Ngā rapanga kua whakatauhia i roto i ngā whārite – te whakatau i ngā wāhanga whārite
1. Ka hangaia he koki 30 ° ki te tuaka-x e te kaha o te 20 Newton. Kimihia ngā wāhanga x me te y o te kaha.
otinga
F x = F cos 30 o = (20)(cos 30 o ) = (20)(0.5 √ 3 ) = 10 √ 3 Ngā Newton
F y = F hara 30 o = (20)(hara 30 o ) = (20)(0.5) = 10 Newtons
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2. Ka hangaia e F 1 = 20 he koki 30 o a Newton me te tuaka y, ā, ka hangaia e F 2 = 30 he koki 60 o a Newton me te tuaka -x. Kimihia ngā wāhanga x me y o F 1 me F 2.
otinga
F 1x = F 1 cos 60 o = (20)(cos 60 o ) = (20)(0.5) = -10 Newton (kino nā te mea he rite te ahunga ki te tuaka -x)
F 2x = F 2 cos 60 o = (30)(cos 60 o ) = (30)(0.5) = -15 Newton (kino nā te mea he rite te ahunga ki te tuaka -x)
F1y =F1 hara 60o = (20)(hara 60o) = (20)(0.5√3) = 10√3 Newton (he pai nā te mea he rite te ahunga ki te tuaka-y)
F2y =F2 hara 60o = (30)(hara 60o) = (30)(0.5√3) = -15√3 Newton (kino nā te mea he rite te ahunga ki te tuaka -y)
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3. F1 = 2 N, F2 = 4 N, F3 = 6 N. Kimihia ngā wāhanga x me y o F1 , F2 me F3!
otinga
F 1x = F 1 cos 60 o = (2)(cos 60 o ) = (2)(0.5) = 1 Newton (he pai nā te mea he rite te ahunga ki te tuaka x)
F2x =F2 cos 30o = (4)(cos 30o) = (4)(0.5√3) = -2√3 Newton (kino nā te mea he rite te ahunga ki te tuaka -x)
F 3x = F 3 cos 60 o = (6)(cos 60 o ) = (6)(0.5) = 3 Newton (he pai nā te mea he rite te ahunga ki te tuaka x)
F1y =F1 hara 60o = (2)(hara 60o) = (2)(0.5√3) = √3 Newton (he pai nā te mea he rite te ahunga ki te tuaka-y)
F 2 y = F 2 sin 3 0 o = (4)(sin 30 o ) = (4)(0.5) = 2 Newton (he pai nā te mea he rite te ahunga ki te tuaka y)
F3y =F3 hara 60o = (6)(hara 60o) = (6)(0.5√3) = -3√3 Newton (kino nā te mea he rite te ahunga ki te tuaka -y)
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