(adsbygoogle = window.adsbygoogle || []).push({}); 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:

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]

- 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

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]

**Physics Forums - The Fusion of Science and Community**

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

Have something to add?

- Similar discussions for: Cross product and dot product of forces expressed as complex numbers

Loading...

**Physics Forums - The Fusion of Science and Community**