# Thread: mechanics, vectors components help

1. ## 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

2. You're going to want to keep the axes as is, but a quick translation shows that you are correct in assuming the $43^{\circ}$ angle.

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

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

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

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

This means that:

$F_x = F_T\cos{\theta}$

Consequently:

$F_y = F_T\sin{\theta}$

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

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

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

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

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

$F_T^2 = F_T^2$

$|F_T| = |F_T|$

And there you go.

The final statement is:

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