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

Question

A vector is inclined at 30° 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.

Tardigrade Admin · Accepted Answer

Here, θ = 30° Let F= magnitude of force. Given, F cos θ = 50 or F = (50/cos 30°) = (50 × 2 /√3) = (100/√3) = 57.74 N Vertical component = F sin θ = 57.74 × sin 30° = 57.74 × (1/2) = 28.87 N

Q. 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

92 N

29 N

36 N

16 N.

Here, $θ = 3 0^{\circ}$
Let F= magnitude of force.
Given, F cos $θ$ = 50 or F = $\frac{50}{cos 3 0 ^{\circ}} = \frac{50 \times 2}{3} = \frac{100}{3} = 57.74 N$
Vertical component = F sin $θ$ = 57.74 $\times$
sin 30 $^{\circ}$ = 57.74 $\times \frac{1}{2}$ = 28.87 N

Q. A vector is inclined at 30∘ to the horizontal. If its rectangular component in the horizontal direction be 50 unit, the magnitude of its vertical component is