# A man is pulling on his dog with a force of 70.0 N directed at an angle of +30.0° to the horizontal. What are the x and y components of this force?

Mar 27, 2018

${F}_{x} = 35 \sqrt{3}$ N
${F}_{y} = 35$ N

#### Explanation:

To put it shortly, any force F making an angle $\theta$ with the horizontal has x and y components $F \cos \left(\theta\right)$ and $F \sin \left(\theta\right)$

$\text{Detailed explanation:}$

He is pulling his dog at an angle of 30 with the horizontal with a force of 70 N

There are an x component and a y component to this force

If we draw this as a vector, then the diagram looks something like this

The Black line is the direction of force and red and green are x and y components respectively. The angle between the Black line and Red line is 30 degrees as given

Since force is a vector, we can move the arrows and rewrite it as

Now since the angle between the Black line and the Red line is 30 degrees and the vector black line has a magnitude of 70 N, we can use trigonometry

$\cos \left(30\right) = {F}_{x} / F$
So, ${F}_{x} i s F \cos \left(30\right)$

$\sin \left(30\right) = {F}_{y} / F$
So, ${F}_{y} = F \sin \left(30\right)$

The x component is $F \cos \left(\theta\right)$ and y component is $F \sin \left(\theta\right)$

So the components are $70 \cos \left(30\right)$ and $70 \sin \left(30\right)$

${F}_{x} = 35 \sqrt{3}$ N
${F}_{y} = 35$ N

Mar 27, 2018

y-direction= 35.0 N
x-direction= 60.6 N

#### Explanation:

You essentially have a right-angled triangle with a 30-degree angle and a hypotenuse with a magnitude 70.0 Newtons.

So the vertical component (y-direction) is given by=
$S \in 30 = \left(\frac{y}{70}\right)$
$70 S \in 30 = y$
$y = 35.0$

The horizontal component (x-direction) is given by
$C o s 30 = \left(\frac{x}{70}\right)$
$70 C o s 30 = x$

$x \approx 60.6$