# mechanics, vectors components help

• Mar 26th 2009, 07:36 AM
Tweety
mechanics, vectors components help
express the force vectors in terms of components parallel and perpendicular to the plane shown.

Can I say the angle the particle make with the plane is also 43 degrees , because they are alternate angles?

how do I make a triangle with the 10N force being the hypotenuse from this diagram ?

thanks
• Mar 26th 2009, 10:24 AM
Aryth
You're going to want to keep the axes as is, but a quick translation shows that you are correct in assuming the $\displaystyle 43^{\circ}$ angle.

Now, you may or may not know that the components of the vectors are scalars, represented by $\displaystyle F_x$ for the force in the $\displaystyle \hat{i}$ direction and $\displaystyle F_y$ for the force in the $\displaystyle \hat{j}$.

We also know that the magnitude of the force (I'll call $\displaystyle F_T$), which is 10N, is the hypotenuse of a right triangle, meaning that we can define $\displaystyle \cos{\theta}$ and $\displaystyle \sin{\theta}$ in terms of the force:

$\displaystyle \cos{\theta} = \frac{F_x}{F_T}$

Because in the unit circle, we define it to be cos = x/hyp.

This means that:

$\displaystyle F_x = F_T\cos{\theta}$

Consequently:

$\displaystyle F_y = F_T\sin{\theta}$

To prove it, we will rederive the magnitude of the force by using the pythagorean theorem, that tells us that:

$\displaystyle F_T^2 = F_x^2 + F_y^2$

$\displaystyle F_T^2 = (F_T\cos{\theta})^2 + (F_T\sin{\theta})^2$

$\displaystyle F_T^2 = F_T^2\cos^2{\theta} + F_T^2\sin^2{\theta}$

We know that $\displaystyle \cos^2{\theta} + \sin^2{\theta} = 1$ so we get:

$\displaystyle F_T^2 = F_T^2$

$\displaystyle |F_T| = |F_T|$

And there you go.

The final statement is:

$\displaystyle \vec{F} = F_x\hat{i} + F_y\hat{j}$