1. He pouaka o papatipu Kei runga i te papa whakarara te 5 kg i te koki 30o. E tautokona ana te pouaka e te taura. Tāutuhia te kaha kume (T) me te kaha noa (N)!
otinga
∑Fx = 0
T – w hara 30o = 0
T = w sin 30o
T = (5 kg)(9.8 m/s)2) hara 30o
T = (49)(0.5)
T = 24.5 Ngā Niutona
∑Fy = 0
N – w cos 30o = 0
N = w cos 30o
N = (49)(0.87)
N = 43 Niutona
2. E rua ngā mea he taumaha m1 = m2 = 2 kg, e honoa ana e te aho kore-papatipu i runga i te pūrei kore-waku. Kimihia te kaha kume T1 a T2.
otinga
(a) Kauwhata tinana-kore mō te mea 1 (b) Kauwhata tinana-kore mō te mea 2
Whakamahia te ture tuatahi a Newton ki te mea 1:
∑Fy = 0
T1 - w1 = 0
T1 =w1 = m1 karamu = (2 kg)(9.8 m/s2) = 19.6 N
Anga Te ture tuatahi a Newton ki te mea 2:
∑Fy = 0
T2 - w2 = 0
T2 =w2 = m2 karamu = (2 kg)(9.8 m/s2) = 19.6 N
T1 =T2 = 19.6 N.
3. He mea nā taimaha wA = 30 N me tētahi mea taumaha wB = 40 N, e herea ana e te taura māmā e whiti ana i runga i te porotaka kore-waku o te papatipu iti noa iho. Whakatauhia te tauwehenga o te mōrahi te waku pumau i waenganui i te wB me te mata hianga, mēnā kei te okioki te pūnaha.
otinga
(a) Kauwhata tinana-kore mō te mea wA (b) Kauwhata tinana-kore mō te mea wB
Whakamahia te ture tuatahi a Newton ki te mea wA i te ahunga poutū (y):
∑Fy = 0 (kāore he whakaterenga i te ahunga poutū)
T – wA = 0
T = wA = 30 Niutona
Whakamahia te ture tuatahi a Newton ki te mea wB i te ahunga poutū (y) :
∑Fy = 0
N – wB whaimana 45o = 0
N = wB whaimana 45o = (40)(0.7) = 28 Ngā Newton
Whakamahia te ture tuatahi a Newton ki te mea wB i te ahunga whakapae (x):
∑Fx = 0
Fk + wB hara 45o – T = 0
μs N + wB hara 45o – T = 0
μs (28) + (40)(0.7) – 30 = 0
μs (28) + 28 – 30 = 0
μs (28) = 30 – 28
μs (28) = 2
μs = 2/28
μs = 0.07
Ko te tauwehenga o te waku pūmau mōrahi i waenga i te wB me te mata piko = 0.07.
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- Ngā matūriki i roto i te taurite kotahi-ahu
- Ngā matūriki i roto i te taurite rua-ahu
- Te taurite o ngā tinana e honoa ana e ngā taura me ngā pūwero
- Te taurite o ngā tinana i runga i te papa whakarara
