UN Fisika 2016 Paket 2 Nomor 9

Perhatikan gambar berikut!

9.gif

Pada batang ABCD yang massanya diabaikan, bekerja tiga gaya. Momen gaya sistem dengan poros titik D adalah …

A.    7 N.m searah jarum jam D. 2 N.m berlawanan arah jarum jam
B.     7 N.m berlawanan arah jarum jam E. 1 N.m berlawanan arah jarum jam
C.     2 N.m searah jarum jam

Penyelesaian

Untuk mengerjakan soal ini, torka searah perputaran jarum jam bernilai – sedangkan torka berlawanan perputaran jarum jam bernilai +.

Total torka sistem dengan poros titik D adalah

\text{ }\!\!\Sigma\!\!\text{ }\tau ={{\tau }_{A}}+{{\tau }_{B}}+{{\tau }_{C}}

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=-\left( {{\text{F}}_{1}}\sin 30{}^\circ \times AD \right)+\left( {{F}_{2}}\times BC \right)-\left( {{F}_{3}}\times CD \right)

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=-\left( 6.\frac{1}{2}\times 3 \right)+\left( 10\times 2 \right)-\left( 4\times 1 \right)

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=-\left( 9 \right)+\left( 20 \right)-\left( 4 \right)

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=+7\text{ }\!\!~\!\!\text{ N}.\text{m}

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=7\text{ }\!\!~\!\!\text{ N}.\text{m }\!\!~\!\!\text{ berlawanan }\!\!~\!\!\text{ perputaran }\!\!~\!\!\text{ jarum }\!\!~\!\!\text{ jam}

jawaban B

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