Question

A sphere is rotating between two rough inclined walls as shown in figure (1). Coefficient of friction between each wall and the sphere is 1/3. If f₁ and f₂ be the friction forces at P and Q. Then $\frac { \mathrm { f } _ { 1 } } { \mathrm { f } _ { 2 } }$ is –

• A. $\frac { 4 } { \sqrt { 3 } }$ + 1
• B. $\frac { 1 } { \sqrt { 3 } }$ + 2
• C. $\frac { 1 } { 2 }$ + $\sqrt { 3 }$
• D. 1 + 2 $\sqrt { 3 }$

Answer 1

Answer: (A)

$\frac { 4 } { \sqrt { 3 } }$ + 1

Answer 2

Let µ be the friction coefficient between sphere and each wall. Free body diagram of sphere is

Net force on the sphere in horizontal direction is zero.

\ N₁ cos 60⁰ + µ N₂ cos 60⁰ = N₂ cos 30º + µ N₁ cos 30⁰

or N₁ + µN₂ = $\sqrt { 3 }$ (N₂ + µN₁)

or N₁ (1 – $\sqrt { 3 }$ µ) = N₂ ( $\sqrt { 3 }$ – µ)

\ $\frac { N _ { 1 } } { N _ { 2 } }$ = $\frac { \sqrt { 3 } - \mu } { 1 - \sqrt { 3 } \mu }$

Substituting µ = $\frac { 1 } { 3 }$ we get,

$\frac { N _ { 1 } } { N _ { 2 } }$ = $\frac { \sqrt { 3 } - \frac { 1 } { 3 } } { 1 - \frac { \sqrt { 3 } } { 3 } }$ = $\frac { 3 \sqrt { 3 } - 1 } { 3 - \sqrt { 3 } }$ = 1 + $\frac { 4 } { \sqrt { 3 } }$

Now $\frac { \mathrm { f } _ { 1 } } { \mathrm { f } _ { 2 } }$ = $\frac { \mu \mathrm { N } _ { 1 } } { \mu \mathrm { N } _ { 2 } }$ = 1 + $\frac { 4 } { \sqrt { 3 } }$