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)

$30^{\circ}$

Answer 2

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 $tan\theta=3\mu$ $=3\times\frac{1}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$ $\therefore\, \theta=30^{\circ}$