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?

2 Answers
Mar 27, 2018

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

http://zonalandeducation.com/mstm/physics/mechanics/forces/forceComponents/forceComponents.htmlhttp://zonalandeducation.com/mstm/physics/mechanics/forces/forceComponents/forceComponents.html

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

http://zonalandeducation.com/mstm/physics/mechanics/forces/forceComponents/forceComponents.htmlhttp://zonalandeducation.com/mstm/physics/mechanics/forces/forceComponents/forceComponents.html

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

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=
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