A vector is inclined at an angle 60° to the horizontal. If its rectangular component in the horizontal direction is 50 N, then its magnitude in the vertical direction is

Question

A vector is inclined at an angle 60° to the horizontal. If its rectangular component in the horizontal direction is 50 N, then its magnitude in the vertical direction is (A) 25 N (B) 75 N (C) 87 N (D) 100 N

Tardigrade Admin · Accepted Answer

Given, vector can be shown below as <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/92b219214047c15a7293314a39efbd93-.png" /> where, θ=60° Then, tan θ=(Ay/Ax) or Ay=Ax tan θ ⇒ Ay=50 tan 60° =50 × √3 (∵ √3=1.732) =86.6 ≃ 87 N

Q. A vector is inclined at an angle $6 0^{\circ}$ to the horizontal. If its rectangular component in the horizontal direction is $50 N$ , then its magnitude in the vertical direction is

$25 N$

$75 N$

$87 N$

$100 N$

Given, vector can be shown below as

where, $θ = 6 0^{\circ}$
Then, $tan θ = \frac{A _{y}}{A _{x}}$ or $A_{y} = A_{x} tan θ$
$\Rightarrow A_{y} = 50 tan 6 0^{\circ} = 50 \times 3$ ( $∵ 3 = 1.732$ )
$= 86.6 ≃ 87 N$

Q. A vector is inclined at an angle 60∘ to the horizontal. If its rectangular component in the horizontal direction is 50N, then its magnitude in the vertical direction is

25N

75N

87N

100N