Concept Card: Friction in Mechanics – Static, Kinetic & Rolling Friction
1. Physical Origin and Nature of Frictional Forces:
Friction is an electromagnetic contact interaction arising from microscopic surface irregularities (asperities) and cold-welding (intermolecular bonding) at real microscopic points of contact.
– Real contact area ($A_{\text{real}}$) is a minute fraction of the macroscopic apparent area ($A_{\text{apparent}}$): $A_{\text{real}} \ll A_{\text{apparent}}$.
– Increasing normal force $N$ deforms asperities, causing $A_{\text{real}} \propto N$.
Total Contact Force Resultant ($\vec{R}$):
$\vec{R} = \vec{N} + \vec{f} \implies R = \sqrt{N^2 + f^2}$
where $\vec{N}$ is the normal reaction perpendicular to the interface and $\vec{f}$ is the frictional force parallel to the interface.
2. Static Friction ($f_s$) and Limiting Friction ($f_L$):
- Self-Adjusting Nature: Acts when there is an impending tendency for relative motion without actual slipping:
$0 \le f_s \le f_{s,\text{max}}$
When applied driving force $F \le f_{s,\text{max}}$, static friction automatically adjusts to match the applied force: $f_s = F_{\text{applied}}$. - Limiting Friction ($f_L = f_{s,\text{max}}$): The threshold of impending motion:
$f_{s,\text{max}} = \mu_s N$
where $\mu_s$ is the dimensionless coefficient of static friction, independent of the macroscopic area of contact.
3. Kinetic (Sliding) Friction ($f_k$):
Operates once relative slipping begins between contacting surfaces:
$$f_k = \mu_k N$$
- Fundamental Inequality: $\mu_k < \mu_s \implies f_k < f_{s,\text{max}}$. Once surfaces slide, microscopic asperities do not have sufficient time to establish full adhesive bonds, making kinetic friction slightly lower than limiting static friction.
- Velocity Independence: Over practical speed ranges, kinetic friction is nearly independent of relative sliding velocity.
4. Rolling Friction ($f_r$):
Arises from localized elastic and plastic deformation of the wheel and contact plane (creating an indentation and frontal mound ahead of the contact patch).
Magnitude: $f_r = \mu_r \frac{N}{R}$ (where $R$ is wheel radius).
Frictional Hierarchy:
$$f_r \ll f_k < f_{s,\text{max}}$$
5. Graph of Frictional Force vs. Applied Force ($f$ vs. $F$):
- Static Zone ($0 \le F \le f_{s,\text{max}}$): Linear 1:1 line with slope $1$ ($f_s = F$). Acceleration is zero.
- Limiting Peak: Maximum value $f_L = \mu_s N$.
- Kinetic Zone ($F > f_{s,\text{max}}$): Drops to constant plateau $f_k = \mu_k N$. Acceleration is $a = \frac{F – \mu_k N}{m}$.
Solved Examples
Example 1 (Self-Adjusting Static vs. Kinetic Friction):
A block of mass $m = 5.0\text{ kg}$ rests on a rough horizontal floor with $\mu_s = 0.50$ and $\mu_k = 0.40$. Taking $g = 10\text{ m/s}^2$:
Find the magnitude of frictional force and acceleration of the block when a horizontal force $F$ applied to it is:
(a) $F = 10.0\text{ N}$
(b) $F = 24.0\text{ N}$
(c) $F = 25.0\text{ N}$
(d) $F = 35.0\text{ N}$
Solution:
$N = mg = 5.0 \times 10 = 50.0\text{ N}$.
Limiting friction: $f_L = \mu_s N = 0.50 \times 50.0 = 25.0\text{ N}$.
Kinetic friction: $f_k = \mu_k N = 0.40 \times 50.0 = 20.0\text{ N}$.
(a) At $F = 10.0\text{ N} < f_L$: Block is stationary. $f_s = F = 10.0\text{ N}, \quad a = 0$.
(b) At $F = 24.0\text{ N} < f_L$: Block is stationary. $f_s = F = 24.0\text{ N}, \quad a = 0$.
(c) At $F = 25.0\text{ N} = f_L$: Limiting equilibrium. $f_s = 25.0\text{ N}, \quad a = 0$.
(d) At $F = 35.0\text{ N} > f_L$: Block accelerates! Frictional force drops to kinetic value $f = f_k = 20.0\text{ N}$.
Acceleration: $a = \frac{F – f_k}{m} = \frac{35.0 – 20.0}{5.0} = \frac{15.0}{5.0} = 3.0\text{ m/s}^2$.
Example 2 (Why Pulling is Easier than Pushing a Roller):
A lawn roller of mass $m = 50.0\text{ kg}$ is moved with uniform velocity along a horizontal surface with $\mu = 0.25$. A force $F$ acts at $\theta = 37^\circ$ to the horizontal ($\cos 37^\circ = 0.8, \sin 37^\circ = 0.6$). Taking $g = 10\text{ m/s}^2$:
(a) Find $F_{\text{pull}}$ if the roller is pulled.
(b) Find $F_{\text{push}}$ if the roller is pushed.
(c) Provide the physical explanation for the difference.
Solution:
(a) When pulled at $37^\circ$ above the horizontal:
$N = mg – F_{\text{pull}}\sin(37^\circ) = 500 – 0.6 F_{\text{pull}}$.
$F_{\text{pull}}\cos(37^\circ) = \mu N \implies 0.8 F_{\text{pull}} = 0.25(500 – 0.6 F_{\text{pull}}) = 125 – 0.15 F_{\text{pull}}$.
$0.95 F_{\text{pull}} = 125 \implies F_{\text{pull}} = \frac{125}{0.95} \approx 131.58\text{ N}$.
(b) When pushed at $37^\circ$ below the horizontal:
$N = mg + F_{\text{push}}\sin(37^\circ) = 500 + 0.6 F_{\text{push}}$.
$F_{\text{push}}\cos(37^\circ) = \mu N \implies 0.8 F_{\text{push}} = 0.25(500 + 0.6 F_{\text{push}}) = 125 + 0.15 F_{\text{push}}$.
$0.65 F_{\text{push}} = 125 \implies F_{\text{push}} = \frac{125}{0.65} \approx 192.31\text{ N}$.
(c) When pulling, the upward vertical component decreases the normal reaction ($N = mg – F\sin\theta$), decreasing friction $\mu N$. When pushing, the downward vertical component increases the normal reaction ($N = mg + F\sin\theta$), increasing friction.
Example 3 (Two-Block System on a Frictionless Floor):
Block $A$ ($m_1 = 2.0\text{ kg}$) rests on top of block $B$ ($m_2 = 4.0\text{ kg}$), which sits on a frictionless horizontal floor. Coefficients between $A$ and $B$ are $\mu_s = 0.40$ and $\mu_k = 0.30$. Taking $g = 10\text{ m/s}^2$:
(a) Find the maximum force $F_{\text{max}}$ applied to $B$ so that both blocks move together without slipping.
(b) If $F = 36.0\text{ N}$ is applied to $B$, find the acceleration of each block.
(c) If $F = 36.0\text{ N}$ is applied to $A$, find the acceleration of each block.
Solution:
(a) Normal force between $A$ and $B$: $N_A = m_1 g = 20.0\text{ N}$.
Maximum static friction on $A$: $f_{s,\text{max}} = \mu_s N_A = 0.40(20.0) = 8.0\text{ N}$.
Maximum acceleration of $A$ without slipping: $a_{\text{max}} = \frac{f_{s,\text{max}}}{m_1} = \frac{8.0}{2.0} = 4.0\text{ m/s}^2$.
Maximum force on $B$: $F_{\text{max}} = (m_1 + m_2) a_{\text{max}} = (2.0 + 4.0)(4.0) = 24.0\text{ N}$.
(b) For $F = 36.0\text{ N}$ on $B$: Since $F > 24.0\text{ N}$, slipping occurs! Kinetic friction operates: $f_k = \mu_k N_A = 0.30(20.0) = 6.0\text{ N}$.
On $A$: $a_A = \frac{f_k}{m_1} = \frac{6.0}{2.0} = 3.0\text{ m/s}^2$.
On $B$: $a_B = \frac{F – f_k}{m_2} = \frac{36.0 – 6.0}{4.0} = \frac{30.0}{4.0} = 7.5\text{ m/s}^2$.
(c) For $F = 36.0\text{ N}$ on $A$: Lower block $B$ can accelerate at most $a_{B,\text{max}} = \frac{f_{s,\text{max}}}{m_2} = \frac{8.0}{4.0} = 2.0\text{ m/s}^2$.
Threshold force on $A$: $F_{\text{threshold}} = (2.0 + 4.0)(2.0) = 12.0\text{ N} < 36.0\text{ N} \implies$ slipping occurs!
On $B$: $a_B = \frac{f_k}{m_2} = \frac{6.0}{4.0} = 1.5\text{ m/s}^2$.
On $A$: $a_A = \frac{F – f_k}{m_1} = \frac{36.0 – 6.0}{2.0} = \frac{30.0}{2.0} = 15.0\text{ m/s}^2$.
Example 4 (Minimum Pulling Force at Optimum Angle):
A block of mass $m$ rests on a rough horizontal surface with coefficient of friction $\mu$. A force $F$ pulls the block at angle $\theta$ above the horizontal.
(a) Derive the expression for $F(\theta)$ to maintain uniform velocity.
(b) Determine the optimum angle $\theta_{\text{opt}}$ to minimize the required force.
(c) Find the minimum force $F_{\text{min}}$.
Solution:
(a) $N = mg – F\sin\theta$. Horizontal: $F\cos\theta = \mu(mg – F\sin\theta) \implies F(\theta) = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$.
(b) Maximize denominator $D(\theta) = \cos\theta + \mu\sin\theta$:
$\frac{dD}{d\theta} = -\sin\theta + \mu\cos\theta = 0 \implies \tan\theta_{\text{opt}} = \mu \implies \theta_{\text{opt}} = \arctan(\mu)$.
(c) $D_{\text{max}} = \sqrt{1 + \mu^2}$. Hence: $F_{\text{min}} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A block of mass $10.0\text{ kg}$ is placed on a rough horizontal floor with $\mu_s = 0.40$ and $\mu_k = 0.30$. If a horizontal force of $20.0\text{ N}$ is applied to the block, the frictional force acting on the block is ($g = 10\text{ m/s}^2$):
(A) $40.0\text{ N}$
(B) $30.0\text{ N}$
(C) $20.0\text{ N}$
(D) Zero
Problem 2 (JEE Main – Single Correct):
A block of mass $m$ resting on a horizontal table is pulled by a force $F$ acting at angle $\theta$ to the horizontal. The force required to just move the block is minimum when:
(A) $\theta = 0^\circ$
(B) $\theta = 45^\circ$
(C) $\theta = \arctan(\mu)$
(D) $\theta = 90^\circ$
Problem 3 (JEE Main – Single Correct):
A block of mass $m$ is pressed against a vertical wall with a horizontal force $F$. If the coefficient of static friction is $\mu$, the minimum horizontal force required to keep the block from sliding down is:
(A) $\mu mg$
(B) $\frac{mg}{\mu}$
(C) $\frac{\mu}{mg}$
(D) $mg$
Problem 4 (JEE Main – Single Correct):
Which of the following correctly orders the magnitudes of frictional forces for given contact surfaces?
(A) Rolling friction < Kinetic friction < Limiting static friction
(B) Limiting static friction < Kinetic friction < Rolling friction
(C) Kinetic friction < Rolling friction < Limiting static friction
(D) Rolling friction < Limiting static friction < Kinetic friction
Problem 5 (JEE Main – Single Correct):
A heavy crate of mass $M$ is pushed along a horizontal concrete floor with a horizontal force $F$ such that it moves with uniform velocity. The magnitude of the total contact force exerted by the floor on the crate is:
(A) $Mg$
(B) $\mu Mg$
(C) $Mg\sqrt{1 + \mu^2}$
(D) $Mg\sqrt{1 – \mu^2}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are correct regarding friction?
(A) Frictional force can never act in the direction of motion of a body.
(B) Static friction is a self-adjusting force in both magnitude and direction.
(C) The work done by kinetic friction on an isolated two-body system is always non-positive (dissipative).
(D) Static friction can perform positive work on a body in a chosen reference frame.
Problem 7 (JEE Advanced – One or More Correct):
A block of mass $m$ rests on a rough horizontal floor with coefficients $\mu_s$ and $\mu_k$ ($\mu_k < \mu_s$). A horizontal force $F$ is gradually increased from zero. Which of the following is/are correct?
(A) Until $F = \mu_s mg$, the frictional force $f = F$.
(B) At $F = \mu_s mg$, the block is in limiting equilibrium.
(C) Once $F > \mu_s mg$, the acceleration is $\frac{F – \mu_k mg}{m}$.
(D) If $F$ is subsequently reduced to a value below $\mu_s mg$ but above $\mu_k mg$, the block continues to accelerate.
Problem 8 (JEE Advanced – One or More Correct):
Block $A$ sits on top of block $B$, which rests on a frictionless horizontal plane. A horizontal force $F$ is applied to block $B$. Which of the following is/are correct?
(A) Block $A$ accelerates exclusively due to the static friction force exerted on it by block $B$.
(B) The maximum possible acceleration of block $A$ without slipping is $\mu_s g$.
(C) The friction force exerted by block $A$ on block $B$ acts opposite to the direction of $F$.
(D) If slipping occurs, the acceleration of block $A$ becomes independent of the magnitude of $F$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A block of mass $m = 4.0\text{ kg}$ rests on a rough horizontal surface with $\mu_s = 0.60$ and $\mu_k = 0.50$. A horizontal force $F = 30.0\text{ N}$ is applied to the block. Taking $g = 10\text{ m/s}^2$, find the acceleration of the block in $\text{m/s}^2$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A block of mass $m = 2.0\text{ kg}$ is held against a rough vertical wall by applying a horizontal force $F = 50.0\text{ N}$ perpendicular to the wall. The coefficient of static friction between the block and the wall is $\mu_s = 0.50$. Taking $g = 10\text{ m/s}^2$, determine the magnitude of the frictional force in Newtons acting on the block.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (C) | 3. (B) | 4. (A) | 5. (C) | 6. (B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 2.5 | 10. 20
Solution 1:
$f_L = \mu_s mg = 0.40(10.0)(10) = 40.0\text{ N}$. Since $F = 20.0\text{ N} < f_L$, static friction balances applied force: $f_s = F = 20.0\text{ N}$.
Correct Answer: (C)
Solution 2:
$F(\theta) = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$. Minimized when $\tan\theta = \mu \implies \theta = \arctan(\mu)$.
Correct Answer: (C)
Solution 3:
$f_s = mg \le \mu N = \mu F \implies F \ge \frac{mg}{\mu}$.
Correct Answer: (B)
Solution 4:
Rolling friction is orders of magnitude smaller than sliding friction, and kinetic friction is less than limiting static friction ($f_r \ll f_k < f_L$).
Correct Answer: (A)
Solution 5:
$R = \sqrt{N^2 + f_k^2} = \sqrt{(Mg)^2 + (\mu Mg)^2} = Mg\sqrt{1 + \mu^2}$.
Correct Answer: (C)
Solution 6:
(A) False: friction can drive forward motion (e.g., walking, car acceleration). (B), (C), and (D) are verified fundamental principles.
Correct Answer: (B, C, D)
Solution 7:
All statements (A, B, C, D) are verified properties of the static-to-kinetic friction transition.
Correct Answer: (A, B, C, D)
Solution 8:
All statements (A, B, C, D) are fundamental results for stacked block friction mechanics.
Correct Answer: (A, B, C, D)
Solution 9:
$f_L = \mu_s mg = 0.60(4.0)(10) = 24.0\text{ N}$. Since $F = 30.0\text{ N} > f_L$, slipping occurs. $f_k = 0.50(4.0)(10) = 20.0\text{ N}$.
$a = \frac{30.0 – 20.0}{4.0} = 2.5\text{ m/s}^2$.
Correct Answer: 2.5
Solution 10:
$f_L = \mu_s N = 0.50(50.0) = 25.0\text{ N}$. Since $f_L > mg = 20.0\text{ N}$, the block does not slide. By equilibrium: $f_s = mg = 20.0\text{ N}$.
Correct Answer: 20