Chapter 5: Laws of Motion

Kinematics tells us how objects move; dynamics, founded on Newton's three laws, tells us why. This chapter is the bedrock of all of mechanics: every problem on collisions, rotation, fluids, oscillations, and even orbital motion eventually reduces to drawing a correct free-body diagram and applying $\vec{F}_{\text{net}} = m\vec{a}$.

Concept Map

• 5.1 — Aristotle's fallacy and the concept of inertia
• 5.2 — Newton's First Law and frames of reference
• 5.3 — Linear momentum
• 5.4 — Newton's Second Law, impulse, impulse–momentum theorem
• 5.5 — Newton's Third Law and action–reaction subtleties
• 5.6 — Conservation of linear momentum
• 5.7 — Equilibrium of a particle; Lami's theorem
• 5.8 — Standard force problems: pulleys, Atwood machine, inclined plane
• 5.9 — Friction: static, kinetic, rolling; angle of repose
• 5.10 — Dynamics of uniform circular motion; banking, conical pendulum, vertical circle, death well
• 5.11 — Pseudo forces and non-inertial frames

5.1 Aristotle's Fallacy and Inertia

Statement

Aristotle held that a body needs a continuous external force to remain in motion. Galileo, through his inclined-plane experiments, showed this is wrong: in the absence of friction a body in motion continues to move with the same velocity.

Derivation (Galileo's thought experiment)

Consider a ball rolling down one incline and up another that meets it at the bottom.

1. With friction present, the ball rises to a slightly smaller height than it started.
2. As friction is reduced, the ball climbs higher — closer to its starting height.
3. In the limit of zero friction, the ball must rise to exactly the same height.

Now make the second incline less and less steep:

$h_{\text{final}} = h_{\text{initial}} \quad \Rightarrow \quad \text{horizontal distance } \to \infty$

If the second surface is horizontal, the ball never reaches that "same height," so it must continue moving forever. Motion is the natural state; force is required to change motion, not maintain it.

Special Cases

• Bodies sliding on ice or air-tracks travel long distances precisely because friction is small.
• A spacecraft in deep space coasts unchanged for years — empirical proof of inertia.

Worked Example

A puck slides on a frictionless air-table at $5\,\text{m/s}$. What net force is needed to keep it moving?

$F_{\text{net}} = 0$

Zero. Inertia takes care of motion at constant velocity.

Common Mistakes

• Equating "moving" with "having a net force on it." Motion at constant velocity needs zero net force.
• Confusing inertia with mass. Mass measures inertia, but inertia is the property of resisting change.

5.2 Newton's First Law and Frames of Reference

Statement

Every body continues in its state of rest or of uniform motion in a straight line unless compelled by an external force to change that state.

Definition (Inertial frame)

A frame of reference in which the first law holds is called an inertial frame. All frames moving with constant velocity relative to an inertial frame are themselves inertial.

Derivation (why the first law is non-trivial)

The first law is not a special case of the second. It defines what counts as an inertial frame. Without specifying the frame, "$\vec{F} = m\vec{a}$" is meaningless because $\vec{a}$ depends on the observer.

Suppose two observers $S$ and $S'$ measure the acceleration of a free particle.

• In $S$ (inertial): $\vec{a} = 0$.
• In $S'$ moving with acceleration $\vec{A}$ relative to $S$: $\vec{a}' = -\vec{A} \neq 0$.

So $S'$ would conclude that a force acts on a force-free particle. The first law thus selects frames in which dynamics is simple.

Special Cases

• A car braking on a highway is non-inertial; the bus passenger lurches forward.
• The Earth's frame is approximately inertial — its rotation gives small Coriolis and centrifugal corrections.

Worked Example

A ball is placed on the smooth floor of a train accelerating at $2\,\text{m/s}^2$. From the ground frame what is the ball's acceleration?

The ground frame is (approximately) inertial. No horizontal force acts on the ball, so

$\vec{a}_{\text{ground}} = 0$

The train moves forward; the ball "appears" to slide backward inside the train.

Common Mistakes

• Treating the Earth as a perfect inertial frame in high-precision experiments.
• Forgetting that "uniform motion" includes rest as a special case ($v = 0$).

5.3 Linear Momentum

Definition

The linear momentum of a particle of mass $m$ moving with velocity $\vec{v}$ is

$\vec{p} = m\vec{v}$

It is a vector with SI unit $\text{kg}\cdot\text{m/s}$.

Derivation (why momentum, not velocity, is the dynamical variable)

A bullet and a truck may have very different velocities, but the momentum tells us "how hard it is to stop them."

For a system of particles,

$\vec{P} = \sum_i m_i \vec{v}_i = M\vec{v}_{\text{cm}}$

where $M = \sum m_i$ and $\vec{v}_{\text{cm}}$ is the velocity of the centre of mass.

Special Cases

• A photon has $p = E/c$ even though its rest mass is zero — quantum extension of the classical idea.
• For variable mass (rockets), $\vec{p}$ is still the fundamental quantity.

Worked Example

A $0.05\,\text{kg}$ bullet leaves a rifle at $400\,\text{m/s}$. Find its momentum.

$p = mv = 0.05 \times 400 = 20\;\text{kg}\cdot\text{m/s}$

Common Mistakes

• Confusing momentum (vector) with kinetic energy (scalar).
• Forgetting that two bodies can have equal kinetic energies but unequal momenta.

5.4 Newton's Second Law

Statement

The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts.

$\vec{F} = \frac{d\vec{p}}{dt}$

Derivation (constant mass form)

Starting from $\vec{p} = m\vec{v}$ with $m$ constant,

$\vec{F} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}$

For variable mass (rockets, conveyor belts),

$\vec{F} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}$

Impulse and the Impulse–Momentum Theorem

Integrate the second law over the duration $\Delta t$ of a force:

$\int_{t_1}^{t_2}\vec{F}\,dt = \int_{\vec{p}_1}^{\vec{p}_2} d\vec{p} = \vec{p}_2 - \vec{p}_1$

Define impulse $\vec{J} = \int \vec{F}\,dt$. Then

$\boxed{\vec{J} = \Delta\vec{p}}$

If $\vec{F}$ is constant, $\vec{J} = \vec{F}\,\Delta t$.

Special Cases

• Large force, short time (cricket ball hitting bat) — impulse calculation avoids the messy details of $F(t)$.
• Why a soft landing hurts less — increasing $\Delta t$ for a fixed $\Delta p$ reduces the average force $F_{\text{avg}} = \Delta p/\Delta t$.

Worked Example

A $150\,\text{g}$ ball strikes a wall at $20\,\text{m/s}$ and rebounds at $15\,\text{m/s}$. If contact lasts $0.01\,\text{s}$, find the average force on the wall.

Take "toward wall" as positive.

$\Delta p = m(v_f - v_i) = 0.15(-15 - 20) = -5.25\;\text{kg}\cdot\text{m/s}$

$F_{\text{avg}} = \frac{|\Delta p|}{\Delta t} = \frac{5.25}{0.01} = 525\;\text{N}$

By Newton's third law, the ball exerts $525\,\text{N}$ on the wall.

Common Mistakes

• Treating rebound as if the ball stops — you must use $v_f - v_i$ as a vector difference.
• Mixing units: time in milliseconds vs seconds.

5.5 Newton's Third Law

Statement

To every action there is an equal and opposite reaction.

If body $A$ exerts a force $\vec{F}_{AB}$ on body $B$, then $B$ exerts $\vec{F}_{BA} = -\vec{F}_{AB}$ on $A$.

Subtleties

1. The two forces act on different bodies. They do not cancel.
2. They are of the same nature (both gravitational, or both contact, etc.).
3. They are simultaneous — there is no "first" or "second."

Derivation (from momentum conservation)

For an isolated two-body system, $\vec{P}_{\text{total}} = \vec{p}_1 + \vec{p}_2 = \text{const}$. Differentiating,

$\frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt} = 0 \quad \Rightarrow \quad \vec{F}_{21} = -\vec{F}_{12}$

This shows that the third law is equivalent to the conservation of momentum in isolated systems.

Worked Example

A horse pulls a cart with $500\,\text{N}$. The cart pulls the horse with $500\,\text{N}$ backward. Why does the cart move?

Because the cart's equation of motion includes:

• Pull from horse: $+500\,\text{N}$ (forward).
• Friction from ground on cart: $< 500\,\text{N}$.

Net force on cart is forward — it accelerates. The "equal and opposite" reaction is on the horse, not the cart.

Common Mistakes

• Believing action–reaction pairs cancel out in $\vec{F}_{\text{net}} = m\vec{a}$ (they act on different bodies).
• Calling the gravitational force on a book and the normal from the table an action–reaction pair. They are not — they act on the same body and are of different natures.

5.6 Conservation of Linear Momentum

Statement

If no external force acts on a system, its total linear momentum remains constant.

Derivation

For a system of particles, total force is the sum of internal and external forces:

$\vec{F}_{\text{net}} = \vec{F}_{\text{ext}} + \vec{F}_{\text{int}}$

By the third law, $\vec{F}_{\text{int}}$ cancels in pairs, so $\vec{F}_{\text{int}} = 0$ for the system. Hence

$\frac{d\vec{P}}{dt} = \vec{F}_{\text{ext}}$

If $\vec{F}_{\text{ext}} = 0$, $\vec{P} =$ constant.

Special Cases

• Recoil of a gun: gun ($M$) + bullet ($m$) initially at rest. After firing,

$0 = mv_b + MV \quad \Rightarrow \quad V = -\frac{m}{M}v_b$

The minus sign shows the gun recoils backward.

• Rocket propulsion: continuous ejection of mass at velocity $u$ leads to the Tsiolkovsky equation $v = u\ln(m_0/m)$ (covered later).

• Explosion: a stationary bomb fragments. The pieces' momenta sum to zero.

Worked Example

A $0.020\,\text{kg}$ bullet leaves a $2.0\,\text{kg}$ gun at $300\,\text{m/s}$. Find the recoil speed.

$V = -\frac{m}{M}v_b = -\frac{0.02}{2}\times 300 = -3\;\text{m/s}$

Common Mistakes

• Applying conservation when an external force acts (e.g., a ball collides with a wall — momentum of the ball is not conserved).
• Forgetting that momentum is a vector — you must apply conservation component by component.

5.7 Equilibrium of a Particle

Definition

A particle is in equilibrium when the net force on it is zero:

$\sum \vec{F}_i = 0$

This is static equilibrium when also $\vec{v} = 0$, and dynamic equilibrium otherwise.

Concurrent Forces and the Triangle Law

If three forces $\vec{F}_1, \vec{F}_2, \vec{F}_3$ keep a body in equilibrium, they can be drawn head-to-tail to form a closed triangle.

Lami's Theorem (Derivation)

For three concurrent coplanar forces in equilibrium,

$\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}$

where $\alpha$ is the angle opposite to $F_1$ (i.e., between $F_2$ and $F_3$), etc.

Proof: Place the three forces head-to-tail forming a closed triangle. The interior angles of the triangle are $\pi - \alpha$, $\pi - \beta$, $\pi - \gamma$. By the sine rule of a triangle,

$\frac{F_1}{\sin(\pi-\alpha)} = \frac{F_2}{\sin(\pi-\beta)} = \frac{F_3}{\sin(\pi-\gamma)}$

Since $\sin(\pi-\theta) = \sin\theta$, the result follows.

Worked Example

A lamp of weight $20\,\text{N}$ hangs from the ceiling by two strings making $30^\circ$ and $60^\circ$ with the ceiling. Find the tensions.

Let $T_1$ be the tension along the $30^\circ$ string, $T_2$ along the $60^\circ$. The angles between the three forces are:

• Between weight (downward) and $T_1$ (up–right at $30^\circ$ above horizontal): $90^\circ + 30^\circ = 120^\circ$.

Wait — let's redo this carefully. If the two strings are at $30^\circ$ and $60^\circ$ to the ceiling, then they make $60^\circ$ and $30^\circ$ to the vertical. The angle between the two strings is $90^\circ$ (since $30^\circ + 60^\circ = 90^\circ$).

Resolve horizontally: $T_1\cos 30^\circ = T_2\cos 60^\circ$, so $T_1\sqrt 3/2 = T_2/2$, giving $T_2 = T_1\sqrt 3$.

Vertically: $T_1\sin 30^\circ + T_2\sin 60^\circ = 20$, so $T_1/2 + T_1\sqrt 3 \cdot \sqrt 3/2 = 20$, i.e., $T_1/2 + 3T_1/2 = 20$, so $2T_1 = 20$, $T_1 = 10\,\text{N}$ and $T_2 = 10\sqrt 3 \approx 17.3\,\text{N}$.

Common Mistakes

• Using wrong angles in Lami's theorem (must be the angle opposite the force).
• Forgetting that Lami applies only to three concurrent coplanar forces.

5.8 Standard Force Problems

Free-Body Diagram (FBD) Recipe

1. Isolate the body.
2. Draw all forces acting on it (not by it).
3. Choose convenient axes.
4. Write $\sum F_x = ma_x$ and $\sum F_y = ma_y$.
5. Solve.

Atwood Machine (Derivation)

Two masses $m_1 > m_2$ are connected by a light inextensible string over a smooth massless pulley.

Let acceleration be $a$, tension $T$.

For $m_1$ (going down): $m_1 g - T = m_1 a$ For $m_2$ (going up): $T - m_2 g = m_2 a$

Adding,

$a = \frac{(m_1 - m_2)g}{m_1 + m_2}$

Substituting,

$T = \frac{2 m_1 m_2 g}{m_1 + m_2}$

Block on a Smooth Inclined Plane (Derivation)

A block of mass $m$ on a smooth incline of angle $\theta$.

Resolve gravity along the incline ($mg\sin\theta$, down-slope) and perpendicular ($mg\cos\theta$, into surface).

• Along incline: $mg\sin\theta = ma \Rightarrow a = g\sin\theta$.
• Perpendicular: $N = mg\cos\theta$.

Two Blocks Connected over a Pulley (one on table, one hanging)

Block $m_1$ on a smooth table, connected via string over a smooth pulley to hanging block $m_2$.

• For $m_2$: $m_2 g - T = m_2 a$
• For $m_1$: $T = m_1 a$

Adding,

$a = \frac{m_2 g}{m_1 + m_2}, \quad T = \frac{m_1 m_2 g}{m_1 + m_2}$

Worked Example

In an Atwood machine, $m_1 = 5\,\text{kg}$, $m_2 = 3\,\text{kg}$, $g = 10\,\text{m/s}^2$. Find $a$ and $T$.

$a = \frac{(5-3)\cdot 10}{5+3} = 2.5\;\text{m/s}^2$ $T = \frac{2\cdot 5\cdot 3\cdot 10}{8} = 37.5\;\text{N}$

Common Mistakes

• Treating the string as massive without justification.
• Forgetting that a frictionless pulley transmits tension; tensions on either side are equal only if the pulley is massless.

5.9 Friction

Definition

Friction is the tangential force at the contact between two surfaces that opposes their relative sliding (or tendency thereof).

Three Regimes

1. Static friction $f_s$ — exists when there is no relative sliding. It self-adjusts to balance applied forces.

$0 \le f_s \le \mu_s N$

2. Kinetic friction $f_k = \mu_k N$ — acts when surfaces slide. Independent of speed (to first approximation).

3. Rolling friction $f_r = \mu_r N$ — much smaller than $\mu_k$. Arises from deformation.

Generally $\mu_s > \mu_k > \mu_r$.

Laws of Friction (Empirical)

1. $f \le \mu N$, independent of contact area.
2. Direction opposes relative motion (or its tendency).
3. $\mu$ depends on the nature of surfaces, not on $N$ (for moderate $N$).

Angle of Friction $\phi$

The angle between the resultant of $N$ and limiting friction $f_s = \mu_s N$, and the normal.

$\tan\phi = \mu_s$

Angle of Repose $\alpha$

The maximum angle of an incline on which a body can rest without sliding.

Derivation: On the verge of sliding,

$mg\sin\alpha = \mu_s mg\cos\alpha \quad \Rightarrow \quad \tan\alpha = \mu_s$

So $\alpha = \phi$.

Block on Rough Incline (Derivation)

For a block of mass $m$ on an incline of angle $\theta > \alpha$:

• Net force along incline (down): $mg\sin\theta - \mu_k mg\cos\theta$.
• $a = g(\sin\theta - \mu_k\cos\theta)$

If $\theta < \alpha$, $a = 0$ and the block rests.

Worked Example

A $5\,\text{kg}$ block on a $30^\circ$ rough incline with $\mu_k = 0.2$, $g = 10\,\text{m/s}^2$. Find acceleration.

$a = 10(\sin 30^\circ - 0.2\cos 30^\circ) = 10(0.5 - 0.173) = 3.27\;\text{m/s}^2$

Common Mistakes

• Writing $f = \mu N$ for static friction. The correct relation is $f \le \mu_s N$.
• Confusing the direction of friction. On a body being pushed, friction is backward; on a body being carried on a moving belt, friction is forward.

5.10 Dynamics of Uniform Circular Motion

Centripetal Acceleration (Derivation)

A particle moving in a circle of radius $r$ at constant speed $v$ has acceleration directed toward the centre.

Position: $\vec{r}(t) = r(\cos\omega t,\,\sin\omega t)$. Then

$\vec{v} = r\omega(-\sin\omega t,\,\cos\omega t)$ $\vec{a} = -r\omega^2(\cos\omega t,\,\sin\omega t) = -\omega^2\vec{r}$

Magnitude $a = \omega^2 r = v^2/r$ since $v = \omega r$.

Centripetal Force

$F_c = \frac{mv^2}{r} = m\omega^2 r$

This is not a new kind of force — it is the net force required, supplied by gravity, tension, friction, normal, etc.

Banking of Roads

Without Friction (Derivation)

A car of mass $m$ on a curve of radius $r$, banking angle $\theta$. Normal $N$ and gravity $mg$ act.

• Vertical: $N\cos\theta = mg$
• Horizontal: $N\sin\theta = \frac{mv^2}{r}$

Dividing,

$\tan\theta = \frac{v^2}{rg} \quad \Rightarrow \quad v = \sqrt{rg\tan\theta}$

This is the design speed.

With Friction (Maximum Speed)

If $v$ exceeds the design speed, friction acts down-slope to provide extra centripetal force. Let $\mu$ be the coefficient of friction.

• Vertical: $N\cos\theta = mg + \mu N\sin\theta$
• Horizontal: $N\sin\theta + \mu N\cos\theta = \frac{mv_{\max}^2}{r}$

Solving,

$v_{\max} = \sqrt{\frac{rg(\tan\theta + \mu)}{1 - \mu\tan\theta}}$

Minimum Speed (similar argument with friction up-slope)

$v_{\min} = \sqrt{\frac{rg(\tan\theta - \mu)}{1 + \mu\tan\theta}}$

Conical Pendulum

A bob of mass $m$ on a string of length $L$ swings in a horizontal circle, the string making angle $\theta$ with the vertical.

• Vertical: $T\cos\theta = mg$
• Horizontal: $T\sin\theta = m\omega^2 r$, with $r = L\sin\theta$.

Dividing,

$\omega^2 = \frac{g}{L\cos\theta} \quad \Rightarrow \quad T = 2\pi\sqrt{\frac{L\cos\theta}{g}}$

Vertical Circle (Derivation)

A ball on a string swings in a vertical circle of radius $r$.

At the top: tension $T_t$ and gravity both point down.

$T_t + mg = \frac{mv_t^2}{r}$

Minimum condition: $T_t = 0$, giving $v_t^{\min} = \sqrt{gr}$.

At the bottom: using energy conservation,

$\tfrac{1}{2}mv_b^2 = \tfrac{1}{2}mv_t^2 + mg(2r)$ $v_b^2 = v_t^2 + 4gr$

So the minimum speed at the bottom is

$v_b^{\min} = \sqrt{5gr}$

Tension at the bottom:

$T_b - mg = \frac{mv_b^2}{r} \quad \Rightarrow \quad T_b = m\!\left(\frac{v_b^2}{r} + g\right)$

Death Well (Wall of Death)

A motorcyclist rides on the inside of a vertical cylindrical wall. The normal $N$ provides the centripetal force, and friction $\mu N$ supports the weight.

$\mu N = mg, \quad N = \frac{mv^2}{r}$

Solving,

$v_{\min} = \sqrt{\frac{gr}{\mu}}$

Worked Example

A car rounds a curve of radius $50\,\text{m}$ banked at $30^\circ$ with $\mu = 0.2$, $g = 10\,\text{m/s}^2$. Find $v_{\max}$.

$v_{\max} = \sqrt{\frac{50\cdot 10\cdot (\tan 30^\circ + 0.2)}{1 - 0.2\tan 30^\circ}} = \sqrt{\frac{500(0.577+0.2)}{1 - 0.115}} = \sqrt{\frac{500\cdot 0.777}{0.885}}$

$= \sqrt{438.98} \approx 20.95\;\text{m/s}$

Common Mistakes

• Treating centripetal force as an additional force on the FBD. It is the net radial force.
• Forgetting that the "minimum speed at top of vertical circle" applies only to string (not rod) constraint.

5.11 Pseudo Forces and Non-Inertial Frames

Definition

In a frame accelerating with $\vec{A}$ relative to an inertial frame, a body of mass $m$ appears to experience an extra "force"

$\vec{F}_{\text{pseudo}} = -m\vec{A}$

This is not a real force — no agent exerts it — but it must be included to make $\vec{F} = m\vec{a}$ work in the non-inertial frame.

Derivation

Let $S$ be inertial, $S'$ accelerate at $\vec{A}$. Position of particle: $\vec{r} = \vec{R} + \vec{r}'$ where $\vec{R}$ is the position of $S'$'s origin. Differentiating twice,

$\vec{a} = \vec{A} + \vec{a}'$

So in $S'$,

$m\vec{a}' = m\vec{a} - m\vec{A} = \vec{F}_{\text{real}} - m\vec{A}$

Worked Example (Accelerating elevator)

An elevator accelerates upward at $a_0$. A spring balance inside shows the apparent weight of a body of mass $m$.

In the elevator frame, pseudo force $ma_0$ acts downward. Effective gravity $g_{\text{eff}} = g + a_0$. Apparent weight:

$W' = m(g + a_0)$

If the elevator falls freely, $a_0 = -g$ and $W' = 0$ — weightlessness.

Centrifugal Force

In a rotating frame at angular velocity $\vec{\omega}$, a particle at distance $r$ from the axis experiences a pseudo centrifugal force $m\omega^2 r$ outward, balancing the (real) centripetal force in that frame.

Common Mistakes

• Treating pseudo forces as real in the inertial frame (they vanish there).
• Forgetting Coriolis force in rotating frames for moving particles.

Solved Problems

Problem 1

A block of mass $2\,\text{kg}$ is pulled by a horizontal force of $20\,\text{N}$ on a rough floor with $\mu_k = 0.3$. Find acceleration. Take $g = 10\,\text{m/s}^2$.

Normal $N = mg = 20\,\text{N}$. Kinetic friction $f_k = \mu_k N = 6\,\text{N}$.

$F_{\text{net}} = 20 - 6 = 14\,\text{N}, \quad a = \frac{14}{2} = 7\;\text{m/s}^2$

Problem 2

A bullet of mass $20\,\text{g}$ moving at $300\,\text{m/s}$ embeds in a wooden block of $1.98\,\text{kg}$ resting on a smooth surface. Find the final velocity.

By conservation of momentum,

$mv = (m+M)V$ $0.02\cdot 300 = 2.0\cdot V \quad \Rightarrow \quad V = 3\,\text{m/s}$

Problem 3

Three blocks of masses $1, 2, 3\,\text{kg}$ are placed in contact on a smooth floor. A $24\,\text{N}$ force pushes the $1\,\text{kg}$ block. Find acceleration and contact forces.

System: $a = \dfrac{24}{1+2+3} = 4\;\text{m/s}^2$.

Force on the $2$ + $3\,\text{kg}$ subsystem from the $1\,\text{kg}$ block: $(2+3)\cdot 4 = 20\,\text{N}$.

Force on the $3\,\text{kg}$ block from the $2\,\text{kg}$ block: $3\cdot 4 = 12\,\text{N}$.

Problem 4

A cyclist negotiates a curve of radius $80\,\text{m}$ at $36\,\text{km/h}$. What is the required angle of leaning if road is unbanked?

$36\,\text{km/h} = 10\,\text{m/s}$.

$\tan\theta = \frac{v^2}{rg} = \frac{100}{80\cdot 10} = 0.125 \quad \Rightarrow \quad \theta \approx 7.1^\circ$

Problem 5

A monkey of mass $20\,\text{kg}$ climbs a vertical rope. Find the tension when (a) it climbs with constant velocity, (b) it climbs with $a = 2\,\text{m/s}^2$ upward, (c) it slides down with $a = 2\,\text{m/s}^2$ downward. Take $g = 10\,\text{m/s}^2$.

(a) $T = mg = 200\,\text{N}$. (b) $T - mg = ma \Rightarrow T = m(g+a) = 240\,\text{N}$. (c) $mg - T = ma \Rightarrow T = m(g-a) = 160\,\text{N}$.

Problem 6

A block of $4\,\text{kg}$ is placed on a $30^\circ$ incline with $\mu_s = 0.6$. Does it slide?

Angle of repose: $\tan\alpha = 0.6 \Rightarrow \alpha \approx 31^\circ > 30^\circ$. Block does not slide.

Problem 7

A particle of mass $m$ moves in a horizontal circle of radius $r$ on a frictionless table, attached to a string passing through a hole in the centre, with another mass $M$ hanging vertically. Find the speed of the particle.

Tension supports $M$: $T = Mg$. Also $T = mv^2/r$. Equate:

$v = \sqrt{\frac{Mgr}{m}}$

JEE/NEET Edge Cases

1. String vs Rod in vertical circle: A rod can push and pull, so the minimum top speed is zero (the rod simply applies a thrust). For a string, $v_{\text{top}}^{\min} = \sqrt{gr}$.
2. Three-block pulley with friction: Always check if static friction prevents motion before applying $a = \dots$.
3. Variable mass: Use $\vec{F} = d\vec{p}/dt$ in full, not $m\vec{a}$.
4. Pulley with mass: Tensions on the two sides differ by an amount related to angular momentum of the pulley (covered in Chapter 7).
5. Inclined plane with the wedge itself free to slide — block-on-wedge problem requires either CM frame or simultaneous Newton's laws for both bodies.
6. Friction on a vehicle's driving wheel points forward (it is what propels the car), while on a wheel that is being braked, it points backward.

Quick Recap

• Force changes motion; constant velocity needs zero net force.
• $\vec{F} = d\vec{p}/dt$, reducing to $m\vec{a}$ for constant mass.
• Action–reaction acts on different bodies.
• Momentum is conserved in absence of external force; key tool for collisions and recoil.
• Friction: $f_s \le \mu_s N$, $f_k = \mu_k N$, $\tan\alpha = \mu_s$.
• Banked road: $\tan\theta = v^2/rg$.
• Minimum top speed in vertical circle: $\sqrt{gr}$; bottom: $\sqrt{5gr}$.
• Pseudo force $-m\vec{A}$ in a frame accelerating at $\vec{A}$.

Formula Sheet