question_answer

Resolve a weight of 10 N in two directions which are parallel and perpendicular to a slope inclined at \[{{30}^{o}}\] to the horizontal.

A)  \[{{w}_{\bot }}=5\sqrt{3}N,\,{{w}_{||}}=5N\]

B)  \[{{w}_{\bot }}=5N,\,{{w}_{||}}=5\sqrt{3}N\]

C)  \[{{w}_{\bot }}=5\sqrt{3},\,{{w}_{||}}=0N\]

D)  \[{{w}_{\bot }}=0N,\,{{w}_{||}}=5N\]

Correct Answer: A

Solution :

Component perpendicular to the plane \[{{w}_{\bot }}=w\,\cos {{30}^{o}}\] \[=10\times \frac{\sqrt{3}}{2}\] \[=5\sqrt{3}N\] and component parallel to the plane \[{{w}_{||}}=w\sin {{30}^{o}}\] \[=10\times \frac{1}{2}\] = 5 N