Question

A vector is inclined at 30$^\circ$ to the horizontal. If its rectangular component in the horizontal direction be 50 unit, the magnitude of its vertical component is

• A. 92 N
• B. 29 N
• C. 36 N
• D. 16 N.

Answer 1

Answer: (B)

29 N

Answer 2

Here, $\theta = 30^\circ$ Let F= magnitude of force. Given, F cos $\theta$ = 50 or F = $\frac{50}{cos\, 30^\circ} = \frac{50 \times 2 }{\sqrt{3}} = \frac{100}{\sqrt{3}} = 57.74 \, N$ Vertical component = F sin $\theta$ = 57.74 $\times$ sin 30$^\circ$ = 57.74 $\times \frac{1}{2}$ = 28.87 N