# Determine the x and y components of F1 and F2

Determine the x and y components of $F_1$ and $F_2$.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will start with force $F_1$.

$\sin45^0\,=\,\dfrac{F_{1x}}{200}$

(Remember sine is opposite over hypotenuse)

$F_{1x}\,=\,200\sin45^0\,=\,141.4$ N

$\cos45^0\,=\,\dfrac{F_{1y}}{200}$

(Remember cosine is adjacent over hypotenuse)

$F_{1y}\,=\,200\cos45^0\,=\,141.4$ N

Now, we will focus on force $F_2$. As before, we can write use trigonometry to solve for the components.

$\cos30^0\,=\,\dfrac{F_{2x}}{-150}$

$F_{2x}\,=\,-150\cos30^0\,=\,-130$ N

(Notice how our force is negative. This is because the x-component of force $F_2$ is going towards the negative x-axis.)

$\sin30^0\,=\,\dfrac{F_{2y}}{150}$

$F_{2y}\,=\,150\sin30^0\,=\,75$ N

$F_{1x}\,=\,141.4$ N
$F_{1y}\,=\,141.4$ N
$F_{2x}\,=\,-130$ N
$F_{2y}\,=\,75$ N