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Calculation of % component of resultant force

  1. Apr 10, 2012 #1
    Hi everyone,
    This appears to be a trivial question but I'm uncertain to the correct answer.

    If I have forces of 3N, 4N and 5N in the x,y and z directions respectively. The resultant force will be the square root of these components squared ≈ 7N. If I want to describe any of the components as a % of the resultant force, how is this done?

    I have seen in a journal paper that from my example the z-component of the force is calculated as 71.4% (5/7*100) of the resultant force, while the y-component was 57% of the resultant force. The addition of these percentages is greater than 100% so this cannot be true.

    If I calculate the z-component as a % of the resultant force by dividing the squares of each number (5^2/7^2*100), then the addition of each % component sums up to be 100% (x - 18%, y - 32%,z - 51%). Is this method correct and if so can someone explain to me why it is necessary to square both vectors.

    Thank you
  2. jcsd
  3. Apr 10, 2012 #2


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    welcome to pf!

    hi soundproof! welcome to pf! :smile:
    this seems a pretty crazy idea to me

    what would be the point? :confused:

    anyway … you have |F| ~ 7, F.i = 3, F.j = 4, F.k = 5

    32 + 42 + 52 ~ 72,

    so of course (3/7)2 + (4/7)2 + (5/7)2 ~ 1 :wink:
  4. Apr 10, 2012 #3


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    the sum of
    The magnitude of the resultant force will be [itex]\sqrt{9+ 16+ 25}= \sqrt{50}= 5\sqrt{2}[/itex] which is, yes, about 7.
    So the x component is about 3/7= 0.429... or about 43% of the whole, the y component is about 4/7= 0.571... or about 57% of the whole, and the z component is about 5/7= 0.714 or about 71% of the whole.

  5. Apr 11, 2012 #4
    Thank you for your replies.

    Just to clarify regarding this statement -
    So, it is not correct/possible to describe a vector in terms of % contributions of x, y and z components? And this is because I am trying to relate a length/scalar to a vector?
  6. Apr 11, 2012 #5


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    you can describe anything you like, but if you want the percentages to add to 100, it's difficult to see how it could be of any practical applicaton

    (we can define the efficiency of a force …

    if we want something to move horizontally, but we pull on it with a rope at an angle, then the efficiency would be the horizontal component divided by the whole force)​
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