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Thread: 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 $\displaystyle \bold{F}=(3\bold{i}+2\bold{j})N$ in the direction of the vector $\displaystyle 2\bold{i}-\bold{j}$.

    Solution:

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

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

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

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


    What do the expressions $\displaystyle \bold{F}\cdot \hat{\bold{a}}$ and $\displaystyle (\bold{F}\cdot \hat{\bold{a}})\cdot \hat{\bold{a}}$ 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 $\displaystyle \bold{F}=(3\bold{i}+2\bold{j})N$ in the direction of the vector $\displaystyle 2\bold{i}-\bold{j}$.

    Solution:

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

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

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

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


    What do the expressions $\displaystyle \bold{F}\cdot \hat{\bold{a}}$ and $\displaystyle (\bold{F}\cdot \hat{\bold{a}})\cdot \hat{\bold{a}}$ 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 $\displaystyle |a| = \sqrt{4+1}=\sqrt{5}$

    Therefore $\displaystyle \hat{\bold{a}}=\frac{1}{\sqrt{5}}(2\bold{i}- \bold{j})$ 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 , $\displaystyle \vec a \text{ and } \vec b$ which include the angle $\displaystyle \theta$ and you know to calculate the length of a vector then the dot product is defined as:

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

    That means $\displaystyle |\vec b | \cdot \cos(\theta)$ 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 ($\displaystyle \vec a^0$) 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 $\displaystyle \vec a$ is perpendicular to $\displaystyle \vec b$ if $\displaystyle \vec a \cdot \vec b = 0$
    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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