Ratio of Wall Height to Stick Length and Friction Force

To find the ratio $h/l$ and the friction force $f$ acting at the bottom end of the stick, we analyze the conditions for translational and rotational equilibrium of the stick.

1. Geometry and Force Analysis

Let the uniform wooden stick have:

• Mass $m = 1.6\text{ kg}$
• Weight $W = mg = 1.6 \times 10 = 16\text{ N}$ acting vertically downward at its center of mass, at a distance $\frac{l}{2}$ from the lower end.
• Angle with the vertical wall $\theta = 30^\circ$, which implies the angle of inclination of the stick with the horizontal floor is: $\phi = 90^\circ - 30^\circ = 60^\circ$

Let $x$ be the distance along the stick from its bottom end to the top edge of the vertical wall. The height of the wall $h$ is related to $x$ by: $h = x \sin 60^\circ = x \frac{\sqrt{3}}{2} \implies x = \frac{2h}{\sqrt{3}}$

The forces acting on the stick are:

1. Weight $W = 16\text{ N}$ vertically downwards at $\frac{l}{2}$.
2. Normal reaction from the floor $N_f$ vertically upwards at the bottom end.
3. Friction force $f$ horizontally towards the wall at the bottom end.
4. Normal reaction force from the smooth top edge of the wall $N_w$, directed perpendicular to the stick.

Resolving the perpendicular force $N_w$ into components:

• Horizontal component (pointing horizontally away from the wall): $N_w \sin 60^\circ$
• Vertical component (pointing vertically upwards): $N_w \cos 60^\circ$

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2. Translational Equilibrium

For horizontal equilibrium: $\Sigma F_x = 0 \implies f = N_w \sin 60^\circ = N_w \frac{\sqrt{3}}{2} \quad \text{--- (1)}$

For vertical equilibrium: $\Sigma F_y = 0 \implies N_f + N_w \cos 60^\circ = W$ $N_f + \frac{N_w}{2} = 16 \quad \text{--- (2)}$

It is given that the magnitude of the reaction force from the wall equals the normal reaction force from the floor ($N_w = N_f$). Substituting $N_f = N_w$ into equation (2): $N_w + \frac{N_w}{2} = 16 \implies \frac{3}{2} N_w = 16 \implies N_w = \frac{32}{3}\text{ N}$

Substituting $N_w = \frac{32}{3}\text{ N}$ into equation (1) yields the friction force: $f = \left(\frac{32}{3}\right) \frac{\sqrt{3}}{2} = \frac{16\sqrt{3}}{3}\text{ N}$

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3. Rotational Equilibrium

Taking torque about the lower end of the stick resting on the floor: $\Sigma \tau_{\text{bottom}} = 0$ $\tau_{N_w} = \tau_W$

The force $N_w$ is perpendicular to the stick at distance $x$, and the weight $W$ acts at distance $\frac{l}{2}$ with lever arm $\frac{l}{2} \cos 60^\circ$: $N_w \cdot x = W \cdot \frac{l}{2} \cos 60^\circ$

Substitute $N_w = \frac{32}{3}\text{ N}$ and $W = 16\text{ N}$: $\left(\frac{32}{3}\right) x = 16 \times \frac{l}{2} \times \frac{1}{2}$ $\left(\frac{32}{3}\right) x = 4l \implies x = \frac{12}{32} l = \frac{3}{8} l$

Using the relation $h = x \sin 60^\circ$: $h = \left(\frac{3}{8} l\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{3\sqrt{3}}{16} l$

Thus, the ratio $\frac{h}{l}$ is: $\frac{h}{l} = \frac{3\sqrt{3}}{16}$

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Conclusion

• $\frac{h}{l} = \frac{3\sqrt{3}}{16}$
• $f = \frac{16\sqrt{3}}{3}\text{ N}$

This corresponds to Option D.