Angle of Repose

The angle of repose is a fascinating concept in physics and engineering, referring to the steepest angle at which a pile of granular material remains stable without sliding. This angle varies depending on factors such as the material's size, shape, and moisture content. Understanding the angle of repose is crucial in fields like civil engineering, where it influences the design of slopes, retaining walls, and embankments to prevent landslides. In everyday life, the angle of repose can be observed in simple activities, such as pouring sugar or sand, where the material naturally forms a mound with a characteristic slope. This concept also metaphorically applies to personal resilience, symbolizing the delicate balance between stability and the pressures we face. Just as materials have their limits, so do individuals, and maintaining that balance is essential for a grounded and stable life.

Angle of Repose

The angle of repose is the maximum angle at which a pile of granular material, such as sand or gravel, can rest on a surface without sliding or collapsing. It is a key concept in understanding the behaviour of loose materials and is influenced by factors such as particle size, shape, and moisture content.

Consider an inclined plane, whose inclination horizontally is gradually increased, till the body placed on its surface just begins to slide down. If $\theta$ is the inclination at which the body just begins to slide down, then $\theta$ is called the angle of repose.

The angle of repose is defined as the angle of the inclined plane horizontally such that the body is placed on it just begins to slide.

Here α is the angle of repose, F is the limiting friction, and R is a normal reaction.

From the figure,
$
R=m g \cos \alpha
$

$\mathrm{F}=\mathrm{mg} \sin \alpha$ and
we know that

$
\begin{aligned}
& \frac{F}{R}=\tan \alpha \\
& \frac{F}{R}=\mu_s=\tan \theta=\tan \alpha
\end{aligned}
$
So $\tan \alpha=\mu_{\mathrm{s}} \Rightarrow \alpha=\tan ^{-1}\left(\mu_{\mathrm{s}}\right)$

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Solved Examples Based on the Angle of Repose

Qu 1. A body will begin to move down an inclined plane if the angle of inclination of the plane is ____________ the angle of friction.
1) equal to

2) less than

3) greater than

4) none of the above

Solution

Angle of Repose

The angle of repose is defined as the angle of the inclined plane horizontally such that the body is placed on it just begins to slide.

Here α is the angle of repose, F is the limiting friction, and R is a normal reaction.

From the figure,

$R=m g \cos \alpha$
$\mathrm{F}=\mathrm{mg} \sin \alpha$ and we know that $\frac{F}{R}=\tan \alpha$ $\frac{F}{R}=\mu_s=\tan \theta=\tan \alpha$ $\therefore \tan \alpha=\mu_{\mathrm{s}} \Rightarrow a=\tan -1\left(\mu_{\mathrm{s}}\right)$

So,

The angle of inclination should be greater than the angle of friction.

Hence, the answer is the option (3).

Qu 2. A body of weight W is placed on an inclined plane. The angle made by the inclined plane with the horizontal, when the body is on the point of moving down is called
1) angle of friction

2) angle of inclination

3) angle of repose

4) angle of limiting friction

Solution

Angle of Repose

The angle of repose is defined as the angle of the inclined plane horizontally such that the body is placed on it just begins to slide.

Here $\alpha$ is the angle of repose, F is the limiting friction, and R is a normal reaction.
From the figure,
$R=m g \cos \alpha$
$\mathrm{F}=\mathrm{mg} \sin \alpha$ and
we know that

$
\frac{F}{R}=\tan \alpha \frac{F}{R}=\mu_s=\tan \theta=\tan \alpha
$

$\therefore \tan \alpha=\mu_{\mathrm{s}} \Rightarrow \alpha=\tan -1\left(\mu_{\mathrm{s}}\right)$

Hence, the answer is the option (3).

Qu 3. A block rolled on a rough surface with a velocity of 8m/s comes to rest after travelling 4m. Compute the coefficient of friction (g=10m/s²)

1) 0.8

2) 0.6

3) 0.4

4) 0.5

Solution

The initial velocity of the block, $u=8 \mathrm{~m} / \mathrm{s}$,
Distance travelled before the block stops, $\mathrm{s}=4 \mathrm{~m}$
Let the coefficient of kinetic friction between the block and surface be $\mu_k$

F.B.D of the block

From F.B.D

$
\begin{aligned}
& N=m g \\
& f_k=\mu_k N \\
& \Rightarrow f_k=\mu_k m g
\end{aligned}
$
Assuming the acceleration of the block to be 'a'.

$
\begin{aligned}
& N=m g \\
& f_k=\mu_k N \\
& \Rightarrow f_k=\mu_k m g \\
& F_{n e t}=m a \\
& -f_k=m a \\
& \Rightarrow a=-\mu_k g
\end{aligned}
$
The block stops $(\mathrm{v}=0)$ after a travelling displacement of 4 m.
Applying 3rd equation of motion

$
\begin{aligned}
& v^2=u^2+2 a s \\
& 0=u^2-2 \mu_k g s \\
& \Rightarrow \mu_k=\frac{u^2}{2 g s}=\frac{8 \times 8}{2 \times 10 \times 4}=0.8
\end{aligned}
$

Hence, the answer is the option (1).

Qu 4. An inclined plane is bent in such a way that the vertical cross-section is given by $y=\frac{x^2}{4}$ where y is in the vertical and x is in the horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction $\mu=0.5$ the maximum height in cm at which a stationary block will not slip downward is $\qquad$ cm.

1) 25

2) 15

3) 20

4) 10

Solution:

At maximum height, the block will experience maximum friction force. Therefore if at this height the slope of the tangent is $\tan \theta$, then $\theta=$ Angle of repose.

$
\begin{aligned}
& \therefore \tan \theta=\frac{d y}{d x}=\frac{2 x}{4}=\frac{x}{2} \\
& \mu=\tan \theta=0.5 \Rightarrow x=1
\end{aligned}
$

and therefore $y=\frac{x^2}{4}=0.25 \mathrm{~m}=25 \mathrm{~cm}$

Hence, the answer is the option (1).

Summary

The angle of repose is the steepest angle at which a pile of granular material remains stable without sliding. It is determined by factors like particle size, shape, and friction. The angle of repose is crucial in engineering, affecting the design of slopes and structures to prevent sliding. It is mathematically expressed as $\alpha=\tan ^{-1}\left(\mu_s\right)$, where $\mu_s$ is the coefficient of static friction. Understanding this concept is key in predicting when a body on an inclined plane will begin to slide, influencing various practical applications and problem-solving scenarios.