Concept Card: Angle of Friction, Angle of Repose & Rough Incline Dynamics
1. Resultant Contact Force & Angle of Friction ($\lambda$):
The angle of friction ($\lambda$) is defined as the angle made by the total resultant contact force $\vec{R}$ with the normal force $\vec{N}$ at the verge of impending relative sliding (limiting equilibrium):
$$\vec{R} = \vec{N} + \vec{f}_L \implies \tan\lambda = \frac{f_L}{N} = \frac{\mu_s N}{N} = \mu_s \implies \lambda = \arctan(\mu_s)$$
- Magnitude of resultant contact force: $R = \sqrt{N^2 + f_L^2} = N\sqrt{1 + \mu_s^2} = \frac{N}{\cos\lambda}$.
- Cone of Friction: The geometric cone of semi-vertical angle $\lambda$ centered around the normal $\vec{N}$. Relative sliding is impossible unless an external applied force tilts the resultant contact force outside this cone.
2. Angle of Repose ($\theta_r$ or $\phi$):
The maximum angle of inclination of a rough plane with the horizontal at which an unassisted body rests in limiting static equilibrium without sliding down under gravity alone:
$$mg\sin\theta_r = \mu_s mg\cos\theta_r \implies \tan\theta_r = \mu_s \implies \theta_r = \arctan(\mu_s)$$
Fundamental Identity:
$$\theta_r = \lambda = \arctan(\mu_s)$$
Regimes of Incline Behavior:
– Case 1 ($\theta < \theta_r$): Static equilibrium ($f_s = mg\sin\theta < f_L$). Acceleration is zero.
– Case 2 ($\theta = \theta_r$): Limiting static equilibrium on the verge of sliding down ($f_s = f_L = \mu_s mg\cos\theta$). Acceleration is zero.
– Case 3 ($\theta > \theta_r$): Active sliding under kinetic friction ($f_k = \mu_k mg\cos\theta$).
3. Dynamics on a Rough Inclined Plane:
- Acceleration Sliding Down ($\theta > \theta_r$):
$a_{\text{down}} = g(\sin\theta – \mu_k\cos\theta)$
Time of descent: $t_{\text{down}} = \sqrt{\frac{2L}{g(\sin\theta – \mu_k\cos\theta)}}$ - Retardation Moving Up (Projected with initial speed $u$):
$a_{\text{up}} = g(\sin\theta + \mu_k\cos\theta)$
Time of ascent: $t_{\text{up}} = \frac{u}{g(\sin\theta + \mu_k\cos\theta)}$ - Ascent vs. Descent Asymmetry:
Because $a_{\text{up}} > a_{\text{down}}$, the time of descent over the same distance is strictly greater than the time of ascent:
$\frac{t_{\text{down}}}{t_{\text{up}}} = \sqrt{\frac{a_{\text{up}}}{a_{\text{down}}}} = \sqrt{\frac{\sin\theta + \mu_k\cos\theta}{\sin\theta – \mu_k\cos\theta}} = \sqrt{\frac{\tan\theta + \mu_k}{\tan\theta – \mu_k}} > 1$ - The $n$-Times Time Ratio Theorem:
If a body takes $n$ times as long to slide down a rough incline as down an identical smooth incline of angle $\theta$:
$\mu_k = \tan\theta\left(1 – \frac{1}{n^2}\right)$ - Optimum Pulling Force Up an Incline:
The minimum force required to drag a crate up a rough incline occurs when pulling at an angle $\alpha = \lambda$ above the inclined surface:
$F_{\text{min}} = Mg\sin(\theta + \lambda)$
Solved Examples
Example 1 (Angle of Repose & Statics-to-Dynamics Regimes):
A block of mass $m = 4.0\text{ kg}$ is on an adjustable incline with $\mu_s = \frac{1}{\sqrt{3}} \approx 0.577$ and $\mu_k = 0.50$. Taking $g = 10\text{ m/s}^2$:
(a) Find the angle of repose $\theta_r$.
(b) If tilted at $\theta = 20^\circ$ ($\sin 20^\circ \approx 0.342$), find the frictional force acting on the block.
(c) If tilted at $\theta = 45^\circ$, find the acceleration of the block down the incline.
Solution:
(a) $\tan\theta_r = \mu_s = \frac{1}{\sqrt{3}} \implies \theta_r = 30^\circ$.
(b) At $\theta = 20^\circ < \theta_r = 30^\circ$: The block remains at rest. Static friction self-adjusts to balance gravity: $f_s = mg\sin(20^\circ) = 4.0 \times 10 \times 0.342 = 13.68\text{ N}$ directed up along the plane ($a = 0$).
(c) At $\theta = 45^\circ > \theta_r = 30^\circ$: The block accelerates down under kinetic friction:
$a_{\text{down}} = g(\sin 45^\circ – \mu_k\cos 45^\circ) = 10\left(\frac{1}{\sqrt{2}} – 0.50\frac{1}{\sqrt{2}}\right) = \frac{5}{\sqrt{2}} \approx 3.54\text{ m/s}^2$.
Example 2 (The $n$-Times Incline Descent Time Theorem):
A block takes $\sqrt{2}$ times as long to slide down a $45^\circ$ rough inclined plane as it does to slide down a frictionless plane of identical length and slope. Taking $g = 10\text{ m/s}^2$:
(a) Derive the expression for $\mu_k$ in terms of $\theta$ and ratio $n$.
(b) Calculate the numerical value of $\mu_k$.
(c) If the rough incline length is $L = 10.0\text{ m}$, calculate the speed of the block at the bottom.
Solution:
(a) $t_{\text{rough}} = n \cdot t_{\text{smooth}} \implies \sqrt{\frac{g\sin\theta}{g(\sin\theta – \mu_k\cos\theta)}} = n \implies \mu_k = \tan\theta\left(1 – \frac{1}{n^2}\right)$.
(b) With $n = \sqrt{2}$ ($n^2 = 2$) and $\theta = 45^\circ$ ($\tan 45^\circ = 1$):
$\mu_k = 1 \times \left(1 – \frac{1}{2}\right) = 0.50$.
(c) $a_{\text{rough}} = 10\left(\frac{1}{\sqrt{2}} – 0.5\frac{1}{\sqrt{2}}\right) = \frac{5}{\sqrt{2}}\text{ m/s}^2$.
$v = \sqrt{2 a_{\text{rough}} L} = \sqrt{2\left(\frac{5}{\sqrt{2}}\right)(10.0)} = \sqrt{100\sqrt{2}} \approx 11.89\text{ m/s}$.
Example 3 (Ascent vs. Descent Cycle on a Rough Incline):
A small block is projected up an incline of slope $\theta = 37^\circ$ with initial speed $u = 10.0\text{ m/s}$. The kinetic friction coefficient is $\mu_k = 0.25$ ($\cos 37^\circ = 0.8, \sin 37^\circ = 0.6$). Taking $g = 10\text{ m/s}^2$:
(a) Find the retardation $a_{\text{up}}$ and the distance $s$ traversed before stopping.
(b) Calculate the time of ascent $t_{\text{up}}$.
(c) Calculate the acceleration $a_{\text{down}}$, time of descent $t_{\text{down}}$, and returning speed $v$.
Solution:
(a) $a_{\text{up}} = g(\sin 37^\circ + \mu_k\cos 37^\circ) = 10[0.6 + 0.25(0.8)] = 10(0.8) = 8.0\text{ m/s}^2$.
Distance: $s = \frac{u^2}{2a_{\text{up}}} = \frac{100}{16} = 6.25\text{ m}$.
(b) Time of ascent: $t_{\text{up}} = \frac{u}{a_{\text{up}}} = \frac{10.0}{8.0} = 1.25\text{ s}$.
(c) Downward driving gravity $mg\sin 37^\circ = 0.6 mg > \mu_s mg\cos 37^\circ = 0.20 mg \implies$ block slides down.
$a_{\text{down}} = g(\sin 37^\circ – \mu_k\cos 37^\circ) = 10[0.6 – 0.2] = 4.0\text{ m/s}^2$.
$t_{\text{down}} = \sqrt{\frac{2s}{a_{\text{down}}}} = \sqrt{\frac{12.5}{4.0}} = \sqrt{3.125} \approx 1.77\text{ s} > t_{\text{up}}$.
Returning speed: $v = \sqrt{2a_{\text{down}}s} = \sqrt{2(4.0)(6.25)} = \sqrt{50.0} = 5\sqrt{2} \approx 7.07\text{ m/s} < 10.0\text{ m/s}$.
Example 4 (Minimum Pulling Force at Optimum Angle on Incline):
A crate of mass $M$ rests on a rough incline of angle $\theta$ with angle of friction $\lambda$. A force $F$ acts at angle $\alpha$ above the inclined plane.
(a) Derive the expression for $F(\alpha)$ to drag the crate up with uniform velocity.
(b) Find the optimum angle $\alpha_{\text{opt}}$.
(c) Determine the minimum pulling force $F_{\text{min}}$.
Solution:
(a) Normal force: $N = Mg\cos\theta – F\sin\alpha$.
Along incline: $F\cos\alpha = Mg\sin\theta + \mu_s(Mg\cos\theta – F\sin\alpha)$.
With $\mu_s = \tan\lambda = \frac{\sin\lambda}{\cos\lambda}$:
$F(\cos\alpha\cos\lambda + \sin\alpha\sin\lambda) = Mg(\sin\theta\cos\lambda + \cos\theta\sin\lambda) \implies F\cos(\alpha – \lambda) = Mg\sin(\theta + \lambda)$.
$F(\alpha) = \frac{Mg\sin(\theta + \lambda)}{\cos(\alpha – \lambda)}$.
(b) Maximize denominator: $\cos(\alpha – \lambda) = 1 \implies \alpha_{\text{opt}} = \lambda = \arctan(\mu_s)$.
(c) Minimum force: $F_{\text{min}} = Mg\sin(\theta + \lambda)$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
If $\mu$ is the coefficient of static friction between a body and a horizontal plane, the angle of friction $\lambda$ is given by:
(A) $\lambda = \arcsin(\mu)$
(B) $\lambda = \arccos(\mu)$
(C) $\lambda = \arctan(\mu)$
(D) $\lambda = \arctan(1/\mu)$
Problem 2 (JEE Main – Single Correct):
A block rests on an inclined plane of inclination $\theta$. As $\theta$ is gradually increased, the block just begins to slide down at $\theta = 30^\circ$. The coefficient of static friction $\mu_s$ between the block and the plane is:
(A) $\frac{1}{2}$
(B) $\frac{1}{\sqrt{3}}$
(C) $\frac{\sqrt{3}}{2}$
(D) $\sqrt{3}$
Problem 3 (JEE Main – Single Correct):
A body of mass $m$ slides down a rough inclined plane of inclination $\theta$ with uniform velocity. The coefficient of kinetic friction is:
(A) $\sin\theta$
(B) $\cos\theta$
(C) $\tan\theta$
(D) $\cot\theta$
Problem 4 (JEE Main – Single Correct):
A block takes twice as long to slide down a $45^\circ$ rough inclined plane as it does to slide down an identical smooth inclined plane. The coefficient of friction of the rough plane is:
(A) $0.25$
(B) $0.50$
(C) $0.75$
(D) $0.80$
Problem 5 (JEE Main – Single Correct):
A particle is projected up a rough inclined plane of angle $\theta$ and then slides back down to the point of projection. If $t_1$ is the time of ascent and $t_2$ is the time of descent, then:
(A) $t_1 = t_2$
(B) $t_1 > t_2$
(C) $t_1 < t_2$
(D) $t_1 t_2 = 1$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are correct regarding the angle of repose $\theta_r$ and the angle of friction $\lambda$?
(A) The angle of repose is numerically equal to the angle of friction for the same pair of surfaces.
(B) If an inclined plane has slope $\theta < \theta_r$, the frictional force on a stationary body is $mg\sin\theta$.
(C) The total contact force exerted by a rough surface on a body in limiting equilibrium is $N\sqrt{1 + \mu_s^2}$.
(D) The direction of the resultant contact force in limiting equilibrium makes an angle $\lambda$ with the normal to the surface.
Problem 7 (JEE Advanced – One or More Correct):
A block of mass $m$ slides down a rough incline of angle $\theta > \theta_r$ with acceleration $a = g(\sin\theta – \mu\cos\theta)$. Which of the following is/are correct?
(A) Mechanical energy of the block is not conserved; the loss in mechanical energy equals work done against friction.
(B) The normal force $N = mg\cos\theta$ does zero work on the block.
(C) The power dissipated by friction at velocity $v$ is $\mu mg v \cos\theta$.
(D) If $\tan\theta = \mu$, the acceleration of the block down the incline is zero.
Problem 8 (JEE Advanced – One or More Correct):
A block of mass $m$ is on a rough inclined plane of angle $\theta$. Let $\mu$ be the coefficient of friction. Which of the following expressions is/are correct?
(A) The minimum force applied parallel to the incline to push the block up is $mg(\sin\theta + \mu\cos\theta)$.
(B) The minimum force applied parallel to the incline to prevent the block from sliding down is $mg(\sin\theta – \mu\cos\theta)$ (for $\tan\theta > \mu$).
(C) The minimum force in any direction required to pull the block up the incline is $mg\sin(\theta + \lambda)$, where $\tan\lambda = \mu$.
(D) The minimum force in any direction must be directed at an angle $\lambda$ above the inclined surface.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A block slides down a rough inclined plane of inclination $\theta = 45^\circ$ with an acceleration of $a = 2.0\text{ m/s}^2$. Taking $g = 10\text{ m/s}^2$ and $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$, determine the value of the coefficient of kinetic friction $\mu_k$ rounded to two decimal places (evaluate $100 \times \mu_k$).
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A block of mass $m = 6.0\text{ kg}$ rests on a rough inclined plane of slope $\theta = 30^\circ$. The coefficient of static friction is $\mu_s = 0.20$. Taking $g = 10\text{ m/s}^2$, find the minimum force $F_{\text{min}}$ in Newtons applied parallel to the incline required to prevent the block from sliding down (round to the nearest integer).
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (B) | 3. (C) | 4. (C) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 72 | 10. 20
Solution 1:
By definition: $\tan\lambda = \frac{f_L}{N} = \mu \implies \lambda = \arctan(\mu)$.
Correct Answer: (C)
Solution 2:
At angle of repose: $\tan\theta_r = \mu_s \implies \mu_s = \tan(30^\circ) = \frac{1}{\sqrt{3}}$.
Correct Answer: (B)
Solution 3:
Uniform velocity means $a = 0 \implies mg\sin\theta – \mu_k mg\cos\theta = 0 \implies \mu_k = \tan\theta$.
Correct Answer: (C)
Solution 4:
$\mu = \tan\theta\left(1 – \frac{1}{n^2}\right) = \tan(45^\circ)\left(1 – \frac{1}{2^2}\right) = 1 – 0.25 = 0.75$.
Correct Answer: (C)
Solution 5:
$a_{\text{up}} = g(\sin\theta + \mu\cos\theta) > a_{\text{down}} = g(\sin\theta – \mu\cos\theta) \implies t_1 < t_2$.
Correct Answer: (C)
Solution 6:
All statements (A, B, C, D) are verified mathematical and physical properties of the angle of repose, angle of friction, and contact force resultant.
Correct Answer: (A, B, C, D)
Solution 7:
All statements (A, B, C, D) are verified kinematic and thermodynamic properties of motion on a rough incline.
Correct Answer: (A, B, C, D)
Solution 8:
All statements (A, B, C, D) are standard verified formulas for forces on a rough inclined plane.
Correct Answer: (A, B, C, D)
Solution 9:
$a = \frac{10}{\sqrt{2}}(1 – \mu_k) = 2.0 \implies 1 – \mu_k = \frac{2\sqrt{2}}{10} \approx 0.2828 \implies \mu_k \approx 0.7172 \approx 0.72$. $100\mu_k = 72$.
Correct Answer: 72
Solution 10:
$F_{\text{min}} = mg\sin(30^\circ) – \mu_s mg\cos(30^\circ) = 30.0 – 0.20(60)\left(\frac{\sqrt{3}}{2}\right) = 30.0 – 10.39 = 19.61\text{ N} \approx 20\text{ N}$.
Correct Answer: 20