
#1
Feb1110, 05:10 AM

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 [tex]\sum{\vec{F} \times \vec{l}} =0[/tex] Where [tex]a \times b = \Re{a} \Im{b}  \Im{a} \Re{b}[/tex] Is that true? Likewise, the force components paralell to the rod is: [tex]\sum{\vec{F} \cdot \hat{\vec{l}}} = 0[/tex] where [tex] a \cdot b = a \overline{b} + b \overline{a} = 2 \Im{a} \Im{b} + 2 \Re{a} \Re{b}[/tex] Is it correct? 3. The attempt at a solution I write the moments around p3. I sum here because:
so the equation for moments: [tex] 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[/tex] so [tex]\lvert F_{2_{n}}\rvert =2.3570226039551 [/tex] which gives [tex]F_{2_{n}} = \lvert F_{2_{n}}\rvert \frac{\mathbf{\imath} p1p3}{\lvert{p1p3}\rvert} = 1.66666666666667 + 1.66666666666667 \mathbf{\imath}[/tex] Now the forces paralell to the rod: We use our unit vector [tex]\hat{l} = \frac{p1p3}{\lvert{p1p3}\rvert}[/tex] , and forget F4 as it is orthogonal to the rod, so the sum: [tex] 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 [/tex] so [tex]\lvert F_{2_{t}}\rvert = 1[/tex] which gives [tex] F_{2_{t}} = 0.707106781186548 + 0.707106781186548 \mathbf{\imath} [/tex] and [tex] F_{2} = F_{2_{n}} + F{2_{t}} = 0.959559885480119 + 2.37377344785321 \mathbf{\imath}[/tex] 



#2
Feb1110, 05:14 AM

P: 125

Well, maybe I should have used [tex]magnitude_{F_{2_{n}}}[/tex] instead of [tex]\lvert F_{2_{n}}\rvert[/tex]...




#3
Feb1110, 10:40 AM

Sci Advisor
HW Helper
P: 2,110

magwas: I got F2n = 2.3570 N, but I got F2t = 0.707 107 N, not 1. You can check your answer by summing forces in the rod tangential direction, to see if the summation equals zero.




#4
Feb1110, 12:51 PM

P: 125

cross product and dot product of forces expressed as complex numbers
I see, [tex]\lvert F_{2_{t}}\rvert \cdot \hat{l} [/tex] was a mistake.
the equation correctly is [tex]F2t + \left ( F_{1} \cdot l \right) + \left ( F_{3} \cdot l \right) = 0[/tex] but it comes down to [tex]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[/tex] which leads to [tex]1.4142135623731 + F2t = 0[/tex], so F2t = 1.4142135623731 Do I have a problem with the definition of complex dot product? Thank you again. 



#5
Feb1110, 05:32 PM

P: 125

I have looked up the definition of vector dot product. Wikipedia tells me that it is
[tex]\sum a_{i} b_{i}[/tex] 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. 


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