The force F=\left\{600i+300j-600k\right\} N acts at the end of the beam. Determine the moment of the force about point A.

#### Solution:

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The moment at A can be found by taking the cross product of r_{AB}\times \vec F. Here, r_{AB} represents a position vector from A to B, in other words, the location of the point where we are calculating the moment and the path from that point to the location where the force is being applied.

Let us write the locations of points A and B in Cartesian vector form.

From the image, we see that the locations of points A and B are at the following places:

B:(0.2i+1.2j+0.4k) m

The position vector, r_{AB} is:

r_{AB}=\left\{0.2i+1.2j+0k\right\}

We can now calculate the moment by taking the cross product. Remember that force F=\left\{600i+300j-600k\right\}.

M_A=\left\{-720i+120j-660k\right\}N\cdot m

#### Final Answer: