# Force Components

• Jan 4th 2008, 11:24 PM
DivideBy0
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
• Jan 5th 2008, 12:38 AM
earboth
Quote:

Originally Posted by DivideBy0
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°.
• Jan 5th 2008, 02:47 AM
DivideBy0
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.
• Jan 5th 2008, 05:41 AM
earboth
Quote:

Originally Posted by DivideBy0
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 $\vec a \cdot \vec b = 0$
• Jan 5th 2008, 09:23 AM
DivideBy0
Thankyou earboth, I think I just need a bit of time to understand