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Math Help - Resultant Forces

  1. #1
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    Resultant Forces

    In the following question, Why can't we resolve horizontally like this:
    6 \cos 55 = 3.44N?

    Can someone explain, how this resolving is done. Thanks in advance.

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  2. #2
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    Every diagonal vector can be broken down into horizontal and vertical components. For vector a, the horizontal component is found by taking the magnitude times cos 55. Thus 3 cos 55 gives you its horizontal force.

    Since vector b is horizontal only, it gets added to 3 cos 55. Is that what you're asking?
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  3. #3
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    Quote Originally Posted by mathdeity View Post
    Every diagonal vector can be broken down into horizontal and vertical components. For vector a, the horizontal component is found by taking the magnitude times cos 55. Thus 3 cos 55 gives you its horizontal force.

    Since vector b is horizontal only, it gets added to 3 cos 55. Is that what you're asking?
    Yes. I was confused earlier as I was told that:

    Horizontal: R \cos \theta
    Vertical: R \sin \theta

    So I though for horizontal it was 6 \cos 55. I now realise the Resultant force is 3N.
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  4. #4
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    Quote Originally Posted by Air View Post
    Yes. I was confused earlier as I was told that:

    Horizontal: R \cos \theta
    Vertical: R \sin \theta

    So I though for horizontal it was 6 \cos 55. I now realise the Resultant force is 3N.
    I think maybe your terminology might be going against you. The "resultant" is the result of your vector addition. This is the 8.1 N at 17.656 degrees.

    To find the components of vector a,
    a_x = | \vec{a} | ~ cos(\theta)
    and
    a_y = | \vec{a} | ~ sin(\theta)
    where | \vec{a} | is the magnitude (size) of vector a and \theta is the angle between the vector and the +x axis.

    Does this help?

    -Dan
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