A vector F1 = 500 N due east and another F2 =250 N due north are on a plane. How do you find F2-F1?

1 Answer
Jul 15, 2016

#vecF_2-vecF_1=(500,250)# has magnitude #250sqrt5N# and it acts at an angle #theta=pi-cos^-1(2/sqrt5)# due East.

Explanation:

We will denote by #vecF_1# the vector F_1.

We are given that, magnitude of #vecF_1=||vecF_1||=500 N#, &, its direction is due East, i.e., the #+ve# direction of the X-axis. The unit vector along this direction is #hati=(1,0)#.

Recall that a #vecx!=vec0# can completely be described as #vecx=||vecx||hatx#, where, #hatx# is a unit vector in the direction of #vecx#.

Hence, #vecF_1=||vecF_1||hati=500(1,0)=(500,0)#.

Similarly, #vecF_2=(0,250)#

Therefore, #vecF_2-vecF_1=vecF#, say #=(0,250)-(500,0)=(-500,250)=250(-2,1)#

#||vecF||=250sqrt{(-2)^2+1^2}=250sqrt5 N#

To find the direction of #vecF#, suppose that #vecF# is making an angle #theta# due East, i.e., with #hati#.

Then, #vecF.hati=||vecF||*||hati||*costheta#

#:. (-500,250).(1,0)=250sqrt5*1*costheta#

#:.-500=250sqrt5*costheta#

#:. costheta=-500/(250sqrt5)=-2/sqrt5#

#:. theta=cos^-1(-2/sqrt5)=pi-cos^-1(2/sqrt5)#

Thus, #vecF_2-vecF_1=(-500,250)# has magnitude #250sqrt5N# and it acts at an angle #theta=pi-cos^-1(2/sqrt5)# due East.

Hope, this will be of Help! Enjoy maths.!