A uniform wooden stick of mass 1.6 kg of length l rests in an inclined manner on a smooth, vertical

Q: Why is the normal reaction from the wall perpendicular to the stick and not horizontal?

Because the stick is resting on the sharp corner of the wall. When a flat surface (the stick) rests on a point (the corner), the normal force is always perpendicular to the flat surface.

Q: Why did we take the torque about the bottom point? Can we take it about the center of mass?

You can take torque about *any* point! However, taking it about the bottom point is a strategic ninja move. Two forces ($N_2$ and $f$) pass right through this point, meaning their lever arms are zero. This instantly eliminates them from the equation, making the math much easier.

Q: How did we get the lever arm for the wall's normal reaction as $h/\cos 30^\circ$?

Look at the right-angled triangle formed by the wall, the floor, and the stick. The height of the wall is $h$, and the angle the stick makes with the vertical wall is $30^\circ$. Using basic trigonometry, $\cos 30^\circ = \text{adjacent} / \text{hypotenuse} = h / \text{distance along stick}$. Rearranging gives the distance as $h/\cos 30^\circ$.

Q: Why is the friction acting towards the left?

The wall's normal reaction $N_1$ pushes the stick up and to the right. To maintain horizontal equilibrium ($\Sigma F_x = 0$), there must be a force pushing to the left. That force is the static friction from the floor.

Q: What if the wall was not smooth?

If the wall were rough, there would be an additional frictional force acting along the stick at the point of contact with the wall. This would add another unknown to our equations, requiring more information (like the coefficient of friction of the wall) to solve.

Question

A uniform wooden stick of mass $1.6 	ext{ kg}$ of 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 $\frac{h}{l}$ and the frictional force $f$ at the bottom of the stick are ($g = 10 	ext{ ms}^{-2}$)

(2016 Adv.)

Options:

(A) $\frac{h}{l} = \frac{\sqrt{3}}{16}, f = \frac{16\sqrt{3}}{3} 	ext{ N}$

(B) $\frac{h}{l} = \frac{3}{16}, f = \frac{16\sqrt{3}}{3} 	ext{ N}$

(C) $\frac{h}{l} = \frac{3\sqrt{3}}{16}, f = \frac{8\sqrt{3}}{3} 	ext{ N}$

(D) $\frac{h}{l} = \frac{3\sqrt{3}}{16}, f = \frac{16\sqrt{3}}{3} 	ext{ N}$

Sigma Percentile · Accepted Answer

### The Geometry of the Leaning Stick

Imagine you are tasked with balancing a heavy wooden stick against a wall. The first step in any mechanics problem is to draw a pristine Free Body Diagram (FBD) and understand the geometry. The problem states that the stick rests on a vertical wall of height $h$ and extends beyond it. This is a subtle but critical detail: the stick is resting on the *corner* of the wall, not flat against it.

Because the stick makes an angle of $30^\circ$ with the vertical wall, it must make an angle of $60^\circ$ with the horizontal floor. This geometric orientation is the foundation upon which we will build our force and torque equations.

### The Arsenal of Forces

Let's break down the forces acting on our stick. First, gravity pulls downwards from the center of mass with a force $mg$. At the bottom, the floor pushes up with a normal reaction $N_2$. To prevent the stick from sliding away, static friction $f$ acts horizontally along the floor.

Now, look at the point of contact with the wall. Because the stick rests on a sharp corner, the normal reaction $N_1$ from the wall is strictly perpendicular to the stick itself, not horizontal! This means $N_1$ has both a vertical and a horizontal component.

### The Master Key: Force Equilibrium

The problem hands us a beautiful constraint: the reaction of the wall equals the reaction of the floor. Mathematically, this means $N_1 = N_2$. We can use this to unlock the force equilibrium equations.

For vertical equilibrium, the upward forces must balance the downward weight:

$$N_1 \sin 30^\circ + N_2 = mg$$

Substituting $N_2 = N_1$, we get:

$$N_1 (0.5 + 1) = mg \implies 1.5 N_1 = mg \implies N_1 = \frac{2}{3} mg$$

For horizontal equilibrium, the leftward friction must balance the rightward push from the wall:

$$f = N_1 \cos 30^\circ$$

Substituting our value for $N_1$ and the given mass $m = 1.6 	ext{ kg}$, we find the frictional force:

$$f = \left(\frac{2}{3} 	imes 16ight) \left(\frac{\sqrt{3}}{2}ight) = \frac{16\sqrt{3}}{3} 	ext{ N}$$

### The Elegance of Torque

Now, we unleash the power of rotational equilibrium to find the ratio $\frac{h}{l}$. By choosing the bottom point of the stick as our pivot, we execute a strategic ninja move. Two forces ($N_2$ and $f$) pass right through this point, meaning their lever arms are zero. This instantly annihilates them from our torque equation!

We only need to balance the torque from the weight $mg$ and the wall's reaction $N_1$. The perpendicular distance of $mg$ from the pivot is $\frac{l}{2} \cos 60^\circ$. The distance of $N_1$ along the stick is found using basic trigonometry as $\frac{h}{\cos 30^\circ}$.

Equating the torques gives us:

$$N_1 \left(\frac{h}{\cos 30^\circ}ight) = mg \left(\frac{l}{2} \cos 60^\circight)$$

Rearranging this elegant equation to isolate our desired ratio yields:

$$\frac{h}{l} = \left(\frac{mg}{N_1}ight) \frac{\cos 60^\circ \cos 30^\circ}{2}$$

Plugging in $\frac{mg}{N_1} = \frac{3}{2}$ and the standard trigonometric values, we arrive at our final, satisfying result:

$$\frac{h}{l} = \frac{3\sqrt{3}}{16}$$

Select Answer:

Visualized Solution

Visualizing the Setup

Identifying the Forces

The Given Condition

Vertical Equilibrium

Solving for Normal Reaction

Horizontal Equilibrium

Calculating Frictional Force

Torque Equilibrium Setup

Calculating Lever Arms

The Torque Equation

Rearranging for Ratio $\frac{h}{l}$

Final Calculation

The Sigma Insight: Torque and Angular Momentum

The Geometry of the Leaning Stick

The Arsenal of Forces

The Master Key

The Elegance of Torque

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