutlo (T) = 6.7 Newtons
9. Boima ba boloko A = 2 kg mme boima ba boloko B = 1 kg. Matla a kgohlano pakeng tsa ntho A le sefofane se rapameng = 2.5 Newtons. Hlokomoloha kgohlano hodima pulley le thapo. Ho potlakisa ha diboloko ka bobedi ke eng?
Tse tsejoang:
Boima ba boloko A (mA) = 2 kg
Boima ba boloko B (m B ) = 1 kg
Matla a khohlano pakeng tsa senotlolo a le sefofane se rapameng (f kA ) = 2.5 Newtons
Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
Boima ba boloko (w B ) = m B g = (1)(10) = 10 Li-Newton
Ho batloa: Ho potlakisa diboloko ka bobedi (a)
Tharollo:
Sebelisa molao oa bobeli oa Newton:
∑ F = ma
w B – f k = (m A + m B ) a
10 – 2.5 = (2 + 1) a
7.5 = 3 a
a = 7.5/3
a = 2.5 m/s 2
10. Boima ba boloko a = 30 kg, phomotse sebakeng se bataletseng se hokahantsweng le boloko B ka boima ba 10 kg hodima pulley. Ho potlaka ha sistimi ke eng? Ho potlaka ka lebaka la matla a khoheli ke 10 ms -2.
Tse tsejoang:
Boima ba boloko A (mA) = 30 kg
Boima ba boloko B (m B ) = 10 kg
Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s 2
boima ba boloko B (w B ) = m B g = (10)(10) = 100 Newtons
Ho batloa: Ho potlaka ha sistimi (a)
Tharollo:
∑ F = ma
w B = (m A + m B ) a
100 = (30 + 10) a
100 = 40 a
a = 100/40
a = 2.5 m/s 2
