We will denote by vecF_1→F1 the vector F_1.
We are given that, magnitude of vecF_1=||vecF_1||=500 N→F1=∣∣∣∣∣∣→F1∣∣∣∣∣∣=500N, &, its direction is due East, i.e., the +ve+ve direction of the X-axis. The unit vector along this direction is hati=(1,0)ˆi=(1,0).
Recall that a vecx!=vec0→x≠→0 can completely be described as vecx=||vecx||hatx→x=∣∣∣∣→x∣∣∣∣ˆx, where, hatxˆx is a unit vector in the direction of vecx→x.
Hence, vecF_1=||vecF_1||hati=500(1,0)=(500,0)→F1=∣∣∣∣∣∣→F1∣∣∣∣∣∣ˆi=500(1,0)=(500,0).
Similarly, vecF_2=(0,250)→F2=(0,250)
Therefore, vecF_2-vecF_1=vecF→F2−→F1=→F, say =(0,250)-(500,0)=(-500,250)=250(-2,1)=(0,250)−(500,0)=(−500,250)=250(−2,1)
||vecF||=250sqrt{(-2)^2+1^2}=250sqrt5 N∣∣∣∣∣∣→F∣∣∣∣∣∣=250√(−2)2+12=250√5N
To find the direction of vecF→F, suppose that vecF→F is making an angle thetaθ due East, i.e., with hatiˆi.
Then, vecF.hati=||vecF||*||hati||*costheta→F.ˆi=∣∣∣∣∣∣→F∣∣∣∣∣∣⋅∣∣∣∣ˆi∣∣∣∣⋅cosθ
:. (-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.!
