In part 1 we saw that it is reasonable to treat the momentum of force as a vector. I promised to show that the cross product is a distributive operation. It means:
a x (b + c) = a x b + a x c
Let b + c = d.
We start with 3 planes formed by vector a and vectors b, c and d. Each product is perpendicular to one of them and has the value equal to the area of the appropriate coloured rectangle. Finally two similar triangles are obtained: ABC and A’B’C’. The first is formed by the components of vectors b, c and d normal to vector a. The second is made of vectors perpendicular to them and a times bigger. (Remember that the length of each cross product is equal to the length of the normal component of c, b or d multiplied by the length of vector a.)
This property expresses an important fact. If you add all the forces applied at a point of a body and then calulate the momentum or you first calculate all the momenta an then sum them up, you get the same result.
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