# Cross product and dot product of forces expressed as complex numbers

 P: 125 1. The problem statement, all variables and given/known data I have came up with an example to illustrate my question. There is a rod, which can turn around p1. p1p2 = (-1+j) m p1p3 = (-3 + 3j) m p1p4 = (1 - j ) m F1 = (1+3j) N F3 = (-1 - 2j ) N F4 = unknown, orthogonal to the rod compute F2_n, orthogonal component of F2 to the rod compute F2_t, paralell component of F2 to the rod 2. Relevant equations The question is actually here: The sum of moments is $$\sum{\vec{F} \times \vec{l}} =0$$ Where $$a \times b = \Re{a} \Im{b} - \Im{a} \Re{b}$$ Is that true? Likewise, the force components paralell to the rod is: $$\sum{\vec{F} \cdot \hat{\vec{l}}} = 0$$ where $$a \cdot b = a \overline{b} + b \overline{a} = 2 \Im{a} \Im{b} + 2 \Re{a} \Re{b}$$ Is it correct? 3. The attempt at a solution I write the moments around p3. I sum here because:all forces are on the same side of the turning point all arms are measured towards the turning point (this is why p1p3 - p1p4) the direction of forces are encoded in their vectors The unit vector normal to the rod is come by dividing a vector along the rod by its length, and multiplying it with j: $$\frac{\mathbf{\imath} p1p3}{\lvert{p1p3}\rvert}$$ so the equation for moments: $$F_{1} \times \left(p1p3 - p1p4\right) + F_{3} \times \left(p1p3 - p1p2\right) + p1p3 \times \left \frac{\mathbf{\imath} p1p3}{\lvert{p1p3}\rvert} \lvert F_{2_{n}}\rvert} = \\  \Im{p1p3} \Im\left(\frac{\lvert F_{2_{n}}\rvert p1p3}{\lvert{p1p3}\rvert}\right) + \Im\left(p1p3 - p1p2\right) \Re{F_{3}} + \Im\left(p1p3 - p1p4\right) \Re{F_{1}} + \Re{p1p3} \Re \left(\frac{\lvert F_{2_{n}}\rvert p1p3}{\lvert{p1p3}\rvert}\right) - \Im{F_{1}} \Re\left(p1p3 - p1p4\right) - \Im{F_{3}} \Re\left(p1p3 - p1p2\right) = \\  10.0 + 4.24264068711929 \lvert F_{2_{n}} \rvert = 0$$ so $$\lvert F_{2_{n}}\rvert =-2.3570226039551$$ which gives $$F_{2_{n}} = \lvert F_{2_{n}}\rvert \frac{\mathbf{\imath} p1p3}{\lvert{p1p3}\rvert} = 1.66666666666667 + 1.66666666666667 \mathbf{\imath}$$ Now the forces paralell to the rod: We use our unit vector $$\hat{l} = \frac{p1p3}{\lvert{p1p3}\rvert}$$ , and forget F4 as it is orthogonal to the rod, so the sum: $$F_{3} \cdot \hat{l} + \lvert F_{2_{t}}\rvert \cdot \hat{l} + F_{1} \cdot \hat{l} = \\ 2 \lvert F_{2_{t}}\rvert \Re{\hat{l}} + 2 \Im{F_{1}} \Im{\hat{l}} + 2 \Im{F_{3}} \Im{\hat{l}} + 2 \Re{F_{1}} \Re{\hat{l}} + 2 \Re{F_{3}} \Re{\hat{l}} = \\ 1.4142135623731 - 1.4142135623731 \lvert F_{2_{t}}\rvert = 0$$ so $$\lvert F_{2_{t}}\rvert = 1$$ which gives $$F_{2_{t}} = -0.707106781186548 + 0.707106781186548 \mathbf{\imath}$$ and $$F_{2} = F_{2_{n}} + F{2_{t}} = 0.959559885480119 + 2.37377344785321 \mathbf{\imath}$$ Attached Thumbnails
 P: 125 Well, maybe I should have used $$magnitude_{F_{2_{n}}}$$ instead of $$\lvert F_{2_{n}}\rvert$$...
 P: 125 Cross product and dot product of forces expressed as complex numbers I see, $$\lvert F_{2_{t}}\rvert \cdot \hat{l}$$ was a mistake. the equation correctly is $$F2t + \left ( F_{1} \cdot l \right) + \left ( F_{3} \cdot l \right) = 0$$ but it comes down to $$F2t + 2 \Im{F_{1}} \Im{l} + 2 \Im{F_{3}} \Im{l} + 2 \Re{F_{1}} \Re{l} + 2 \Re{F_{3}} \Re{l} = 0$$ which leads to $$1.4142135623731 + F2t = 0$$, so F2t = -1.4142135623731 Do I have a problem with the definition of complex dot product? Thank you again.
 P: 125 I have looked up the definition of vector dot product. Wikipedia tells me that it is $$\sum a_{i} b_{i}$$ for vectors a=(a1,...,an) and b=(b1,...bn). So a . b must be re(a)re(b)+im(a)im(b), not twice that. In this way I get the same result as you, I believe.