Exempla quaestionum de determinandis componentibus vectoribus

Duo exempla quaestionum de determinandis componentibus vectoribus

1. Vector vis F = 20 Newton angulum 30 ^° cum axem x positivum format. Componentes vectoris F in axem x (F _{_x} ) et axem y (F_{_y} ) determina.

Disputatio
F_x = F cos 30^o = (20)(cos 30)^o) = (20)(0,5√³) = 10√³ Newtoni
F_y = F sin 30^o = (20)(sin 30^o) = (20)(0,5) = 10 Newtoni

2. Vector F₁ ₌ 20 Newton angulum 30 ^° cum axe y positivo et F₂ ₌ 30 Newton angulum 60 ^° cum axe x negativo format. Componentes vectorum F₁ _et F₂ _in axe x et in axe y determina.

Disputatio
F_1x F =₁ XXX cos^o = (20)(cos 60)^o) = (20)(0,5) = -10 Newton (negativum quia in directione x negativa est)
F_2x F =₂ XXX cos^o = (30)(cos 60)^o) = (30)(0,5) = -15 Newton (negativum quia in directione x negativa est)
F_1y F =₁ peccatum 60^o = (20)(sin 60^o) = (20)(0,5√³) = 10√³ Newton (positivum quia in directione y positiva est)
F_2y F =₂ peccatum 60^o = (30)(sin 60^o) = (30)(0,5√3) = -15√3 Newton (negativum quia in directione y negativa est)

3. F₁ ₌ 2 N, F₂ ₌ 4 N, F₃ ₌ 6 N. Componentes vectorum F₁ _, F₂ _et F₃ _in axibus x et y determina!

Disputatio
F_1x F =₁ XXX cos^o = (2)(cos 60)^o) = (2)(0,5) = 1 Newtoni (positivum quia in directione positiva x est)
F_2x F =₂ XXX cos^o = (4)(cos 30)^o) = (4)(0,5√3) = -2√3 Newton (negativum quia in directione x negativa est)
F_3x F =₃ XXX cos^o = (6)(cos 60)^o) = (6)(0,5) = 3 Newtoni (positivum quia in directione positiva x est)
F_1y F =₁ peccatum 60^o = (2)(sin 60^o) = (2)(0,5√3) = √3 Newton (positivum quia in directione y positiva est)
F_2y F =₂ peccatum 30^o = (4)(sin 30^o) = (4)(0,5) = 2 Newtoni (positivum quia in directione y positiva est)
F_3y F =₃ peccatum 60^o = (6)(sin 60^o) = (6)(0,5√3) = -3√3 Newton (negativum quia in directione y negativa est)

Quaestiones de determinandis componentibus vectoribus

1. Vector vis F = 8 Newton angulum 60° format.^o ad axem x positivum. Determina componentes vectoris F in axe x (F_x) et axis y (F_y).
2. Vector F₁ = 8 Newtoni angulum 45 formant^o ad axem y positivum et F₂ = 10 Newtoni angulum 30 formant^o ad axem x negativum. Determina componentes vectoris F₁ et F.₂ in axe x et in axe y.
3. F₁ = 4 N, F₂ = 8 N, F₃ = 12 N. Componentes vectoris F determina.₁F₂ et F.₃ in axibus x et axibus y!

[Anglice: Solvendo problemata vectoria – determinando componentes vectores ]