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

## Answer This Question

## Related Questions

- Engineering/Physics - A force F = 7^i + 6^j - 5^k lb. acts at the origin. Find ...
- physics - A particle moves through an xyz coordinate system while a force acts ...
- physics - A particle moves through an xyz coordinate system while a force acts ...
- Engineering - Calculate the moment about a point (5; 4; 3) ft. caused by the ...
- Engineering - Calculate the moment about a point (5; 4; 3) ft. caused by the ...
- Physics/ Engineering - Calculate the moment about a point (5; 4; 3) ft. caused ...
- Physics or Math - A force F = 0.8 i + 2.5 j N is applied at the point x = 3.0 m...
- physics - A force 1 of magnitude 5.40 units acts at the origin in a direction ...
- ph - A coin slides over a frictionless plane and across an xy coordinate system ...
- Physics - A coin slides over a frictionless plane and across an xy coordinate ...

More Related Questions