In kinematics, we described the motion of a particle in space quantitatively using velocity and acceleration without discussing its cause. Dynamics investigates the factors that govern and change the motion of bodies. A force is required to put a stationary body in motion, or to retard/stop a moving body. This force is provided by an external agency, which may be in contact (e.g., hands, wind, friction) or act at a distance (e.g., gravitational, magnetic, or electrostatic forces).
The Greek philosopher Aristotle (384 B.C. – 322 B.C.) postulated that:
"An external force is required to keep a body in motion."
The Flaw in Aristotelian Mechanics: Aristotle's view is flawed because it fails to account for opposing environmental forces such as friction (for solids) and viscous drag (for fluids) present in the natural world. In practice, an applied force is only required to counter these opposing forces, making the net external force acting on the body zero. If friction and viscous forces were completely absent, a body in uniform motion would continue to move indefinitely without any external force.
Historical Chronicle: Ideas on Motion in Ancient Indian Science
Galileo Galilei studied the motion of objects on inclined planes to understand the fundamental nature of force and motion, leading to the formulation of the law of inertia .
Definition of Inertia: The inherent property of a body by virtue of which it resists any change in its state of rest or uniform motion in a straight line. Inertia literally means 'resistance to change' .
Isaac Newton built upon Galileo's insights to lay the foundation of classical mechanics. The first law of motion formalizes the concept of inertia.
Statement:
"Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise."
Mathematically, if the net external force $\vec{F}_{ext} = 0$, then the acceleration $\vec{a} = 0$. Conversely, acceleration can be non-zero only if there is a net external force acting on the body.
Mathematical Problem Statement: Example 4.1: An astronaut accidentally gets separated out of his small spaceship accelerating in interstellar space at a constant rate of 100 m s -2 . What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume there are no nearby stars to exert gravitational force on him.)
Step-by-Step Algorithmic Solution Trace:
- Identify the forces acting on the astronaut once he is outside the spaceship. Since there are no nearby stars, gravitational forces are negligible.
- The spaceship's acceleration does not affect the astronaut once contact is broken, as the spaceship exerts no gravitational or contact force on him.
- According to Newton's first law, since the net external force $\vec{F}_{ext} = 0$, the acceleration $\vec{a}$ of the astronaut must be zero.
Final Evaluation Output: $a = 0$
The second law of motion quantitatively relates the net external force acting on a body to its rate of change of momentum.
The linear momentum $\vec{p}$ of a body is defined as the product of its mass $m$ and velocity $\vec{v}$:
$$\vec{p} = m \vec{v}$$
Momentum is a vector quantity. Its SI unit is kg m s -1 (or N s ), and its dimensional formula is $[MLT^{-1}]$.
Statement:
"The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts."
$$\vec{F} \propto \frac{d\vec{p}}{dt} \implies \vec{F} = k \frac{d\vec{p}}{dt}$$
For a body of constant mass $m$:
$$\frac{d\vec{p}}{dt} = \frac{d}{dt}(m\vec{v}) = m\frac{d\vec{v}}{dt} = m\vec{a}$$
$$\vec{F} = k m \vec{a}$$
In SI units, the constant of proportionality $k$ is chosen to be 1 , defining the unit of force (Newton):
$$\vec{F} = \frac{d\vec{p}}{dt} = m\vec{a}$$
Definition of Newton (N):
One Newton is the force that produces an acceleration of
1 m s
-2
when applied to a mass of
1 kg
.
1 N = 1 kg m s
-2
.
Mathematical Problem Statement: Example 4.2: A bullet of mass 0.04 kg moving with a speed of 90 m s -1 enters a heavy wooden block and is stopped after a distance of 60 cm . What is the average resistive force exerted by the block on the bullet?
Step-by-Step Algorithmic Solution Trace:
- Convert all quantities to SI units: mass $m = 0.04\text{ kg}$, initial velocity $u = 90\text{ m s}^{-1}$, final velocity $v = 0$, distance $s = 60\text{ cm} = 0.6\text{ m}$.
- Use the kinematic equation to find constant retardation $a$:
$$v^2 = u^2 + 2as \implies 0 = 90^2 + 2a(0.6)$$
$$a = -\frac{90 \times 90}{1.2} = -6750 \text{ m s}^{-2}$$- Apply Newton's second law to find the retarding force $F$:
$$F = ma = 0.04 \text{ kg} \times (-6750 \text{ m s}^{-2}) = -270 \text{ N}$$Final Evaluation Output: Average resistive force is $270 \text{ N}$ (opposing the motion).
Mathematical Problem Statement: Example 4.3: The motion of a particle of mass $m$ is described by $y = ut + \frac{1}{2}gt^2$. Find the force acting on the particle.
Step-by-Step Algorithmic Solution Trace:
- Differentiate the position equation $y$ with respect to time $t$ to find velocity $v$:
$$v = \frac{dy}{dt} = \frac{d}{dt}\left(ut + \frac{1}{2}gt^2\right) = u + gt$$- Differentiate velocity $v$ with respect to time $t$ to find acceleration $a$:
$$a = \frac{dv}{dt} = \frac{d}{dt}(u + gt) = g$$- Apply Newton's second law:
$$F = ma = mg$$Final Evaluation Output: $F = mg$ (force due to gravity acting vertically downward).
When a large force acts on a body for a very short duration of time, it produces a finite change in its momentum. This product of force and time duration is defined as impulse .
$$\text{Impulse} = \vec{F} \times \Delta t = \Delta \vec{p}$$
Impulse is equal to the total change in momentum of the body. It is a vector quantity with SI units of N s or kg m s -1 .
Mathematical Problem Statement: Example 4.4: A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s -1 . If the mass of the ball is 0.15 kg , determine the impulse imparted to the ball. (Assume linear motion of the ball.)
Step-by-Step Algorithmic Solution Trace:
- Define the coordinate system: Let the direction from the batsman to the bowler be positive.
- Initial velocity $u = -12\text{ m s}^{-1}$ (towards the batsman), final velocity $v = +12\text{ m s}^{-1}$ (towards the bowler).
- Calculate the change in momentum $\Delta p$:
$$\Delta p = mv - \mu = m(v - u)$$
$$\Delta p = 0.15 \text{ kg} \times [12 - (-12)] \text{ m s}^{-1} = 0.15 \times 24 = 3.6 \text{ kg m s}^{-1}$$- Since Impulse = Change in Momentum, the impulse is $3.6\text{ N s}$.
Final Evaluation Output: $\text{Impulse} = 3.6 \text{ N s}$ (directed from the batsman to the bowler).
Newton's third law of motion describes the mutual interactions between two bodies, stating that forces always occur in matched pairs.
Statement:
"To every action, there is always an equal and opposite reaction."
If body $A$ exerts a force $\vec{F}_{AB}$ on body $B$, then body $B$ simultaneously exerts an equal and opposite force $\vec{F}_{BA}$ on body $A$:
$$\vec{F}_{AB} = -\vec{F}_{BA}$$
Mathematical Problem Statement: Example 4.5: Two identical billiard balls strike a rigid wall with the same speed $u$ but at different angles, and get reflected without any change in speed, as shown in Fig. 4.6. Case (a): Ball strikes normally. Case (b): Ball strikes at an angle of 30° to the normal. What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall?
Step-by-Step Algorithmic Solution Trace:
- Case (a) - Normal Impact:
Initial momentum $p_{x,\text{initial}} = \mu$, $p_{y,\text{initial}} = 0$.
Final momentum $p_{x,\text{final}} = -\mu$, $p_{y,\text{final}} = 0$.
Change in momentum along x-axis: $\Delta p_x = -\mu - \mu = -2mu$.
Impulse along y-axis: $\Delta p_y = 0$.
Impulse vector $\vec{I}_a = -2mu \hat{i}$. Force on the wall is along the $+x$ direction (normal to the wall).- Case (b) - Angled Impact (30° to normal):
Initial momentum components: $p_{x,\text{initial}} = -\mu\cos 30^\circ$, $p_{y,\text{initial}} = -\mu\sin 30^\circ$.
Final momentum components: $p_{x,\text{final}} = \mu\cos 30^\circ$, $p_{y,\text{final}} = -\mu\sin 30^\circ$.
Change in momentum along x-axis: $\Delta p_x = -\mu\cos 30^\circ - \mu\cos 30^\circ = -2mu\cos 30^\circ$.
Change in momentum along y-axis: $\Delta p_y = 0$.
Impulse vector $\vec{I}_b = -2mu\cos 30^\circ \hat{i}$. Force on the wall is along the $+x$ direction (normal to the wall).- Ratio of Impulse Magnitudes:
$$\frac{I_a}{I_b} = \frac{2mu}{2mu\cos 30^\circ} = \frac{1}{\cos 30^\circ} = \frac{2}{\sqrt{3}} \approx 1.2$$Final Evaluation Output: (i) Force on the wall is normal to the wall in both cases. (ii) Ratio of impulses is $1.2$.
The second and third laws of motion lead directly to the law of conservation of momentum .
Consider two bodies $A$ and $B$ with initial momenta $\vec{p}_A$ and $\vec{p}_B$. They collide for a time interval $\Delta t$ and emerge with final momenta $\vec{p}'_A$ and $\vec{p}'_B$.
By Newton's second law:
$$\vec{F}_{AB}\Delta t = \vec{p}'_A - \vec{p}_A$$
$$\vec{F}_{BA}\Delta t = \vec{p}'_B - \vec{p}_B$$
By Newton's third law, $\vec{F}_{AB} = -\vec{F}_{BA}$. Therefore:
$$\vec{p}'_A - \vec{p}_A = -(\vec{p}'_B - \vec{p}_B)$$
$$\vec{p}'_A + \vec{p}'_B = \vec{p}_A + \vec{p}_B$$
Statement:
"The total momentum of an isolated system of interacting particles is conserved."
Equilibrium of a particle in mechanics refers to the state where the net external force acting on the particle is zero.
$$\sum \vec{F} = 0$$
For a particle acted upon by three concurrent forces $\vec{F}_1$, $\vec{F}_2$, and $\vec{F}_3$, equilibrium requires:
$$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$$
This \implies that the vector sum of any two forces must be equal and opposite to the third force. In component form:
$$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0$$
Mathematical Problem Statement: Example 4.6: A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint $P$ of the rope. What is the angle the rope makes with the vertical in equilibrium? (Take $g = 10\text{ m s}^{-2}$. Neglect the mass of the rope.)
Step-by-Step Algorithmic Solution Trace:
- Analyze the lower half of the rope: The tension $T_2$ must balance the weight of the suspended mass $W = mg = 6\text{ kg} \times 10\text{ m s}^{-2} = 60\text{ N}$. Thus, $T_2 = 60\text{ N}$.
- Analyze the equilibrium of midpoint $P$ under three forces: horizontal force $F = 50\text{ N}$, downward tension $T_2 = 60\text{ N}$, and tension $T_1$ in the upper rope making an angle $\theta$ with the vertical.
- Resolve forces at $P$ horizontally and vertically:
$$T_1 \sin\theta = 50 \text{ N}$$
$$T_1 \cos\theta = T_2 = 60 \text{ N}$$- Divide the two equations to find $\tan\theta$:
$$\tan\theta = \frac{50}{60} = \frac{5}{6}$$
$$\theta = \tan^{-1}\left(\frac{5}{6}\right) \approx 40^\circ$$Final Evaluation Output: $\theta \approx 40^\circ$
Forces in mechanics can be broadly classified into contact forces and non-contact forces (forces acting at a distance like gravity).
When two bodies are in contact, friction is the component of the contact force parallel to the surfaces in contact that opposes impending or actual relative motion between them.
| Friction Type | Mathematical Domain | Active Formulas | Systemic Role & Limits |
|---|---|---|---|
| Static Friction | $0 \le f_s \le \mu_s N$ | $(f_s)_{\text{max}} = \mu_s N$ | Self-adjusting; prevents relative motion up to a threshold. |
| Kinetic Friction | Constant value | $f_k = \mu_k N$ | Opposes active relative sliding; dissipates energy as heat. |
| Rolling Friction | Very small | $f_r = \mu_r \frac{N}{r}$ | Opposes rolling; minimized using ball bearings. |
Mathematical Problem Statement: Example 4.7: Determine the maximum acceleration of a train in which a box lying on its floor will remain stationary, given that the coefficient of static friction between the box and the train's floor is 0.15 .
Step-by-Step Algorithmic Solution Trace:
- The force accelerating the box is the static friction $f_s$ between the box and the floor: $f_s = ma$.
- For the box to remain stationary relative to the train, the acceleration $a$ must satisfy:
$$ma \le (f_s)_{\text{max}} = \mu_s N$$- Since $N = mg$ on a horizontal floor:
$$ma \le \mu_s mg \implies a \le \mu_s g$$- Calculate maximum acceleration $a_{\text{max}}$:
$$a_{\text{max}} = \mu_s g = 0.15 \times 10 \text{ m s}^{-2} = 1.5 \text{ m s}^{-2}$$Final Evaluation Output: $a_{\text{max}} = 1.5 \text{ m s}^{-2}$
Mathematical Problem Statement: Example 4.8: A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle of 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?
Step-by-Step Algorithmic Solution Trace:
- Resolve the weight $mg$ of the block along and perpendicular to the inclined plane:
Component parallel to incline: $mg \sin\theta$ (tends to slide the block down).
Component perpendicular to incline: $mg \cos\theta$ (balanced by normal reaction $N$).- At the verge of sliding, static friction reaches its maximum value:
$$f_s = mg \sin\theta$$
$$N = mg \cos\theta$$- Substitute $(f_s)_{\text{max}} = \mu_s N$:
$$\mu_s (mg \cos\theta) = mg \sin\theta \implies \mu_s = \tan\theta$$- Calculate $\mu_s$ for $\theta = 15^\circ$:
$$\mu_s = \tan 15^\circ \approx 0.27$$Final Evaluation Output: $\mu_s = 0.27$
Mathematical Problem Statement: Example 4.9: What is the acceleration of the block and trolley system shown in Fig. 4.12(a), if the coefficient of kinetic friction between the trolley and the surface is 0.04 ? What is the tension in the string? (Take $g = 10\text{ m s}^{-2}$. Neglect the mass of the string.)
Step-by-Step Algorithmic Solution Trace:
- Identify system parameters: mass of trolley $M = 20\text{ kg}$, mass of hanging block $m = 3\text{ kg}$, coefficient of kinetic friction $\mu_k = 0.04$.
- Write the equation of motion for the hanging block:
$$mg - T = ma \implies 30 - T = 3a \quad \text{--- (Equation 1)}$$- Write the equation of motion for the trolley:
$$T - f_k = Ma \implies T - \mu_k N = 20a$$
Since $N = Mg = 20 \times 10 = 200 \text{ N}$:
$$f_k = 0.04 \times 200 = 8 \text{ N}$$
$$T - 8 = 20a \quad \text{--- (Equation 2)}$$- Add Equation 1 and Equation 2 to solve for $a$:
$$(30 - T) + (T - 8) = 3a + 20a \implies 22 = 23a$$
$$a = \frac{22}{23} \approx 0.96 \text{ m s}^{-2}$$- Substitute $a$ back into Equation 1 to find $T$:
$$T = 30 - 3(0.96) = 30 - 2.88 = 27.12 \text{ N}$$Final Evaluation Output: $a = 0.96 \text{ m s}^{-2}$, $T = 27.1 \text{ N}$
An object in uniform circular motion of radius $R$ with speed $v$ experiences a centripetal acceleration $a_c = v^2/R$ directed towards the center. The force providing this acceleration is the centripetal force :
$$f_c = \frac{mv^2}{R}$$
On a flat, level circular road, the centripetal force is provided solely by static friction between the tires and the road:
$$\frac{mv^2}{R} \le f_s \le \mu_s N$$
Since $N = mg$:
$$\frac{mv^2}{R} \le \mu_s mg \implies v \le \sqrt{\mu_s g R}$$
The maximum safe speed on a level circular road is:
$$v_{\text{max}} = \sqrt{\mu_s g R}$$
To reduce reliance on friction and prevent tire wear, roads are banked at an angle $\theta$ to the horizontal. The normal force $N$ is tilted, and its horizontal component contributes to the centripetal force.
From force equilibrium equations:
$$N \cos\theta = mg + f \sin\theta$$
$$N \sin\theta + f \cos\theta = \frac{mv^2}{R}$$
Setting $f = f_s = \mu_s N$ to find the maximum speed limit $v_{\text{max}}$ yields:
$$v_{\text{max}} = \left[ Rg \left( \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta} \right) \right]^{1/2}$$
Optimum Speed ($v_o$): The speed at which no frictional force is needed at all ($\mu_s = 0$):
$$v_o = \sqrt{Rg \tan\theta}$$
Mathematical Problem Statement: Example 4.10: A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing speed. The coefficient of static friction between the tires and the road is 0.1 . Will the cyclist slip while taking the turn?
Step-by-Step Algorithmic Solution Trace:
- Convert speed to SI units: $v = 18\text{ km/h} = 18 \times \frac{5}{18} = 5\text{ m s}^{-1}$.
- Calculate the minimum coefficient of friction required to prevent slipping:
$$v^2 \le \mu_s g R \implies \mu_{s,\text{req}} = \frac{v^2}{gR}$$- Substitute the given values ($v = 5\text{ m s}^{-1}$, $R = 3\text{ m}$, $g = 9.8\text{ m s}^{-2}$):
$$\mu_{s,\text{req}} = \frac{5^2}{9.8 \times 3} = \frac{25}{29.4} \approx 0.85$$- Compare with the actual coefficient of static friction ($\mu_s = 0.1$): Since $\mu_{s,\text{req}} = 0.85 > 0.1$, the available friction is insufficient.
Final Evaluation Output: Yes, the cyclist will slip.
Mathematical Problem Statement: Example 4.11: A circular racetrack of radius 300 m is banked at an angle of 15° . If the coefficient of friction between the wheels of a race-car and the road is 0.2 , what is the (a) optimum speed of the race-car to avoid wear and tear on its tires, and (b) maximum permissible speed to avoid slipping?
Step-by-Step Algorithmic Solution Trace:
- (a) Optimum Speed ($v_o$):
$$v_o = \sqrt{Rg \tan\theta}$$
Substitute $R = 300\text{ m}$, $g = 9.8\text{ m s}^{-2}$, $\theta = 15^\circ$ ($\tan 15^\circ \approx 0.268$):
$$v_o = \sqrt{300 \times 9.8 \times 0.268} = \sqrt{787.92} \approx 28.1 \text{ m s}^{-1}$$- (b) Maximum Permissible Speed ($v_{\text{max}}$):
$$v_{\text{max}} = \left[ Rg \left( \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta} \right) \right]^{1/2}$$
Substitute $\mu_s = 0.2$:
$$v_{\text{max}} = \left[ 300 \times 9.8 \times \left( \frac{0.2 + 0.268}{1 - 0.2 \times 0.268} \right) \right]^{1/2}$$
$$v_{\text{max}} = \left[ 2940 \times \left( \frac{0.468}{0.9464} \right) \right]^{1/2} = \sqrt{1453.7} \approx 38.1 \text{ m s}^{-1}$$Final Evaluation Output: (a) Optimum speed is $28.1 \text{ m s}^{-1}$. (b) Maximum speed is $38.1 \text{ m s}^{-1}$.
To solve complex mechanical problems systematically, use the Free-Body Diagram (FBD) method.
Mathematical Problem Statement: Example 4.12: A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and cylinder together go down with an acceleration of 0.1 m s -2 . What is the action of the block on the floor (a) before and (b) after the floor yields? (Take $g = 10\text{ m s}^{-2}$. Identify action-reaction pairs.)
Step-by-Step Algorithmic Solution Trace:
- (a) Before the floor yields (System is at rest):
Mass of block $m = 2\text{ kg}$.
Forces on block: downward gravity $W = mg = 20\text{ N}$, upward normal reaction $R$.
Since $a = 0$, $R = W = 20\text{ N}$.
By Newton's third law, the action of the block on the floor is equal and opposite to $R$.- (b) After the floor yields (System accelerates downward):
Combined mass $M = 2 + 25 = 27\text{ kg}$. Downward acceleration $a = 0.1\text{ m s}^{-2}$.
Forces on combined system: downward gravity $W' = Mg = 270\text{ N}$, upward normal reaction $R'$.
Apply Newton's second law:
$$Mg - R' = Ma \implies 270 - R' = 27 \times 0.1$$
$$R' = 270 - 2.7 = 267.3 \text{ N}$$
The action of the system on the floor is equal and opposite to $R'$.Final Evaluation Output: (a) Action on the floor is $20 \text{ N}$ vertically downward. (b) Action on the floor is $267.3 \text{ N}$ vertically downward.