Question

A uniform wooden stick of mass $1.6kg$ and length $l$ rests in an inclined manner on a smooth, vertical wall of height $h(<l)$ such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of ${30}^{\circ }$ with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio of $h/l$ and the frictional force $f$ at the bottom of the stick are respectively
(take, $g=10m{s}^{-2}$)

• A. $\frac{h}{l}=\frac{\sqrt{3}}{16},f=\frac{16\sqrt{3}}{3}N$
• B. $\frac{h}{l}=\frac{3}{16},f=\frac{16\sqrt{3}}{3}N$
• C. $\frac{h}{l}=\frac{3\sqrt{3}}{16},f=\frac{8\sqrt{3}}{3}N$
• D. $\frac{h}{l}=\frac{3\sqrt{3}}{16},f=\frac{16\sqrt{3}}{3}N$

Answer 1

The correct option is D $\frac{h}{l}=\frac{3\sqrt{3}}{16},f=\frac{16\sqrt{3}}{3}N$

Let $x$ lenght of stick is in contact with floor.
For equlibrium condition,
$\Sigma F_y=0\Rightarrow N+N\times sin\left({30}^o\right)=mgN+\frac{N}{2}=mgN=\frac{2mg}{3}$
and, $\Sigma F_X=0$
So, $Ncos\left({30}^o\right)\Rightarrow \frac{N\sqrt{3}}{2}=\frac{16\sqrt{3}}{3}N$
Balancing torque about the lowest point, we get
Torque balance equation $\Rightarrow {\tau }_{net}=0$
$N\times x=mg\times l/2\times sin\left({30}^o\right)$
$\Rightarrow \frac{2mg}{3}\times x=mg\times l/2\times sin\left({30}^o\right)x=\frac{3l}{8}$
By analysing the right angle tringle,
$cos\left({30}^o\right)=\frac{h}{x}\Rightarrow \frac{\sqrt{3}}{2}=\frac{8h}{3l}\Rightarrow \frac{h}{l}=\frac{3\sqrt{3}}{16}$
Hence, answer is option $D.$