Question

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?

Answer 1

`F_x=35sqrt3` 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 `Fcos(theta)` and `Fsin(theta)`

`"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(30)=F_x/F`
So, `F_x is Fcos(30)`

`sin(30)=F_y/F`
So, `F_y=Fsin(30)`

The x component is `Fcos(theta)` and y component is `Fsin(theta)`

So the components are `70cos(30)` and `70sin(30)`

`F_x=35sqrt3` N
`F_y=35` N

Answer 2

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=
`Sin30=(y/70)`
`70Sin30=y`
`y=35.0`

The horizontal component (x-direction) is given by
`Cos30=(x/70)`
`70Cos30=x`

`x approx 60.6`