Question

A uniform solid sphere of mass $1kg$ and radius $10cm$ is kept stationary on a rough inclined plane by fixing a highly dense particle at $B$. Inclination of plane is ${37}^0$ with horizontal and $AB$ is the diameter of the sphere which is parallel to the plane, as shown in the figure. Then,

• A. mass of the particle fixed at B is 4 Kg
• B. mass of the particle fixed at B is 3 Kg
• C. minimum required coefficient of friction between sphere and plane to keep sphere in equilibrium is ${\mu }_{min}$ = $0.75$
• D. minimum required coefficient of friction between sphere and plane to keep sphere in equilibrium is ${\mu }_{min}$ = $0.5$

Answer 1

Answer: (B), (C)

The correct options are
B minimum required coefficient of friction between sphere and plane to keep sphere in equilibrium is ${\mu }_{min}$ = $0.75$
C mass of the particle fixed at B is 3 Kg

Let us assume the mass of the particle to be m. Let us draw the free body diagram of the system of sphere and particle. Since system is in static equilibrium, torque of forces acting on the system should be zero. Taking torque about point $A$ of all of the forces acting on the system
$\left(Mg\sin {37}^0\right)+\left(mg\sin {37}^0\right)R$ = $\left(mg\cos {37}^0\right)R$ $\Rightarrow$ $3kg$
Considering forces normal the plane,
$N$ = $Mg\cos {37}^0+mg\cos {37}^0$ = $32$ $N$
The friction force between sphere and plane is static nature.
Now considering forces along the plane
$f$ = $mg\sin {37}^0+mgsin{37}^0$ = $24$ $N$
But $f$ $\le$ $\frac{f}{n}$ or
${\mu }_{min}$ = $0.75$