Question

A car is moving on a plane inclined at $30^{\circ}$ to the horizontal with an acceleration of $10\, ms ^{-2}$ parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is ___
$\left(\right.$ Take $\left.g =10 \,ms ^{-2}\right)$

Answer 1

$\tan (30+\theta)=\frac{ mg \sin 30^{\circ}+ ma }{ mg \cos 30^{\circ}}$
$\tan (30+\theta)=\frac{5+10}{5 \sqrt{3}}=\frac{1+2}{\sqrt{3}}$
$\frac{\tan \theta+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}} \tan \theta}=\sqrt{3}$
$\sqrt{3} \tan \theta+1=3-\sqrt{3} \tan \theta$
$2 \sqrt{3} \tan \theta=2$
$\tan \theta=\frac{1}{\sqrt{3}}$
$\theta=30^{\circ}$