Roboguru

Tiga buah vektor gaya masing-masing F1 = 30 N, F2 = 70 N, dan F3 = 30 N, disusun seperti pada gambar di atas. Besar resultan ketiga vektor tersebut adalah ...

Pertanyaan

Tiga buah vektor gaya masing-masing F1 = 30 N, F2 = 70 N, dan F3 = 30 N, disusun seperti pada gambar di atas. Besar resultan ketiga vektor tersebut adalah ... space

  1. 0 Nspace

  2. 70 Nspace

  3. 85square root of 3 Nspace

  4. 85 Nspace

  5. 100 Nspace

Pembahasan Video:

Pembahasan Soal:

Diketahui

F subscript 1 equals 30 space straight N F subscript 2 equals 70 space straight N F subscript 3 equals 30 space straight N theta equals 60 degree

Ditanya : besar resultan ketiga vektor ?

Penyelesaian

Uraikan vektor terhadap sumbu x dan sumbu y

F subscript 1 x end subscript equals F subscript 1 cos 60 degree F subscript 1 x end subscript equals 30 times 1 half F subscript 1 x end subscript equals 15 space straight N  F subscript 1 y end subscript equals F subscript 1 sin 60 degree F subscript 1 y end subscript equals 30 times 1 half square root of 3 F subscript 1 y end subscript equals 15 square root of 3 space straight N  F subscript 2 x end subscript equals 70 space straight N F subscript 2 y end subscript equals 0 space straight N  F subscript 3 x end subscript equals F subscript 3 c o s 60 degree F subscript 3 x end subscript equals 30 times 1 half F subscript 3 x end subscript equals 15 space straight N  F subscript 3 y end subscript equals negative F subscript 3 s i n 60 degree F subscript 3 y end subscript equals negative 30 times 1 half square root of 3 F subscript 3 y end subscript equals negative 15 square root of 3 space straight N

Tentukan besar komponen sumbu x dan sumbu y

capital sigma F subscript x equals F subscript 1 x end subscript plus F subscript 2 x end subscript plus F subscript 3 x end subscript capital sigma F subscript x equals 15 plus 70 plus 15 capital sigma F subscript x equals 100 space straight N  capital sigma F subscript y equals F subscript 1 y end subscript plus F subscript 2 y end subscript plus F subscript 3 y end subscript capital sigma F subscript y equals 15 square root of 3 plus 0 plus open parentheses negative 15 square root of 3 close parentheses capital sigma F subscript y equals 0 space N

maka besar resultan ketiga vektor adalah

F subscript R equals square root of F subscript x squared plus F subscript y squared end root F subscript R equals square root of 100 squared plus 0 squared end root F subscript R equals square root of 10000 F subscript R equals 100 space straight N

Dengan demikian, besar resultan ketiga vektor adalah 100 N.

Jadi, jawaban yang benar adalah E. space

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 30 Agustus 2021

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Pertanyaan yang serupa

Sebuah balok ditarik oleh dua gaya seperti pada gambar berikut. Tentukan resultan gaya tersebut!

Pembahasan Soal:

Resultan gaya vektor dihitung menggunakan metode analitis adalah:

Hitung dahulu resultan gaya komponen x

table attributes columnalign right center left columnspacing 0px end attributes row cell begin inline style sum for blank of end style F subscript x end cell equals cell F subscript 1 times cos space 120 degree plus F subscript 2 end cell row blank equals cell 100 times open parentheses negative 1 half close parentheses plus 100 end cell row blank equals cell 50 space straight N end cell end table

Hitung resultan gaya komponen y

table attributes columnalign right center left columnspacing 0px end attributes row cell begin inline style sum for blank of end style F subscript y end cell equals cell space F subscript space 1 end subscript sin space 30 degree plus F subscript 2 end cell row blank equals cell 100 times 1 half plus 0 end cell row blank equals cell 50 space straight N end cell end table

Resultan gaya nya adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell begin inline style sum for blank of end style F end cell equals cell square root of open parentheses 50 close parentheses squared plus open parentheses 50 close parentheses squared end root end cell row blank equals cell square root of 2500 plus 2500 end root end cell row blank equals cell 70 comma 7 space straight N end cell end table

Sehingga besar resultan gaya nya dalah 70,7 N

0

Roboguru

Perhatikan gambar gaya-gaya di bawah ini! Besar resultan ketiga gaya tersebut adalah ....

Pembahasan Soal:

Diketahui:

F subscript 1 equals 3 space straight N F subscript 2 equals 3 space straight N F subscript 3 equals 6 space straight N  

Ditanya: R equals... ? 

Penyelesaian:

Uraikan ketiga gaya arah sumbu x dan sumbu y.

Ke kanan dan ke atas bernilai positif, sedangkan ke kiri dan ke bawah bernilai negatif.

F subscript 1 x end subscript equals F subscript 1 cos 60 degree equals 3.1 half equals 1 comma 5 space straight N F subscript 1 y end subscript equals F subscript 1 siin 60 degree equals 3.1 half square root of 3 equals 1 comma 5 square root of 3 space straight N F subscript 2 x end subscript equals negative F subscript 2 equals negative 3 space straight N F subscript 2 y end subscript equals 0 space straight N F subscript 3 x end subscript equals F subscript 3 cos 60 degree equals 6.1 half equals 3 space straight N F subscript 3 y end subscript equals negative F subscript 3 sin 60 degree equals negative 6.1 half square root of 3 equals negative 3 square root of 3 space straight N  

Gaya arah sumbu x:

table attributes columnalign right center left columnspacing 0px end attributes row cell F subscript x end cell equals cell F subscript 1 x end subscript plus F subscript 2 x end subscript plus F subscript 3 x end subscript end cell row blank equals cell 1 comma 5 minus 3 plus 3 end cell row blank equals cell 1 comma 5 space straight N end cell end table  

Gaya arah sumbu y:

table attributes columnalign right center left columnspacing 0px end attributes row cell F subscript y end cell equals cell F subscript 1 y end subscript plus F subscript 2 y end subscript plus F subscript 3 y end subscript end cell row blank equals cell 1 comma 5 square root of 3 plus 0 minus 3 square root of 3 end cell row blank equals cell negative 1 comma 5 square root of 3 space straight N end cell end table  

Resultan gaya:

table attributes columnalign right center left columnspacing 0px end attributes row R equals cell square root of F subscript x squared plus F y squared end root end cell row blank equals cell square root of open parentheses 1 comma 5 close parentheses squared plus open parentheses negative 1 comma 5 square root of 3 close parentheses squared end root end cell row blank equals cell square root of 2 comma 25 plus 6 comma 75 end root end cell row blank equals cell square root of 9 end cell row blank equals cell 3 space straight N end cell end table  

Dengan demikian, besar resultan ketiga gaya tersebut adalah 3,0 N.

Oleh karena itu, jawaban yang tepat adalah C.

0

Roboguru

Ditentukan 2 vektor yang sama besarnya. Jika perbandingan antara besar jumlah dan besar selisih kedua vektor sama dengan , sudut yang dibentuk antara kedua vektor itu adalah ...

Pembahasan Soal:

Penyelesaian soal menggunakan metode analitis. Metode analitis yaitu cara menjumlahkan vektor dengan memproyeksikan vektor-vektor pada sumbu-X dan sumbu-Y diagram cartesius, lalu komponen-komponen vektor pada masing-masing sumbu dijumlahkan secara biasa.

2 vektor yang sama besar diasumsikan sebagai F, maka:
F1 = F
F2 = F

Perbandingan ketika vektor dijumlahkan dan diselisihkan = square root of 3

Ketika vektor dijumlahkan
text R(+)= end text square root of text a end text to the power of 2 plus text b end text to the power of 2 plus 2 times text a·b·cos end text space theta end root R (+) equals square root of F subscript 1 to the power of 2 plus F subscript 2 to the power of 2 plus 2 times F subscript 1 times F subscript 2 times cos space theta end root R (+) equals square root of F to the power of 2 plus F to the power of 2 plus 2 times F times F times c o s space theta end root R (+) equals square root of 2 F to the power of 2 plus 2 F to the power of 2 times c o s space theta end root R (+) equals square root of 2 F to the power of 2 left parenthesis 1 plus c o s space theta right parenthesis end root

 

Ketika vektor diselisihkan:
R text (-) end text equals square root of text a end text to the power of text 2 end text end exponent text +b end text to the power of text 2 end text end exponent text -2·a·b·cos θ end text end root R text (-)= end text square root of F subscript 1 to the power of 2 plus F subscript 2 to the power of 2 minus 2 times F subscript 1 times F subscript 2 times c o s space theta end root R (-) equals square root of F to the power of 2 plus F to the power of 2 minus 2 times F times F times c o s space theta end root R (-) equals square root of 2 F minus 2 2 F to the power of 2 times c o s space theta end root R (-) equals square root of 2 F to the power of 2 left parenthesis 1 minus c o s space theta right parenthesis end root

 

Mencari sudut
fraction numerator R (+) over denominator R (-) end fraction equals square root of 3 fraction numerator square root of 2 F to the power of 2 left parenthesis 1 plus c o s space theta right parenthesis end root over denominator square root of 2 F to the power of 2 left parenthesis 1 minus c o s space theta right parenthesis end root end fraction equals square root of 3 fraction numerator 2 F to the power of 2 left parenthesis 1 plus c o s space theta right parenthesis over denominator 2 F to the power of 2 left parenthesis 1 minus c o s space theta right parenthesis end fraction equals 3 fraction numerator 1 plus c o s space theta over denominator 1 minus c o s space theta end fraction equals 3 1 plus c o s space theta equals 3 open parentheses 1 minus c o s space theta close parentheses 1 plus c o s space theta equals 3 minus 3 c o s space theta c o s space theta plus 3 c o s space theta equals 3 minus 1 4 c o s space theta equals 2 c o s space theta equals 2 over 4 c o s space theta equals 1 half

Nilai sudut dari dua vektor adalah cos space theta equals 1 half. Nilai cos yang mempunyai hasil 1 half adalah 60.

Dengan demikian, jawaban yang tepat adalah D.space

0

Roboguru

Perhatikan gambar percobaan vektor gaya resultan dengan menggunakan tiga neraca pegas beriku! Gambar yang sesuai dengan rumus vektor gaya resultan secara analisis adalah ...

Pembahasan Soal:

Rumus resultan vektor secara analisis adalah sebagai berikut:

R space equals space square root of A squared plus space B squared plus space 2 A B space cos space theta end root

Maka gambar yang sesuai dengan rumus vektor gaya resultan secara analisis adalah

Gambar 1 

table attributes columnalign right center left columnspacing 0px end attributes row cell R space end cell equals cell space square root of A squared plus space B squared plus space 2 A B space cos space theta end root end cell row blank equals cell space square root of 5 squared plus 5 squared plus 2 left parenthesis 5 right parenthesis left parenthesis 5 right parenthesis cos space left parenthesis 120 degree right parenthesis end root end cell row blank equals cell space square root of 5 squared plus 5 squared plus 2 left parenthesis 5 right parenthesis squared open parentheses negative 1 half close parentheses end root end cell row blank equals cell space square root of left parenthesis 5 squared right parenthesis end root end cell row blank equals cell 5 space straight N end cell end table

maka gambar 1 Benar.

Gambar 2

tan space beta equals 4 over 3 cos space beta equals 3 over 5 A r c space cos space 3 over 5 equals space 53 degree

maka, 180º - 53º = 127 º

table attributes columnalign right center left columnspacing 0px end attributes row cell R space end cell equals cell space square root of A squared plus space B squared plus space 2 A B space cos space theta end root end cell row blank equals cell space square root of 3 squared plus 5 squared plus 2 left parenthesis 3 right parenthesis left parenthesis 5 right parenthesis cos space left parenthesis 127 degree right parenthesis end root end cell row blank equals cell space square root of 9 plus 25 plus 30 left parenthesis negative 0 comma 6 right parenthesis end root end cell row blank equals cell space square root of 34 minus 18 end root end cell row blank equals cell square root of 16 space straight N end cell row blank equals cell 4 space straight N end cell end table

maka gambar 2 Benar.

Gambar 3

table attributes columnalign right center left columnspacing 0px end attributes row cell R space end cell equals cell space square root of A squared plus space B squared plus space 2 A B space cos space theta end root end cell row blank equals cell space square root of 8 squared plus 8 squared plus 2 left parenthesis 8 right parenthesis left parenthesis 8 right parenthesis cos space left parenthesis 90 degree right parenthesis end root end cell row blank equals cell space square root of 8 squared plus 8 squared plus 2 left parenthesis 8 right parenthesis left parenthesis 8 right parenthesis left parenthesis 0 right parenthesis end root end cell row blank equals cell space square root of 2 left parenthesis 8 squared right parenthesis end root end cell row blank equals cell 8 square root of 2 space straight N end cell end table

maka gambar 3 Salah

Oleh karena itu, jawabannya adalah C. space

0

Roboguru

Resultan dari penjumlahan dua buah vektor yang sama besarnya 3 satuan. Jika kedua vektor tersebut membentuk sudut  , nilai masing-masing vektor tersebut adalah ….

Pembahasan Soal:

open vertical bar straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top close vertical bar equals square root of open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared plus vertical line straight b with rightwards harpoon with barb upwards on top vertical line squared plus 2 open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar open vertical bar straight b with rightwards harpoon with barb upwards on top close vertical bar cos open parentheses straight theta close parentheses end root 3 equals square root of open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared plus open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared plus 2 open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared cos open parentheses 60 to the power of straight o close parentheses end root 9 equals open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared plus open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared plus 2 open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared left parenthesis 1 half right parenthesis 9 equals 3 open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared equals 3 open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar equals square root of 3

Jadi, vektor tersebut masing-masing besarnya adalah square root of 3 satuan

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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