Wednesday
October 1, 2014

Homework Help: Engineering

Posted by Aaron on Sunday, February 12, 2012 at 1:07am.

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+5j-7k?

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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