Question

A car is moving on a plane inclined at $30^{\circ }$ to the horizontal with an acceleration of $10m{s}^{-2}$ 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 ___
$($ Take $g=10m{s}^{-2})$

Answer 1

Answer: 30

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