Question

A sphere is placed rotating with its centre initially at rest in a corner as shown in the figures (a) and (b). Coefficient of friction between all surfaces and the sphere is $\frac{1}{3}$. Find the ratio of the friction forces $\frac{f_a}{f_b}$ by ground in situations (a) and (b).

• A. $1$
• B. $\frac{9}{10}$
• C. $\frac{10}{9}$
• D. $noneofthese$

Answer 1

The correct option is B $\frac{9}{10}$

From the FBD of figure a, we get the translational equilibrium in vertical direction,
$\mu N_1+N_2=mg$ ....(i)
In horizontal directions,
$N_1=\mu N_2$....(ii)
Solving, Eqs. (i) and (ii), $N_2=\frac{mg}{1+{\mu }^2}$
The required friction is
$\mu N_2=f_a=\frac{3}{10}mg$
Now in case of figure b, we see that the sphere has the tendency to move towards the right and thus there would be no interaction between the sphere and the vertical surface. Thus the normal reaction in the horizontal direction would be zero. Thus we get
$N_1=0;N_2=mg$
$f_b=\mu N_2=\frac{mg}{3}$
which gives $\frac{f_a}{f_b}=\frac{9}{10}$