Question

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel to and normal to the plane that makes an angle of StartFraction pi Over 3 EndFractionwith the positive​ x-axis. Show that the total force is the sum of the two component forces.

Answer 1

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

`\theta=(\pi)/(3)`

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=
`r=cos(\pi)/(3)i-sin(\pi)/(3)j=(1)/(2)i-(\sqrt 3)/(2)j`

By using
`cos(\pi)/(3)=(1)/(2),sin(\pi)/(3)=(\sqrt 3)/(2)`

Force along the plane will be=
`\mid F_x\mid=F\cdot r`

Force along the plane will be =
`\mid F_x\mid=F\cdot ((1)/(2)i-(\sqrt 3)/(2)j)=-8j\cdot((1)/(2)i-(\sqrt 3)/(2)j)=8* (\sqrt 3)/(2)=4\sqrt 3`N

By using
`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`

Therefore, force along the plane=
`\mid F_x\mid((1)/(2)i-(\sqrt 3)/(2)j)=4\sqrt 3((1)/(2)i-(\sqrt 3)/(2)j)`

2.The vector perpendicular to the plane=
`r=-sin(\pi)/(3)-cos(\pi)/(3)=-(\sqrt 3)/(2)i-(1)/(2)j`

The force perpendicular to the plane=
`\mid F_y\mid=F\cdot r=-8j(-(\sqrt 3)/(2)i-(1)/(2)j)`

The force perpendicular to the plane=
`4`N

Therefore,
`F_y=4(-(\sqrt 3)/(2)i-(1)/(2)j)`

Sum of two component of force=
`F_x+F_y=4\sqrt 3((1)/(2)i-(\sqrt 3)/(2)j)+4(-(\sqrt 3)/(2)i-(1)/(2)j)`

Sum of two component of force=
`2\sqrt 3i-6j-2\sqrt3 i-2j=-8j`

Hence,sum of two component of forces=Total force.