UN Fisika 2016 Paket 3 Nomor 9

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9

Batang AB yang massanya diabaikan dipengaruhi tiga gaya {{F}_{A}}=10~N, {{F}_{C}}={{F}_{B}}=20~N seperti gambar. Besar momen gaya terhadap titik P …

A.    12 N.m D. 21 N.m
B.     15 N.m E. 24 N.m
C.     18 N.m

Penyelesaian

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

Total torka sistem dengan poros titik P adalah

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

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=-\left( {{F}_{A}}\times PA \right)+\left( {{F}_{C}}\sin 30{}^\circ \times PC \right)-\left( {{F}_{B}}\times PB \right)

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=-\left( 10\times 0,3 \right)+\left( 20\left( \frac{1}{2} \right)\times 0,3 \right)-\left( 20\times 0,9 \right)

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=-\left( 3 \right)+\left( 3 \right)-\left( 18 \right)

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=-18\text{ }\!\!~\!\!\text{ N}.\text{m}

\text{ }\!\!\Sigma\!\!\text{  }\!\!\tau\!\!\text{ }=18\text{ }\!\!~\!\!\text{ N}.\text{m }\!\!~\!\!\text{ searah }\!\!~\!\!\text{ perputaran }\!\!~\!\!\text{ jarum }\!\!~\!\!\text{ jam}

jawaban C

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