A plane is inclined at angle of 30 degree with horizontal. Find the component of a force vector F=-10 K N perpendicular to the plane. Given that z- direction is vertically upward.

Dear Student,

Please find below the solution to the asked query:

In which direction the force is applied. If the force is applied in the vertically upward (z) direction, then the angle with the vertical is also 30⁰. Therefore, the component of force perpendicular to the inclined plane.


So, the perpendicular component is,

$$\begin{gathered} F\text{'}=F\times \cos {30}^0=-10\times \frac{\sqrt{3}}{2} \\ \Rightarrow F\text{'}=-5\sqrt{3} N \end{gathered}$$

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards