Question

A body is placed on rough $\left(\mu =\frac{1}{3\sqrt{3}}\right)$ inclined plane. $A$ force $F$ is needed to stop this body to slide downward. $A$ force $2F$ is needed so that the body is just about to move upwards. Slope of inclined plane is

• A. $30^{\circ }$
• B. $45^{\circ }$
• C. $60^{\circ }$
• D. $20^{\circ }$

Answer 1

Answer: (A)

Let $\theta$ be angle of inclination and $m$ be the mass of a body
For motion down the plane.

The equation of motion is
$F+f-mgsin\theta =0$
$F+\mu N-mgsin\theta =0$
$F+\mu mgcos\theta -mgsin\theta =0\dots \left(i\right)$
For motion up the plane

The equation of motion is
$2F-f-mgsin\theta =0$
$2F-\mu N-mgsin\theta =0$
$2F-\mu mgcos\theta -mgsin\theta =0\dots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get
$3F=2mgsin\theta \dots \left(iii\right)$
Subtracting $\left(i\right)$ from $\left(ii\right)$, we get
$F=2\mu mgcos\theta \dots \left(iv\right)$
Dividing $\left(iii\right)$ by $\left(iv\right)$, we get
$\frac{3F}{F}=\frac{2mgsin\theta }{2\mu mgcos\theta }$
or $tan\theta =3\mu$
$=3\times \frac{1}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\therefore \theta =30^{\circ }$