Question

Resolve a weight of 20 N in two directions which are parallel and perpendicular to a slope inclined at 30° to the horizontal.

• A. W\textsubscript{⊥} = 10 N; W\textsubscript{||} = 10$\sqrt{3}$ N
• B. W\textsubscript{⊥} = 10$\sqrt{3}$ N; W\textsubscript{||} = 10 N
• C. W\textsubscript{⊥} = 10 N; W\textsubscript{||} = 10 N
• D. W\textsubscript{⊥} = 10$\sqrt{3}$ N; W\textsubscript{||} = 10$\sqrt{3}$ N

Answer 1

Answer: (B)

W\textsubscript{⊥} = 10$\sqrt{3}$ N; W\textsubscript{||} = 10 N

Answer 2

The weight $W = 20$ N acts vertically downwards. The slope is inclined at an angle $\theta = 30°$ to the horizontal. The component of weight perpendicular to the slope is given by $W_{⊥} = W \cos(\theta)$. The component of weight parallel to the slope is given by $W_{||} = W \sin(\theta)$.

Substituting the given values: $W_{⊥} = 20 \text{ N} \times \cos(30°) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$ N. $W_{||} = 20 \text{ N} \times \sin(30°) = 20 \times \frac{1}{2} = 10$ N.

Therefore, $W_{⊥} = 10\sqrt{3}$ N and $W_{||} = 10$ N.