Question

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

(i) Mass of the particle fixed at B

• A. 2 kg, 0.50
• B. 3kg, 0.75
• C. 4 kg, 0.75
• D. 4 kg, 0.50

Answer 1

The correct option is B

3kg, 0.75

Let assume the mass of the particle be m. let us draw the free body diagram of the system of sphere and particle. Since system is in static equilibrium, therefore, torque of forces acting on the system should be zero.

Taking torque about point P of all of the forces acting on the system,

$R\left(Mgsin{37}^{\circ }\right)+\left(Mgsin{37}^{\circ }\right)R=\left(Mgcos{37}^{\circ }\right)R\Rightarrow$ $m=3kg$

Considering forces normal the plane, N = Mg cos ${37}^{\circ }$ + mg cos ${37}^{\circ }$ = 32 N

The friction force between sphere and plane is static nature, Now considering forces along the plane,

f= mg sin ${37}^{\circ }$ + mg sin 37 = 24 N

But $f\le \mu N$ where $\mu$ is coefficient of friction, which gives $\mu \ge \frac{f}{N}$ or ${\mu }_{min}$ = 0.75.