Pembahasan
Diketahui
Fa = 500 N
Fb = 100 N
Ï´ = 37°
Ditanya
a. Penjumlahan dari vektor tersebut
b. Selisih kedua vektor tersebut
Dijawab
a. Resultannya \begin{aligned} R&=&\ \sqrt{F_a^2+F_b^2+2F_aF_bcos\theta}\\ R&=&\ \sqrt{{500}^2+{100}^2+2\bullet500\bullet100cos37°}\\ R&=&\ \sqrt{250000+10000+2\bullet500\bullet100cos\left(0,8\right)}\\ R&=&\ \sqrt{260000+80000}\\ R&=&\ 100\sqrt{34}N\\ \end{aligned} b. Selisihnya \begin{aligned} R&=&\ \sqrt{F_a^2+F_b^2-2F_aF_bcos\theta}\\ R&=&\ \sqrt{{500}^2+{100}^2-2\bullet500\bullet100cos37°}\\ R&=&\ \sqrt{250000+10000-2\bullet500\bullet100cos\left(0,8\right)}\\ R&=&\ \sqrt{260000-80000}\\ R&=&\ 300\sqrt2N\\ \end{aligned}
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