1. Kei runga i te papa piko tētahi pouaka he 5 kg te taumaha i te koki 30 ° . E tautokona ana te pouaka e tētahi taura. Tātaihia te kaha kume (T) me te kaha noa (N)!

otinga
∑Fx = 0
T – w sin 30 o = 0
T = w sin 30 o
T = (5 kg)(9.8 m/s 2 ) sin 30 o
T = (49)(0.5)
T = 24.5 Ngā Niutona
∑ F y = 0
N – w cos 30 o = 0
N = w cos 30 o
N = (49)(0.87)
N = 43 Niutona
[irp]
2. E rua ngā mea he taumaha m 1 = m 2 = 2 kg, e honoa ana e te aho kore taumaha i runga i te pūrei kore waku. Kimihia ngā kaha kume T 1 me T 2.

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:
∑ F y = 0
T 1 – w 1 = 0
T 1 = w 1 = m 1 karamu = (2 kg)(9.8 m/s 2 ) = 19.6 N
Whakamahia te ture tuatahi a Newton ki te mea 2:
∑ F y = 0
T 2 – w 2 = 0
T 2 = w 2 = m 2 karamu = (2 kg)(9.8 m/s 2 ) = 19.6 N
T 1 = T 2 = 19.6 N.
[irp]
3. He mea taumaha w A = 30 N me te mea taumaha w B = 40 N, e herea ana e te taura māmā e whiti ana i runga i te pūrei kore-waku he iti noa te papatipu. Whakatauhia te tauwehenga o te waku pūmau mōrahi i waenga i te w B me te mata piko, mena kei te okioki te pūnaha.

otinga

(a) Kauwhata tinana-kore mō te mea w A (b) Kauwhata tinana-kore mō te mea w B
Whakamahia te ture tuatahi a Newton ki te mea w A i te ahunga poutū (y):
∑ F y = 0 (kāore he whakaterenga i te ahunga poutū)
T – w A = 0
T = w A = 30 Newton
Whakamahia te ture tuatahi a Newton ki te mea w B i te ahunga poutū (y) :
∑ F y = 0
N – w B cos 45 o = 0
N = w B cos 45 o = (40)(0.7) = 28 Ngā Newton
Whakamahia te ture tuatahi a Newton ki te mea w B i te ahunga whakapae (x):
∑ F x = 0
F k + w B sin 45 o – T = 0
μ s N + w B hara 45 o – T = 0
μ s (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 w B me te mata piko = 0.07.
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