Soal dan Pembahasan Vektor
kelas 11 Kurikulum Merdeka
Soal Nomor 1
Dua buah vektor besarnya 4 N dan 7 N. Tentukanlah :
a. Besar resultan paling besar yang mungkin
b. Besar resultan paling kecil yang mungkin
c. Mungkinlah resultan bernilai : (i) 2 N, (ii) 3 N, (iii) 4 N, (iv) 7 N, (v) 11 N, (vi) 12 N
Jawab :
Soal a
Kemungkinan besar resultan terbesar, saat sudut apit kedua vektornya 0°, sehingga:
\begin{aligned} \left| R \right| &= \left|F_1 + F_2 \right| \\ &=\left|4 + 7 \right| \\ &= 11 \ N \end{aligned}
Soal b
Kemungkinan besar resultan terkecil, saat sudut apit kedua vektornya 180°, sehingga:
\begin{aligned} \left| R \right| &= \left|F_1 - F_2 \right| \\ &= \left|4 - 7 \right|\\ &= 3 \ N \end{aligned}
Soal c
Kemungkinan nilai resultan dia vektor memenuhi persamaan :
\left|F_1 - F_2 \right| ≤ R ≤ \left|F_1 + F_2 \right|, sehingga
Nilai resultan dua vektor tersebut yang mungkin 3 ≤ R ≤ 11.Jadi nilai resultan yang mungkin antara 3 N hingga 11 N.
Soal Nomor 2a. Besar resultan paling besar yang mungkin
b. Besar resultan paling kecil yang mungkin
c. Mungkinlah resultan bernilai : (i) 2 N, (ii) 3 N, (iii) 4 N, (iv) 7 N, (v) 11 N, (vi) 12 N
Jawab :
Soal a
\begin{aligned} \left| R \right| &= \left|F_1 + F_2 \right| \\ &=\left|4 + 7 \right| \\ &= 11 \ N \end{aligned}
Soal b
\begin{aligned} \left| R \right| &= \left|F_1 - F_2 \right| \\ &= \left|4 - 7 \right|\\ &= 3 \ N \end{aligned}
Soal c
Kemungkinan nilai resultan dia vektor memenuhi persamaan :
\left|F_1 - F_2 \right| ≤ R ≤ \left|F_1 + F_2 \right|, sehingga
Nilai resultan dua vektor tersebut yang mungkin 3 ≤ R ≤ 11.Jadi nilai resultan yang mungkin antara 3 N hingga 11 N.
(i) 2 N (tidak mungkin)
(ii) 3 N (mungkin)
(iii) 4 N (mungkin)
(iv) 7 N (mungkin)
(v) 11 N (mungkin)
(vi) 12 N (tidak mungkin)
(ii) 3 N (mungkin)
(iii) 4 N (mungkin)
(iv) 7 N (mungkin)
(v) 11 N (mungkin)
(vi) 12 N (tidak mungkin)
Dua buah vektor sebidang masing-masing besarnya 6 satuan dan 8 satuan, dan sudut apitnya 60°. Tentukan :
a. Besar dan arah resultan
b. Besar dan Arah selisih resultan
Diketahui :
\vec{A}=6 \ satuan
\vec{B}=8 \ satuan
\alpha = 60°
Ditanya :
a. Besar dan arah resultan
b. Besar dan Arah selisih resultan
Jawab:
Soal a
Besar resultan kedua vektor
\begin{aligned} \left | R \right | &=\sqrt{{\left | \vec{A} \right |}^2+{\left | \vec{B} \right |}^2+2 \cdot \left | \vec{A} \right | \cdot \left | \vec{B} \right | \cdot cos \alpha }\\ &=\sqrt{{\left | 6 \right |}^2+{\left | 8 \right |}^2+2 \cdot \left | 6 \right | \cdot \left | 8 \right | \cdot cos \ 60°} \\ &= \sqrt{36+64+\left(2 \cdot 6 \cdot 8 \cdot \frac{1}{2} \right)} \\ &= \sqrt{100+48} \\ &= \sqrt{148} \\ &= 2 \sqrt{37} \\ &= 12,16 \ satuan \end{aligned}
Arah resultan
\begin{aligned} \frac{\left | \vec{R} \right |}{sin \ \alpha} &= \frac{\left | \vec{B} \right |}{sin \ \beta} \\ \frac{\left | 2 \sqrt{37} \right |}{sin \ 60°} &= \frac{\left | 8 \right |}{sin \ \beta} \\ \frac{2 \sqrt{37}}{\frac{1}{2} \sqrt{3}} &= \frac{8}{sin \ \beta} \\ sin \beta &= \frac{\frac{1}{2} \sqrt{2} \cdot 8}{2 \sqrt{37}} \\ &= 0,569 \\ \beta &= 34,715° \end{aligned}
Soal b
Besar selisih kedua vektor
\begin{aligned} \left | R \right | &=\sqrt{{\left | \vec{A} \right |}^2+{\left | \vec{B} \right |}^2-2 \cdot \left | \vec{A} \right | \cdot \left | \vec{B} \right | \cdot cos \ \alpha }\\ &=\sqrt{{\left | 6 \right |}^2+{\left | 8 \right |}^2-2 \cdot \left | 6 \right | \cdot \left | 8 \right | \cdot cos \ 60° }\\ &= \sqrt{36+64-\left(2 \cdot 6 \cdot 8 \cdot \frac{1}{2} \right)} \\ &= \sqrt{100-48} \\ &= \sqrt{52} \\ &= 2 \sqrt{13} \\ &= 7,21 \ satuan \end{aligned}
atau
\begin{aligned} \left | R \right | &=\sqrt{{\left | \vec{A} \right |}^2+{\left | \vec{B} \right |}^2+2 \cdot \left | \vec{A} \right | \cdot \left | \vec{B} \right | \cdot cos \alpha }\\ &=\sqrt{{\left | 6 \right |}^2+{\left | 8 \right |}^2+2 \cdot \left | 6 \right | \cdot \left | 8 \right | \cdot cos \ 120° }\\ &= \sqrt{36+64+\left(2 \cdot 6 \cdot 8 \cdot -\frac{1}{2} \right)} \\ &= \sqrt{100-48} \\ &= \sqrt{52} \\ &= 2 \sqrt{13} \\ &= 7,21 \ satuan \end{aligned}
Arah resultan
\begin{aligned} \frac{\left | \vec{R} \right |}{sin \ 120°} &= \frac{\left | \vec{B} \right |}{sin \ \beta} \\ \frac{\left | 2 \sqrt{13} \right |}{\frac{1}{2} \sqrt{3}} &= \frac{\left | 8 \right |}{sin \ \beta} \\ sin \beta &= \frac{\frac{1}{2} \sqrt{3} \cdot 8}{2 \sqrt{13}} \\ &= 0,961 \\ \beta &= 73,9° \end{aligned}
Soal Nomor 3a. Besar dan arah resultan
b. Besar dan Arah selisih resultan
Diketahui :
\vec{A}=6 \ satuan
\vec{B}=8 \ satuan
\alpha = 60°
Ditanya :
a. Besar dan arah resultan
b. Besar dan Arah selisih resultan
Jawab:
Soal a
\begin{aligned} \left | R \right | &=\sqrt{{\left | \vec{A} \right |}^2+{\left | \vec{B} \right |}^2+2 \cdot \left | \vec{A} \right | \cdot \left | \vec{B} \right | \cdot cos \alpha }\\ &=\sqrt{{\left | 6 \right |}^2+{\left | 8 \right |}^2+2 \cdot \left | 6 \right | \cdot \left | 8 \right | \cdot cos \ 60°} \\ &= \sqrt{36+64+\left(2 \cdot 6 \cdot 8 \cdot \frac{1}{2} \right)} \\ &= \sqrt{100+48} \\ &= \sqrt{148} \\ &= 2 \sqrt{37} \\ &= 12,16 \ satuan \end{aligned}
Arah resultan
\begin{aligned} \frac{\left | \vec{R} \right |}{sin \ \alpha} &= \frac{\left | \vec{B} \right |}{sin \ \beta} \\ \frac{\left | 2 \sqrt{37} \right |}{sin \ 60°} &= \frac{\left | 8 \right |}{sin \ \beta} \\ \frac{2 \sqrt{37}}{\frac{1}{2} \sqrt{3}} &= \frac{8}{sin \ \beta} \\ sin \beta &= \frac{\frac{1}{2} \sqrt{2} \cdot 8}{2 \sqrt{37}} \\ &= 0,569 \\ \beta &= 34,715° \end{aligned}
Soal b
\begin{aligned} \left | R \right | &=\sqrt{{\left | \vec{A} \right |}^2+{\left | \vec{B} \right |}^2-2 \cdot \left | \vec{A} \right | \cdot \left | \vec{B} \right | \cdot cos \ \alpha }\\ &=\sqrt{{\left | 6 \right |}^2+{\left | 8 \right |}^2-2 \cdot \left | 6 \right | \cdot \left | 8 \right | \cdot cos \ 60° }\\ &= \sqrt{36+64-\left(2 \cdot 6 \cdot 8 \cdot \frac{1}{2} \right)} \\ &= \sqrt{100-48} \\ &= \sqrt{52} \\ &= 2 \sqrt{13} \\ &= 7,21 \ satuan \end{aligned}
atau
\begin{aligned} \left | R \right | &=\sqrt{{\left | \vec{A} \right |}^2+{\left | \vec{B} \right |}^2+2 \cdot \left | \vec{A} \right | \cdot \left | \vec{B} \right | \cdot cos \alpha }\\ &=\sqrt{{\left | 6 \right |}^2+{\left | 8 \right |}^2+2 \cdot \left | 6 \right | \cdot \left | 8 \right | \cdot cos \ 120° }\\ &= \sqrt{36+64+\left(2 \cdot 6 \cdot 8 \cdot -\frac{1}{2} \right)} \\ &= \sqrt{100-48} \\ &= \sqrt{52} \\ &= 2 \sqrt{13} \\ &= 7,21 \ satuan \end{aligned}
Arah resultan
\begin{aligned} \frac{\left | \vec{R} \right |}{sin \ 120°} &= \frac{\left | \vec{B} \right |}{sin \ \beta} \\ \frac{\left | 2 \sqrt{13} \right |}{\frac{1}{2} \sqrt{3}} &= \frac{\left | 8 \right |}{sin \ \beta} \\ sin \beta &= \frac{\frac{1}{2} \sqrt{3} \cdot 8}{2 \sqrt{13}} \\ &= 0,961 \\ \beta &= 73,9° \end{aligned}
Besar resultan dua buah gaya adalah 6 \sqrt{2} \ N dan mengapit
sudut 45° terhadap salah satu gaya yang besarnya (6+6 \sqrt{2}) \ N.
Hitunglah :
a. Besar gaya lainnya
b. Sinus sudut apit kedua gaya tersebut
Diketahui :
\vec{R}=6 \sqrt{2} \ N
\beta = 45°
\vec{F_1}=(6+6 \sqrt{2}) \ N
Ditanya :
a. \vec{F_1}= .... \ Nb. \alpha = ...
Jawab :
Soal a
\begin{aligned} \vec{F_2} &= \vec{F_1} - \vec{R} \\ \left
| \vec{F_2} \right | &=\sqrt{{\left | \vec{F_1} \right |}^2+{\left |
\vec{R} \right |}^2-2 \cdot \left | \vec{F_1} \right | \cdot \left |
\vec{R} \right | \cdot cos \beta }\\ &=\sqrt{\left (6+6\sqrt{2}
\right )^2+\left ( 6 \sqrt{2} \right )^2-\left [ 2 \cdot \left
(6+6\sqrt{2} \right ) \cdot \left ( 6 \sqrt{2}\right ) \cdot cos \ 45°
\right ]} \\ &=\sqrt{\left (6+6\sqrt{2} \right )^2+\left ( 6
\sqrt{2} \right )^2-\left [ 2 \cdot \left (6+6\sqrt{2} \right ) \cdot
\left (6 \sqrt{2}\right ) \cdot \left (\frac{1}{2} \sqrt{2} \right )
\right ]} \\ &=\sqrt{\left (36+72\sqrt{2}+72 \right )+72 -\left (72+72\sqrt{2} \right )} \\ &=\sqrt{36+72\sqrt{2}+72 +72 -72-72\sqrt{2}} \\ &=\sqrt{36+72} \\ &=\sqrt{108} \\ &=6\sqrt{3} \\ &=10,392 \ N \end{aligned}
soal b
Sudut apit kedua gaya tersebut
\begin{aligned} \vec{R} &= \vec{F_1} + \vec{F_2} \\ \left | \vec{R} \right | &=\sqrt{{\left | \vec{F_1} \right |}^2+{\left | \vec{F_2} \right |}^2+2 \cdot \left | \vec{F_1} \right | \cdot \left | \vec{F_2} \right | \cdot cos \ \alpha }\\ 6 \sqrt{2} &=\sqrt{\left (6+6 \sqrt{2} \right )^2+{\left (6\sqrt{3} \right )}^2+\left [2 \cdot\left ( 6+6 \sqrt{2} \right ) \cdot \left ( 6\sqrt{3} \right ) \cdot cos \ \alpha \right ] }\\ \left ( 6 \sqrt{2} \right )^2 &=\left ( \sqrt{\left (6+6 \sqrt{2} \right )^2+{\left (6\sqrt{3} \right )}^2+\left [2 \cdot\left ( 6+6 \sqrt{2} \right ) \cdot \left ( 6\sqrt{3} \right ) \cdot cos \ \alpha \right ] } \right )^2 \\ 72 &= \left (6+6 \sqrt{2} \right )^2+{\left (6\sqrt{3} \right )}^2+\left [2 \cdot\left ( 6+6 \sqrt{2} \right ) \cdot \left ( 6\sqrt{3} \right ) \cdot cos \ \alpha \right ] \\ 72 &= \left (6+6 \sqrt{2} \right )^2+{\left (6\sqrt{3} \right )}^2+\left [2 \cdot\left ( 6+6 \sqrt{2} \right ) \cdot \left ( 6\sqrt{3} \right ) \cdot cos \ \alpha \right ] \\ 72 &= 209,823 +108 + \left (301,071 \cdot cos \ \alpha \right ) \\ 72 -317,823 &= 301,071 \cdot cos \ \alpha \\ -245,823 &=301,071 \cdot cos \ \alpha \\ cos \ \alpha &= \frac{-245,823}{301,071}\\ &= -0,816 \\ \alpha &= 144,735° \end{aligned}
Sehingga besar sinus sudut apit kedua gaya tersebut
\begin{aligned} sin \ 144,735° = 0,577 \end{aligned}
Soal Nomor 4Hitunglah :
a. Besar gaya lainnya
b. Sinus sudut apit kedua gaya tersebut
Diketahui :
\vec{R}=6 \sqrt{2} \ N
\beta = 45°
\vec{F_1}=(6+6 \sqrt{2}) \ N
Ditanya :
a. \vec{F_1}= .... \ Nb. \alpha = ...
Jawab :
Soal a
soal b
\begin{aligned} \vec{R} &= \vec{F_1} + \vec{F_2} \\ \left | \vec{R} \right | &=\sqrt{{\left | \vec{F_1} \right |}^2+{\left | \vec{F_2} \right |}^2+2 \cdot \left | \vec{F_1} \right | \cdot \left | \vec{F_2} \right | \cdot cos \ \alpha }\\ 6 \sqrt{2} &=\sqrt{\left (6+6 \sqrt{2} \right )^2+{\left (6\sqrt{3} \right )}^2+\left [2 \cdot\left ( 6+6 \sqrt{2} \right ) \cdot \left ( 6\sqrt{3} \right ) \cdot cos \ \alpha \right ] }\\ \left ( 6 \sqrt{2} \right )^2 &=\left ( \sqrt{\left (6+6 \sqrt{2} \right )^2+{\left (6\sqrt{3} \right )}^2+\left [2 \cdot\left ( 6+6 \sqrt{2} \right ) \cdot \left ( 6\sqrt{3} \right ) \cdot cos \ \alpha \right ] } \right )^2 \\ 72 &= \left (6+6 \sqrt{2} \right )^2+{\left (6\sqrt{3} \right )}^2+\left [2 \cdot\left ( 6+6 \sqrt{2} \right ) \cdot \left ( 6\sqrt{3} \right ) \cdot cos \ \alpha \right ] \\ 72 &= \left (6+6 \sqrt{2} \right )^2+{\left (6\sqrt{3} \right )}^2+\left [2 \cdot\left ( 6+6 \sqrt{2} \right ) \cdot \left ( 6\sqrt{3} \right ) \cdot cos \ \alpha \right ] \\ 72 &= 209,823 +108 + \left (301,071 \cdot cos \ \alpha \right ) \\ 72 -317,823 &= 301,071 \cdot cos \ \alpha \\ -245,823 &=301,071 \cdot cos \ \alpha \\ cos \ \alpha &= \frac{-245,823}{301,071}\\ &= -0,816 \\ \alpha &= 144,735° \end{aligned}
Sehingga besar sinus sudut apit kedua gaya tersebut
\begin{aligned} sin \ 144,735° = 0,577 \end{aligned}
Dua buah vektor memiliki besar yang sama. Berapakah sudut apit kedua
vektor tersebut jika hasil bagi selisih dan resultan kedua vektor
tersebut adalah \frac{1}{3} \sqrt{3}
Diketahui :
\left | \vec{F_1} \right | = F
\left | \vec{F_2} \right | = F
\frac{\left | \vec{F_1}-\vec{F_2} \right |}{\left | \vec{F_1}+\vec{F_2} \right |} = \frac{1}{3} \sqrt{3}
Ditanya :
\alpha = ...?
Jawab:
\begin{aligned} \frac{\sqrt{{\left | \vec{F_1} \right |}^2+{\left | \vec{F_2} \right |}^2-2 \cdot \left | \vec{F_1} \right | \cdot \left | \vec{F_2} \right | \cdot cos \ \alpha }}{\sqrt{{\left | \vec{F_1} \right |}^2+{\left | \vec{F_2} \right |}^2+2 \cdot \left | \vec{F_1} \right | \cdot \left | \vec{F_2} \right | \cdot cos \ \alpha }} &= \frac{1}{3} \sqrt{3} \\ \frac{\sqrt{F^2+F^2- \left( 2 \cdot F \cdot F \cdot cos \ \alpha \right )}}{\sqrt{F^2+F^2+ \left( 2 \cdot F \cdot F \cdot cos \ \alpha \right )}}&=\frac{1}{3} \sqrt{3} \\ \frac{\sqrt{F^2+F^2- \left( 2 \cdot F^2 \cdot cos \ \alpha \right ) }}{\sqrt{F^2+F^2+ \left( 2 \cdot F^2 \cdot cos \ \alpha \right )}}&=\frac{1}{3} \sqrt{3} \\ \frac{\left [ \sqrt{F^2+F^2- \left( 2 \cdot F^2 \cdot cos \ \alpha \right ) } \right ]^2}{\left [ \sqrt{F^2+F^2+ \left( 2 \cdot F^2 \cdot cos \ \alpha \right )} \right ]^2}&=\left [ \frac{1}{3} \sqrt{3} \right ]^2\\ \frac{F^2+F^2- \left( 2 \cdot F^2 \cdot cos \ \alpha \right)}{F^2+F^2+ \left( 2 \cdot F^2 \cdot cos \ \alpha \right)}&=\frac{3}{9} \\ \frac{2 \ F^2- \left( 2 \ F^2 \cdot cos \ \alpha \right)}{2 \ F^2+ \left( 2 \ F^2 \cdot cos \ \alpha \right)}&=\frac{1}{3} \\ 2 \ F^2+ 2 \ F^2 \cdot cos \ \alpha &= 6 \ F^2- 6 \ F^2 \cdot cos \ \alpha \\ 2 \ F^2 \cdot cos \ \alpha + 6 \ F^2 \cdot cos \ \alpha &= 6 \ F^2-2 \ F^2 \\ 8 \ F^2 \cdot cos \ \alpha &= 4 \ F^2-2 \ F^2\\ cos \ \alpha &= \frac{4 \ F^2}{8 \ F^2} \\ &= \frac{1}{2} \\ \alpha &= 60° \end{aligned}
Diketahui :
\left | \vec{F_1} \right | = F
\left | \vec{F_2} \right | = F
\frac{\left | \vec{F_1}-\vec{F_2} \right |}{\left | \vec{F_1}+\vec{F_2} \right |} = \frac{1}{3} \sqrt{3}
Ditanya :
\alpha = ...?
Jawab:
\begin{aligned} \frac{\sqrt{{\left | \vec{F_1} \right |}^2+{\left | \vec{F_2} \right |}^2-2 \cdot \left | \vec{F_1} \right | \cdot \left | \vec{F_2} \right | \cdot cos \ \alpha }}{\sqrt{{\left | \vec{F_1} \right |}^2+{\left | \vec{F_2} \right |}^2+2 \cdot \left | \vec{F_1} \right | \cdot \left | \vec{F_2} \right | \cdot cos \ \alpha }} &= \frac{1}{3} \sqrt{3} \\ \frac{\sqrt{F^2+F^2- \left( 2 \cdot F \cdot F \cdot cos \ \alpha \right )}}{\sqrt{F^2+F^2+ \left( 2 \cdot F \cdot F \cdot cos \ \alpha \right )}}&=\frac{1}{3} \sqrt{3} \\ \frac{\sqrt{F^2+F^2- \left( 2 \cdot F^2 \cdot cos \ \alpha \right ) }}{\sqrt{F^2+F^2+ \left( 2 \cdot F^2 \cdot cos \ \alpha \right )}}&=\frac{1}{3} \sqrt{3} \\ \frac{\left [ \sqrt{F^2+F^2- \left( 2 \cdot F^2 \cdot cos \ \alpha \right ) } \right ]^2}{\left [ \sqrt{F^2+F^2+ \left( 2 \cdot F^2 \cdot cos \ \alpha \right )} \right ]^2}&=\left [ \frac{1}{3} \sqrt{3} \right ]^2\\ \frac{F^2+F^2- \left( 2 \cdot F^2 \cdot cos \ \alpha \right)}{F^2+F^2+ \left( 2 \cdot F^2 \cdot cos \ \alpha \right)}&=\frac{3}{9} \\ \frac{2 \ F^2- \left( 2 \ F^2 \cdot cos \ \alpha \right)}{2 \ F^2+ \left( 2 \ F^2 \cdot cos \ \alpha \right)}&=\frac{1}{3} \\ 2 \ F^2+ 2 \ F^2 \cdot cos \ \alpha &= 6 \ F^2- 6 \ F^2 \cdot cos \ \alpha \\ 2 \ F^2 \cdot cos \ \alpha + 6 \ F^2 \cdot cos \ \alpha &= 6 \ F^2-2 \ F^2 \\ 8 \ F^2 \cdot cos \ \alpha &= 4 \ F^2-2 \ F^2\\ cos \ \alpha &= \frac{4 \ F^2}{8 \ F^2} \\ &= \frac{1}{2} \\ \alpha &= 60° \end{aligned}
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