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The moment of force, for F (+j + k)N actingat points (-2, 3, 4) about point (1, 2, 3), is(1) 4j k(3) 4(j -3k)(2)(4)4 (j -k)4 (j + k) |
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Answer» moment = F×ds and here ds = (-2-1)i+(3-2)j+(4-3)k = -3i+j+k so, M = -[ (i+j+k)×(-3i+j+k)]=> M = -[i(0) -j(1+3) +k(1+3)]= +4j-4k = 4(j-k) |
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