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

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

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

Answer 2

Given \(\mu = \frac{1}{3\sqrt{3}}\)

1. Forces Acting on the Body:
- Weight mg acting vertically downward.
- Normal force N acting perpendicular to the inclined plane.
- Friction force f opposing the motion parallel to the plane.

2. Equations of Motion:
- Downward Motion:
\(F + f - mg \sin \theta = 0\)
\(F + \mu N - mg \sin \theta = 0\)
\(F + \mu mg \cos \theta - mg \sin \theta = 0\)
Upward Motion:
\(2F - f - mg \sin \theta = 0\)
\(2F - \mu N - mg \sin \theta = 0\)
\(2F - \mu mg \cos \theta - mg \sin \theta = 0\)
Adding equations (i) and (ii) to eliminate N gives:
\(3F = 2mg \sin \theta\)
- Subtracting equation (i) from equation (ii) to find F gives:
\(F = 2\mu mg \cos \theta\)
- Dividing the resultant equations and simplifying leads to:
\(\tan \theta = 3\mu\)

\(\tan \theta = 3 \times \frac{1}{3\sqrt{3}} = \frac{1}{\sqrt{3}}\)

\(\theta = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = 30^\circ\)

So, the correct option is (A): \(30\degree\)