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
To find the moment of the force about the origin, you need to calculate the cross product between the position vector (4i+5j-7k) and the force vector (26/√3 i + 26/√3 j + 26/√3 k).
The cross product between two vectors A and B is given by the formula:
A x B = (AyBz - AzBy)i + (AzBx - AxBz)j + (AxBy - AyBx)k
Let's calculate it step by step:
1. Take the position vector, A = 4i + 5j - 7k, and the force vector, B = (26/√3)i + (26/√3)j + (26/√3)k.
2. Calculate the cross product term by term:
AxBy = 4 * (26/√3) = (104/√3) i
AyBx = 5 * (26/√3) = (130/√3) j
AzBx = (-7) * (26/√3) = (-182/√3) k
3. Substitute the values in the formula to find the cross product:
A x B = [(130/√3) - (-182/√3)]i + [(-182/√3) - (104/√3)]j + [(104/√3) - (130/√3)]k
= (312/√3)i + (-286/√3)j + (-26/√3)k
Therefore, the moment of the force about the origin is (312/√3)i + (-286/√3)j + (-26/√3)k.
Don't worry about the seemingly complex numbers - it's just the result of the mathematical computations.