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Finding magnitude and direction of resultant force

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A rocket fires two engines simultaneously. One produces a thrust of 725 N directly forward, while the other gives a 513-N thrust at 32.4° above the forward direction. Find the magnitude and direction (relative to the forward direction) of the resultant force that these engines exert on the rocket.



    3. The attempt at a solution
    I tried finding the magnitude by doing 725+513(cos(32.4°)) and got 1158 N but for the answer it says its 1190 N. Did I do something wrong or is the book wrong? And I am not sure how to find the direction after I get the magnitude.
     
  2. jcsd
  3. Sep 1, 2008 #2
    Hi Azn4ly, Well the way I would approch this is break everything up in to components, and then add the componets of the vectors together and apply pythags therom. So think about it, break the 725 focre in to components and then the 513 force in to components. Try using that method :-)
     
  4. Sep 1, 2008 #3

    LowlyPion

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    You only calculated the X component of velocity. You need to factor the Y component in as Galadirith suggested by adding its value through the square root of the squares of the components.
     
  5. Sep 11, 2008 #4
    I am still having problems with this problem. What formula would I use?
     
  6. Sep 11, 2008 #5

    LowlyPion

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    Pythagorean Theorem.

    RSS - The Root Sum of the Squares.

    The x and y components are at right angles and hence for the sides of a right triangle.
    The relationship of the sides to the hippopotamus is
    a2 + b2 = c2
     
  7. Sep 1, 2009 #6
    hope this helps
     

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