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Force F=(3i4j+12k)N acts at point A(1,2,3). Find -

(i) Moment of force about origin

(ii) Moment of force about point B(2,1,2)m..

1 Answer
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Given :- F=3i4j+12krA=i2j+3krB=2i+j+2k

To find:-

Mo,MB

Solutions:

Mo=rA×F

=|ijk1233412|

=(24+12)i(129)j+(4+6)kMo=12i3j+2k   NmrAB=rArB=(i2j+3k)(2i+j+2k)rAB=i3j+kMB=rAB×F

=|ijk1313412|

=(36+4)i(123)j+(4+9)kMB=32i+15j+13k   Nm

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