Author Topic: Component Forces (Read 1327 times) Tweet Share

/0 · « **on:** January 11, 2008, 03:54:18 pm »

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

The book said to find the unit vector $\hat{\bold{a}}=\frac{1}{\sqrt{5}}(2\bold i - \bold j)$ and somehow go $(\bold{F}\cdot \hat{\bold{a}})( \hat{\bold{a}})$ but I don't understand how that is suppsoed to give the answer??

Thanks...

Collin Li · « **Reply #1 on:** January 11, 2008, 04:42:32 pm »

Generally speaking:

Given a vector $\textbf{a}$ , to find the component of the force in the direction of $\textbf{b}$ , draw a right angled triangle.

Firstly, I have chosen $\lambda\textbf{b}$ because I don't care how long the vector $\textbf{b}$ is, I only want the component of $\textbf{a}$ in its direction.

Now, using trigonometry: $\cos\theta = \frac{\lambda\|\textbf{b}\|}{\|\textbf{a}\|}$

By the dot product definition: $\textbf{a}\cdot\textbf{b}= \|\textbf{a}\|\|\textbf{b}\|\cos\theta \implies \cos\theta = \frac{\textbf{a}\cdot\textbf{b}}{\|\textbf{a}\|\|\textbf{b}\|} = \frac{\lambda\|\textbf{b}\|}{\|\textbf{a}\|}$

$\implies \mbox{distance of vector }\textbf{a}\mbox{ in the direction }\textbf{b} = \lambda\|\textbf{b}\| = \frac{\textbf{a}\cdot\textbf{b}}{\|\textbf{b}\|} = \textbf{a}\cdot\hat{\textbf{b}}$

This is known as the scalar resolute of $\textbf{a}$ in the direction of $\textbf{b}$

Therefore, to find the vector resolute, multiply this magnitude by the unit vector of $\textbf{b}$ (a vector of length 1 in the correct direction).

General formula for vector resolute: $\left(\textbf{a}\cdot\hat{\textbf{b}}\right)\hat{\textbf{b}}$

/0 · « **Reply #2 on:** January 11, 2008, 06:48:40 pm »

Wow! Thanks so much coblin, that's the best explanation I could have hoped for!

It all fits together nicely now

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