Laws of Motion • Topic 3 of 3

Friction & Circular Dynamics

Friction is the force that opposes relative motion between surfaces in contact. It arises from the interlocking of microscopic irregularities and acts parallel to the surface, opposite to the direction of motion (or attempted motion). Without it we could not walk, drive or even hold a pencil.

  • Static friction ($f_s$) acts when the body is still. It is self-adjusting — it grows to match the applied force up to a maximum (limiting) value: $f_s \le \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ the normal reaction.
  • Kinetic friction ($f_k$) acts once the body slides, and is nearly constant: $f_k=\mu_k N$. In general $\mu_k < \mu_s$, which is why it is harder to start an object moving than to keep it moving.
  • Friction depends on the nature of the surfaces and the normal reaction, but (to a good approximation) not on the area of contact.

Motion on an inclined plane. For a block on a plane inclined at angle $\theta$, the weight resolves into $mg\sin\theta$ down the slope and $mg\cos\theta$ perpendicular to it, so $N=mg\cos\theta$. The block is on the verge of sliding when $mg\sin\theta=\mu_s mg\cos\theta$, giving the angle of repose $\theta=\tan^{-1}\mu_s$. Once sliding, its acceleration is $a=g(\sin\theta-\mu_k\cos\theta)$.

Dynamics of circular motion. A body moving in a circle of radius $r$ at speed $v$ is always accelerating toward the centre — the centripetal acceleration $a_c=\frac{v^2}{r}$. A real force must supply this; the required centripetal force is $F=\frac{mv^2}{r}$. For a car on a level road this is provided by friction; the maximum safe speed is $v_{max}=\sqrt{\mu_s rg}$.

Banking of roads. To turn at higher speed safely, roads are banked (tilted inward at angle $\theta$) so the horizontal component of the normal reaction supplies the centripetal force, reducing reliance on friction. For the ideal (frictionless) banking speed, $\tan\theta=\frac{v^2}{rg}$. A conical pendulum (a bob whirled in a horizontal circle on a string making angle $\theta$ with the vertical) obeys the same relation: the tension's horizontal component gives the centripetal force while its vertical component balances gravity, so again $\tan\theta=\frac{v^2}{rg}$.

Block on an incline: weight components, normal reaction and frictionthetammgNf (friction)mg sin thetaN = mg cos theta; angle of repose: tan theta = mu_s
Solved Examples
1
Worked Example
A block of mass 10 kg rests on a horizontal surface with coefficient of static friction $\mu_s=0.4$. Find the maximum (limiting) static friction. (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: On a horizontal surface $N=mg=10 \times 10=100\ \text{N}$.
  2. Step 2: Limiting static friction $f_s=\mu_s N$.
  3. Step 3: Substitute: $f_s=0.4 \times 100$.
  4. Step 4: Compute: $f_s=40\ \text{N}$.

Answer: $f_s=40\ \text{N}$

2
Worked Example
A 5 kg block slides on a floor with kinetic friction coefficient $\mu_k=0.2$. Find the kinetic friction force. (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: Normal reaction $N=mg=5 \times 10=50\ \text{N}$.
  2. Step 2: Use $f_k=\mu_k N$.
  3. Step 3: Substitute: $f_k=0.2 \times 50$.
  4. Step 4: Compute: $f_k=10\ \text{N}$.

Answer: $f_k=10\ \text{N}$

3
Worked Example
Find the angle of repose for a surface whose coefficient of static friction is $\mu_s=\frac{1}{\sqrt{3}}$.
Solution
  1. Step 1: At the angle of repose $\tan\theta=\mu_s$.
  2. Step 2: Substitute: $\tan\theta=\frac{1}{\sqrt{3}}$.
  3. Step 3: Therefore $\theta=30^{\circ}$.

Answer: Angle of repose $=30^{\circ}$

4
Worked Example
A block slides down a frictionless incline of $30^{\circ}$. Find its acceleration. (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: On a frictionless incline $a=g\sin\theta$.
  2. Step 2: Substitute: $a=10 \times \sin 30^{\circ}=10 \times 0.5$.
  3. Step 3: Compute: $a=5\ \text{m/s}^2$.

Answer: $a=5\ \text{m/s}^2$ down the incline.

5
Worked Example
A car of mass 800 kg takes a circular turn of radius 50 m at a speed of 10 m/s. Find the centripetal force required.
Solution
  1. Step 1: Use $F=\frac{mv^2}{r}$.
  2. Step 2: Substitute: $F=\frac{800 \times (10)^2}{50}=\frac{800 \times 100}{50}$.
  3. Step 3: Compute: $F=\frac{80000}{50}=1600\ \text{N}$.

Answer: $F=1600\ \text{N}$ (directed toward the centre).

6
Worked Example
A road of radius 90 m is to be banked for a design speed of 30 m/s. Find the banking angle. (Take $g=10\ \text{m/s}^2$.)
Solution
  1. Step 1: For ideal banking $\tan\theta=\frac{v^2}{rg}$.
  2. Step 2: Substitute: $\tan\theta=\frac{(30)^2}{90 \times 10}=\frac{900}{900}$.
  3. Step 3: So $\tan\theta=1$, giving $\theta=45^{\circ}$.

Answer: Banking angle $\theta=45^{\circ}$

Key Points

  • Static friction is self-adjusting up to a maximum: $f_s\le\mu_s N$; kinetic friction is $f_k=\mu_k N$, with $\mu_k<\mu_s$.
  • On an incline $N=mg\cos\theta$; the angle of repose is $\theta=\tan^{-1}\mu_s$ and sliding acceleration is $a=g(\sin\theta-\mu_k\cos\theta)$.
  • Circular motion needs a centripetal force $F=\frac{mv^2}{r}$ directed toward the centre ($a_c=\frac{v^2}{r}$).
  • Ideal banking of roads: $\tan\theta=\frac{v^2}{rg}$, so the normal reaction supplies the centripetal force.
  • A conical pendulum obeys the same relation $\tan\theta=\frac{v^2}{rg}$, with tension providing the centripetal force.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The maximum static friction is given by:
Explanation: Static friction is self-adjusting and at most $\mu_s N$, so $f_s\le\mu_s N$.
Q2.The centripetal acceleration of a body moving at speed $v$ in a circle of radius $r$ is:
Explanation: Centripetal acceleration is $a_c=\frac{v^2}{r}$, directed toward the centre.
Q3.For ideal (frictionless) banking of a road, the banking angle satisfies:
Explanation: $\tan\theta=\frac{v^2}{rg}$ for the ideal banking speed.
Q4.The angle of repose is related to the coefficient of static friction by:
Explanation: At the angle of repose $\tan\theta=\mu_s$, so $\theta=\tan^{-1}\mu_s$.
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