Results 1 to 4 of 4

Math Help - Vector Problem

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    69

    Vector Problem

    Im a bit stuck on this question.

    In the first part, i worked out that a vector had the component form:

    61.3 j + 56.1 i

    For the next part, i need to write it in magnitude and direction form.
    Im unsure on how to do this.

    I know that:

    61.3 = m sin d
    56.1 = m cos d

    m is the magnitude in newtons
    d is the direction in degrees

    How could i solve this simultaniously?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    629
    Hello, nugiboy!

    You should be familiar with certain formulas . . .


    In the first part, i worked out that a vector had the component form: . 61.3 j + 56.1 i

    For the next part, i need to write it in magnitude and direction form.
    Im unsure on how to do this.

    I know that: . \begin{array}{ccc}61.3 &= &m\sin d \\ 56.1 & = & m\cos d\end{array}

    m is the magnitude in newtons
    d is the direction in degrees
    If we don't know the formulas, we can still work it out . .

    Divide the two equations: . \frac{m\sin d}{m\cos d} \:=\:\frac{61.3}{56.1}

    . . This gives us: . \tan d \:=\:1.092691622

    Hence: . d \;=\;\tan^{-1}(1.092591622) \;=\;47.53614518^o

    . . Therefore: . \boxed{d \;\approx\;47.5^o}


    For the magnitude, think of the Pythagorean Theorem.

    . . m^2 \:=\:(61.3)^2 + (56.1)^2 \;=\;6904.9

    Hence: . m \;=\;\sqrt{6904.9} \;=\;83.09572793

    . . Therefore: . \boxed{m \;\approx\;83.1}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2007
    Posts
    69
    Quote Originally Posted by Soroban View Post
    Hello, nugiboy!

    You should be familiar with certain formulas . . .

    If we don't know the formulas, we can still work it out . .
    Thanks for the reply!
    I understand how you got the answer, but i was just wondering what you meant by i should be familiar with certain formulas. Is there a simple formula to go from component form to magnitude and angle form, or is the method you showed me the only way?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,710
    Thanks
    629
    Hello again, nugiboy!

    Well, if you learn a few conversion formulas, it'll save a lot of time.
    . . But you seem to be familiar with them already.


    Going from (r,\,\theta) to (x,\,y)\!:\;\;\begin{Bmatrix}x & = & r\!\cdot\!\cos\theta \\ y & = & r\!\cdot\!\sin\theta \end{Bmatrix}


    Going from (x,y) to (r,\,\theta)\!:\;\;\begin{Bmatrix}r & = &\sqrt{x^2+y^2} \\ \theta & =\ & \arctan\left(\frac{y}{x}\right)\end{Bmatrix}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Vector problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2010, 03:03 PM
  2. Vector Problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 22nd 2010, 12:26 AM
  3. 3D vector problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 7th 2010, 08:22 PM
  4. Vector problem
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: April 24th 2009, 11:00 AM
  5. Vector Problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: August 28th 2008, 10:58 AM

Search Tags


/mathhelpforum @mathhelpforum