Dua vektor gaya masing-masing 2N dan 20N diapit oleh sudut alfa sebesar 120°.
Tentukan besar resultan gaya kedua vektor gaya tersebut!
# bonus poin besar awal tahun #
Resultan dari Kedua Vektor Gaya Tersebut Sebesar 2√91 N.
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Diketahui:
F1 = 2 N
F2 = 20 N
a = 120°
(cos a = -1/2)
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Ditanya:
Resultan?
[tex] \bf = \sqrt{ f_{1} {}^{2} + f_{2} {}^{2} + 2. f_{1}. f_{2}cos \: \alpha } \\ \\ \bf = \sqrt{ {2}^{2} + {20}^{2} + 2 \times 2 \times 20 \times cos\: 120 {}^{ \circ} } \\ \\ \bf = \sqrt{2(2) + 20(20) + 4 \times 20 \times (- \frac{1} {2} )} \\ \\ \bf = \sqrt{4 + 400 + ( \frac{80 -1}{2} )} \\ \\ \bf = \sqrt{404 + (-40) } \\ \\ \bf = \sqrt{404 - 40} \\ \\ \bf = \sqrt{364 } \: n \\ \\ \bf = √4 × √91 = 2^{\cancel2}√91 \\ \\ \boxed{\bf = 2√91 \: N}[/tex]
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→ brainly.co.id/tugas/9994732
→ https://brainly.co.id/tugas/18737678
→ https://brainly.co.id/tugas/19911985
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(Koreksi Kak, Terimakasih)
PENYELESAIAN:
Vektor 1 (F1)=2 N
Vektor 2 (F2)=20 N
R= Resultan
maka, R adalah:
[tex]R = \sqrt{F1 + F2 + 2.F1.F2.cos( \alpha )} \\ = \sqrt{ {2}^{2} + {20}^{2} + 2.2.20.cos( {120}^{ \circ}) } \\ = \sqrt{4 + 400 + 4.20. - \frac{1}{2} } \\ = \sqrt{404 + 80. - \frac{1}{2} } \\ = \sqrt{404 + ( - 40)} \\ = \sqrt{364} \\ = \boxed{ \bf2 \sqrt{91} \: N }[/tex]
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