Vector Components vs Sine and Cosine Law...

Discussion in 'Physics & Math' started by Captain Covalency, Feb 26, 2015.

  1. Captain Covalency Registered Member

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    When adding two non-perpendicular vectors (Example: a puck travels 4.2m [S 38 degrees W], then travels 2.7 [E 25 degrees north]), when would you split the two vectors into vector components to solve for the total displacement, and when would you just use the Sine and Cosine law?
     
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  3. Spellbound Banned Valued Senior Member

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    It would be no different to add two non-perpendicular vectors as it is to add two perpendicular ones because the x and y components simplify each vector by making the relationship between the components of each vector perpendicular. So 4.2 meters (S 38 degrees West) would be 4.2 Sin 38 degrees = x meters. And 4.2 Cos 38 degrees = y meters. Then you just add them to yield total displacement. It's the same method as it is for perpendicular vectors because the components simplify the model of the vector despite the angles.

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  5. James R Just this guy, you know? Staff Member

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    The sine and cosine laws are ok if you're just adding two vectors. Splitting into components works for any number of vectors, so it's a much more general procedure. I think that it usually requires less thought, too: the procedure is the same for all problems.
     
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  7. Captain Covalency Registered Member

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    Yes, true. Thank you very much. It definitely makes logical sense that, when given only two vectors, either procedure should work equally well. The only issue I find, is when adding two vectors using the sine and cosine laws, it is easy to obtain the magnitude of the resultant vector, however it becomes more difficult to obtain the direction of the given vector.
     
  8. rpenner Fully Wired Valued Senior Member

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    This is similar to, but distinct from the US Army notation for bearing, so I assume the first letter is a reference direction, the second letter is a direction perpendicular to the first and the number is how far away from the reference direction in the direction of the second we go.

    Other ways to write this: is 4.2 m (bearing 180+38°) + 2.7 m ( bearing 90-25°) = 4.2 m (bearing 218°) + 2.7 m ( bearing 65°)

    The component in the North direction (bearing 0°) would be 4.2 m × cosine 218° + 2.7 m × cosine 65° = -2.16857585844834374040986478686170550967467739837066621857368... m
    The component in the East direction (bearing 90°) would be 4.2 m × sine 218° + 2.7 m × sine 65° = -0.13874717136880987385411441670669778383108557238904873078984... m

    The total length would be gotten by the Pythagorean theorem or 2.173009901359820514354921965398885364351415371918016070487032... = 3/10 sqrt(277-252 cos(27°))
    The bearing (expressed as an angle clockwise from N) is obtained via the inverse tangent. Since the mathematical arctangent is computed counterclockwise from the + X direction (or E on a map), swapping the arguments is the correct thing to do for the two argument form: arctan( - 0.138.. , -2.168... ) =
    -93.6608387482601393686507842989043585540288178787473191165320...°

    If you don't have the two argument inverse tangent available to you, then you have to use the single argument version on the ratio and figure out which quadrant it belongs to.


    Or 2.173 m bearing 266.34° or 2.173 m [S 86.34° W] or 2.173 m [W 3.66° S]

    But if all you want is the total (unsigned) displacement from the sum of two vectors, the law of cosines is enough and is much less work IF you keep your angle conventions straight.

    \(\sqrt{ 4.2^2 + 2.7^2 - 2 \times 4.2 \times 2.7 \times \cos (180^\circ + 65^\circ - 218^\circ) } = \frac{3}{10} \sqrt{ 14^2 + 9^2 - 2 \times 14 \times 9 \times \cos (27^\circ) } = \frac{3}{10} \sqrt{ 277 - 252 \times \cos (27^\circ) } \approx 2.173 \, \textrm{m}\)
     
  9. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    To apply the law of sines, we must always keep track of side length A,B,C; bearing with respect to N, α,β,γ; and angle between the other two sides, a, b, c.
    Since A is the vector sum of B and C, it follows that the orientation of A (the bearing α) is 180° different than the direction for A + B + C to form a closed loop.

    Then we have a = 180° - | β - γ mod 360° |, b = 180° - | 180° + α - γ mod 360° |, c = 180° - | 180° + α - β mod 360° |, a + b + c = 180°

    So with A = 2.173 m, B = 4.2 m, C = 2.7 m, β = 218°, γ = 65°, a = 27°
    We add c = 153° - b, b = 180° - | α + 115° mod 360° |, c = 180° - | α - 38° mod 360° |

    sin b = sin | α + 115° |

    A/sin a = B/sin b means sin b = (B/A) sin a so we look at arcsin [ (14 / sqrt(277-252 cos(27°)) ) sin 27° ] = 61.34°
    But because sin 180° - x = sin x, then we don't know if b is 61.34° or 118.66° yet.
    If b is 61.34° then c = 91.66° and if b = 118.66° then c = 34.34°

    Now if we look at arcsin [ (C/A) sin a ] = arcsin [ (9 / sqrt(277-252 cos(27°)) ) sin 27° ] = 34.34° ( meaning c = 34.34° or 145.66° )

    Since a = 27°, b = 118.66°, c = 34.34° are the only self-consistent answers we know to pick them.

    So now we have | α + 115° mod 360° | = 180° - 118.66° = 61.34° and | α - 38° mod 360° | = 180° - 34.34° = 145.66° to solve.

    We could guess α + 115° = 61.34° or α + 115° = - 61.34° (all mod 360° ) with solutions for α being 306.34 and 183.66.
    We could guess α - 38° = 145.66° or α - 38° = -145.66° with solutions for α being 183.66 and 252.34
    So α = 183.66°

    Alternately using the cosine addition and subtraction formula we can turn this into a pair of linear equations

    cos( α + 115° ) = (cos α)(cos 115°) - (sin α)(sin 115°)= cos(61.34°)
    cos( α - 38° ) = (cos α)(cos 38°) + (sin α)(sin 38°) = cos(145.66°)

    with solution cos α = -0.997960427132662, sin α = -0.0638356160555814 or α = 183.66°
     
  10. rpenner Fully Wired Valued Senior Member

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    4,833
    oops, I mean arctan( -2.168..., - 0.138.. ) = 183.6608387482601393686507842989043585540288178787473191165320...°

    Or 2.173 m bearing 183.66° or 2.173 m [S 3.66° W]

    Always check your work.

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  11. James R Just this guy, you know? Staff Member

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    Yes. I always end up having to draw a diagram when I use the sine and cosine rules. If you use components, there's less room for error, in my opinion. Keeping track of the angles is tricky using the sine and cosine rules.
     

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