Question

A block of mass $1\text{kg}$ is kept on an inclined accelerating conveyor belt. Acceleration of the belt is $1{\text{m/s}}^2$ as shown in figure. Minimum coefficient of static friction between the belt and block to prevent slipping of the block will be: [Take $g=10{\text{m/s}}^2$]

• A. $0.4$
• B. $0.3$
• C. $0.2$
• D. $0.69$

Answer 1

The correct option is D $0.69$

$m=1\text{kg}$
Acceleration of belt, $a=1{\text{m/s}}^2$
Considering ${\mu }_{\text{min}}$, limiting friction $(f={\mu }_{\text{min}}N)$ should be sufficient to ensure that the block moves with same acceleration as the bellt, to avoid slipping.

$FBD$ of block:


Applying equation of dynamics along direction of acceleration,
$f-mg\sin {30}^{\circ }=ma$
$\Rightarrow f=ma+mg\sin {30}^{\circ }$
$\Rightarrow f=(1\times 1)+1\times 10\times \frac{1}{2}=6\text{N}...\left(i\right)$
From equilibrium condition along perpendicular to the inclined surface,
$N=mg\cos {30}^{\circ }$
$\therefore N=1\times 10\times \frac{\sqrt{3}}{2}=5\sqrt{3}\text{N}...\left(ii\right)$
Substituting for limiting friction $\left(f\right)$,
$f={\mu }_{\text{min}}N$
$\Rightarrow 6={\mu }_{\text{min}}\times 5\sqrt{3}$
i.e ${\mu }_{\text{min}}=\frac{6}{5\sqrt{3}}=\frac{2}{5}\sqrt{3}=0.69$