Engineering
posted by Aaron .
A force of 26 lb. acts through a point (4; 5; 7) ft. and is equally inclined
to the positive ends of the orthogonal axes. What is the moment of this
force about the origin
So i'm assuming vector r is 4i+5j7k?
And using the fact that cos^2α + cos^2β + cos^2γ = 1
therefore l=m=n=√1/3
I understand the force vector is equivalent to (26l + 26m + 26n) which equates to: (26/√3 i + 26/√3 j + 26/√3 k)
Thus the moment of force about the origin is (4i+5j+7k) X (26/√3 i + 26/√3 j + 26/√3 k)
This just seems very wrong to me as the cross products would yield very ugly numbers.
I have close to zero confidence in my answer
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