Cross product and dot product of forces expressed as complex numbers

In summary, the conversation discusses a problem involving a rod that can turn around a pivot point. The goal is to compute the orthogonal and parallel components of a given force acting on the rod. The conversation also discusses the equations and methods used to solve the problem, including the use of complex numbers and dot product. Ultimately, the solution is found to be F2n = 2.3570 N and F2t = 0.707107 N.
  • #1
magwas
125
0

Homework Statement



I have came up with an example to illustrate my question.

There is a rod, which can turn around p1.

attachment.php?attachmentid=23657&stc=1&d=1265886809.png


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, parallel component of F2 to the rod

Homework 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 parallel 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?

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 parallel 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]
 

Attachments

  • example.png
    example.png
    2.2 KB · Views: 582
Last edited:
Physics news on Phys.org
  • #2
Well, maybe I should have used [tex]magnitude_{F_{2_{n}}}[/tex] instead of [tex]\lvert F_{2_{n}}\rvert[/tex]...
 
  • #3
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
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
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.
 

What is the difference between the cross product and dot product of forces expressed as complex numbers?

The cross product of two complex numbers represents the vector perpendicular to both of the original vectors, while the dot product represents the scalar projection of one vector onto the other.

How do you calculate the cross product and dot product of forces expressed as complex numbers?

The cross product of two complex numbers, z1 and z2, is given by z1 x z2 = Im(z1)*Re(z2) - Re(z1)*Im(z2). The dot product is calculated by taking the conjugate of one of the complex numbers and multiplying it by the other, then taking the real part of the result.

What do the results of the cross product and dot product represent?

The result of the cross product is a complex number that represents the magnitude and direction of the perpendicular vector to the original vectors. The result of the dot product is a real number that represents the scalar projection of one vector onto the other.

Under what circumstances would you use the cross product instead of the dot product?

The cross product is useful for calculating torque in physics and determining the direction of rotation in mechanics, while the dot product is useful for calculating work and determining if two vectors are parallel or perpendicular to each other.

Can the cross product and dot product of forces expressed as complex numbers be used in any other contexts?

Yes, both operations can be applied in any situation where complex numbers and vectors are involved, such as in electrical engineering, signal processing, and computer graphics.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
808
  • Introductory Physics Homework Help
Replies
5
Views
774
  • Introductory Physics Homework Help
Replies
12
Views
833
  • Introductory Physics Homework Help
Replies
5
Views
918
  • Introductory Physics Homework Help
Replies
11
Views
168
  • Introductory Physics Homework Help
Replies
10
Views
1K
Replies
5
Views
906
  • Introductory Physics Homework Help
Replies
28
Views
261
  • Linear and Abstract Algebra
Replies
8
Views
664
  • Introductory Physics Homework Help
Replies
9
Views
857
Back
Top