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  4. 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

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Q. A vector is inclined at an angle $60^{\circ}$ to the horizontal. If its rectangular component in the horizontal direction is $50 \,N$, then its magnitude in the vertical direction is

Motion in a Plane

Solution:

Given, vector can be shown below as
image
where, $\theta=60^{\circ}$
Then, $\tan \theta=\frac{A_{y}}{A_{x}}$ or $A_{y}=A_{x} \tan \theta$
$\Rightarrow A_{y}=50 \tan 60^{\circ} =50 \times \sqrt{3}$ ($\because \sqrt{3}=1.732$)
$=86.6 \simeq 87\, N$