Question

It is found that a body on an inclined plane just starts sliding down if inclination is ${\sin }^{-1}\left(3/5\right)$, then the angle of repose (friction) is

• A. $si{n}^{-1}3/5$
• B. $si{n}^{-1}3/4$
• C. $ta{n}^{-1}3/5$
• D. $ta{n}^{-1}3/4$

Answer 1

Answer: (A), (D)

$ta{n}^{-1}3/4$
$si{n}^{-1}3/5$

We know, angle of repose $=$ angle of friction
Also, $\mu =tan\theta$
$\theta =si{n}^{-1}\left(\frac{3}{5}\right)$
Also, by trigonometry:
$\theta =ta{n}^{-1}\left(\frac{3}{4}\right)$