Answer with Step-by-step explanation:
We are given that
F=<0,-8>=0i-8j=-8j
[tex]\theta=\frac{\pi}{3}[/tex]
The component of force is divided into two direction
1.Along the plane
2.Perpendicular to the plane
1.The vector parallel to the plane will be=[tex]r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j[/tex]
By using [tex] cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}[/tex]
Force along the plane will be=[tex]\mid F_x\mid=F\cdot r[/tex]
Force along the plane will be =[tex]\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3[/tex]N
By using [tex]i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0[/tex]
Therefore, force along the plane=[tex]\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)[/tex]
2.The vector perpendicular to the plane=[tex]r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j[/tex]
The force perpendicular to the plane=[tex]\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)[/tex]
The force perpendicular to the plane=[tex]4[/tex]N
Therefore, [tex]F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)[/tex]
Sum of two component of force=[tex]F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)[/tex]
Sum of two component of force=[tex]2\sqrt 3i-6j-2\sqrt3 i-2j=-8j[/tex]
Hence,sum of two component of forces=Total force.
