Question

What will be the component of a $10N$ weight along the surface of an inclined plane, inclined at an angle ${30}^{\circ }$ to the horizontal?

Answer 1

Component perpendicular to the plane

$W_{\perp }=W\cos {30}^{\circ }=\left(10\right)\frac{\sqrt{3}}{2}=5\sqrt{3}N$

and Component parallel to the plane

$W_{\parallel }=W\sin {30}^{\circ }=\left(10\right)\left(\frac{1}{2}\right)=5N$