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Math Help - Force Components

  1. #1
    Senior Member DivideBy0's Avatar
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    Force Components

    I don't understand the solution to this question:

    Find the component of the force \bold{F}=(3\bold{i}+2\bold{j})N in the direction of the vector 2\bold{i}-\bold{j}.

    Solution:

    Let \bold{a}=2\bold{i}-\bold{j}. Then the unit vector in the direction of \bold{a} is \hat{\bold{a}}=\frac{1}{\sqrt{5}}(2\bold{i}-\bold{j}).

    \bold{F}\cdot \hat{\bold{a}}=(3\bold{i}+2\bold{j})\cdot \frac{1}{\sqrt{5}}(2\bold{i}-\bold{j})=\frac{4\sqrt{5}}{5}

    and (\bold{F}\cdot \hat{\bold{a}})\cdot \hat{\bold{a}}=\frac{4\sqrt{5}}{5}\times \frac{1}{\sqrt{5}}(2\bold{i}-\bold{j})=\frac{4}{5}(2\bold{i}-\bold{j})

    Hence the component of \bold{F} in the direction of 2\bold{i}-\bold{j} is \frac{4}{5}(2\bold{i}-\bold{j})N


    What do the expressions \bold{F}\cdot \hat{\bold{a}} and (\bold{F}\cdot \hat{\bold{a}})\cdot \hat{\bold{a}} mean? How in general do you solve these problems? Thanks
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  2. #2
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    Quote Originally Posted by DivideBy0 View Post
    I don't understand the solution to this question:

    Find the component of the force \bold{F}=(3\bold{i}+2\bold{j})N in the direction of the vector 2\bold{i}-\bold{j}.

    Solution:

    Let \bold{a}=2\bold{i}-\bold{j}. Then the unit vector in the direction of \bold{a} is \hat{\bold{a}}=\frac{1}{\sqrt{5}}(2\bold{i}-\bold{j}).

    \bold{F}\cdot \hat{\bold{a}}=(3\bold{i}+2\bold{j})\cdot \frac{1}{\sqrt{5}}(2\bold{i}-\bold{j})=\frac{4\sqrt{5}}{5}

    and (\bold{F}\cdot \hat{\bold{a}})\cdot \hat{\bold{a}}=\frac{4\sqrt{5}}{5}\times \frac{1}{\sqrt{5}}(2\bold{i}-\bold{j})=\frac{4}{5}(2\bold{i}-\bold{j})

    Hence the component of \bold{F} in the direction of 2\bold{i}-\bold{j} is \frac{4}{5}(2\bold{i}-\bold{j})N


    What do the expressions \bold{F}\cdot \hat{\bold{a}} and (\bold{F}\cdot \hat{\bold{a}})\cdot \hat{\bold{a}} mean? How in general do you solve these problems? Thanks
    Hello,

    1. use the definition of the dot product of vectors
    2. make a sketch (see attachment)

    Use |a| = \sqrt{4+1}=\sqrt{5}

    Therefore \hat{\bold{a}}=\frac{1}{\sqrt{5}}(2\bold{i}- \bold{j}) is a vector in the direction of a which has the length (value) of 1 (blue vector). In Germany such a vector is labeled by a.
    Attached Thumbnails Attached Thumbnails Force Components-force_component.gif  
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  3. #3
    Senior Member DivideBy0's Avatar
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    Thanks earboth... I understand what a unit vector is, but I don't know why the dot product was used, and how it helped. All I know is the dot product helps determine how 'parallel' two vectors are.
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  4. #4
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    Quote Originally Posted by DivideBy0 View Post
    Thanks earboth... I understand what a unit vector is, but I don't know why the dot product was used, and how it helped. All I know is the dot product helps determine how 'parallel' two vectors are.
    Hello,

    I'm going to try to explain the geometric meaning of a dot product.

    If you have 2 vectors , \vec a \text{ and } \vec b which include the angle \theta and you know to calculate the length of a vector then the dot product is defined as:

    \vec a \cdot \vec b = |\vec a| \cdot |\vec b | \cdot \cos(\theta)

    That means |\vec b | \cdot \cos(\theta) is the "shadow" of vector b on the vector a (if you use a light with parallel rays perpendicular to vector a)

    The value of the dot product is a real number. This number represents the area of a rectangle (painted red in my sketch)

    If you use a unit vector of a ( \vec a^0) then the value of the area (in square units) is as great as the "shadow" of b (in length units) with respect to the vector a. (compare the rectangle painted blue)

    That means: With the use of an unit vector you have introduced a kind of metric and you can calculate the length of components with respect to given directions.

    And now it must be quite clear why \vec a is perpendicular to \vec b if \vec a \cdot \vec b = 0
    Attached Thumbnails Attached Thumbnails Force Components-geom_vektprod.gif  
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  5. #5
    Senior Member DivideBy0's Avatar
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    Thankyou earboth, I think I just need a bit of time to understand
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