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Cross product and dot product of forces expressed as complex numbers

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magwas
#1
Feb11-10, 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:
  • 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: [tex]\frac{\mathbf{\imath} p1p3}{\lvert{p1p3}\rvert} [/tex]
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]
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magwas
#2
Feb11-10, 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]...
nvn
#3
Feb11-10, 10:40 AM
Sci Advisor
HW Helper
P: 2,121
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.

magwas
#4
Feb11-10, 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.
magwas
#5
Feb11-10, 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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