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

Answer:

#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.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.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

Answer:

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#