Laws of Motion — JEE Advanced Physics

01

Introduction to Dynamics

From Kinematics to Dynamics

In the previous chapters we studied kinematics, the branch of mechanics that describes motion - how position, velocity, and acceleration change with time. We learned to measure these changes using equations and graphs.

We now turn to dynamics, the study of why motion changes. What causes objects to speed up, slow down, or change direction? This chapter deals with the causes of acceleration.

Kinematics describes; dynamics explains.

The Concept of Inertia

Every object resists a change in its state of motion. This property of matter is called inertia.

An object at rest, such as a heavy textbook on a desk, tends to stay at rest. An object already in motion, such as a hockey puck gliding across smooth ice, tends to keep moving in a straight line at constant speed (ignoring friction for a moment). Inertia is the natural tendency of an object to maintain its current velocity.

Different objects resist changes in motion by different amounts. For example, a truck opposes a change in its state of motion more strongly than a bicycle does. In other words, the truck has more inertia than the bicycle.

Mass: The Measure of Inertia

Mass is the quantitative measure of an object's inertia. It is denoted by the symbol $m$ . The more massive an object, the greater its inertia, and the harder it is to change its velocity. So,

Inertia is the tendency of an object to oppose a change in its state of motion. Mass is the mathematical measure of the inertia of the object.

Concept Check: the same force acts on a 1 kg cart and a 4 kg cart. After one second, how much further has the light one travelled? Think about the answer first, then study the figure.

True or False: An object with greater mass has less inertia, meaning it's easier to change its state of motion.

What is Force?

A force is an interaction that tends to change an object's state of motion, that is, to produce acceleration.

Objects resist changes in motion because of their inertia. What, then, makes a resting textbook start sliding, or a moving ball slow down and stop? The cause is a force. We commonly experience forces as a push or a pull exerted by one object on another, such as pushing a door open or pulling a wagon.

If an agent pushes or pulls an object, we say that the agent applies a force on the object.

Is 'force' just another name for an object's inertia or mass?

A puck is sliding across frictionless ice at a steady 4 m/s. What horizontal force is needed to keep it going at 4 m/s?

Key Takeaways

Dynamics studies why motion changes (causes of acceleration)
Inertia is the tendency to resist changes in velocity
Mass quantifies inertia - greater mass means greater resistance to acceleration
Force is a push or pull that causes acceleration
Understanding the interplay between mass and force is central to dynamics

02

Fundamental Forces in Nature

The Four Fundamental Forces

All forces in nature are forms of a few basic interactions. We experience many pushes and pulls in daily life - friction, the pull of a rope, the force of a spring, the force from a surface - but each of them arises from one of just four known fundamental forces:

Gravitational Force: An attractive force that exists between any two objects with mass. It is the weakest fundamental force but can act over large distances, dominating large-scale structures like planets, stars, and galaxies.
Electromagnetic Force: Acts between electrically charged particles. It can be attractive or repulsive and is responsible for holding atoms and molecules together. It is much stronger than gravity and underlies most everyday forces like friction, normal force, and tension.
Strong Nuclear Force: The strongest of the four forces, but acts only over extremely short distances (within atomic nuclei). It binds protons and neutrons together in the nucleus, overcoming the electromagnetic repulsion between protons.
Weak Nuclear Force: Responsible for certain types of radioactive decay (like beta decay) and interactions involving subatomic particles. It also acts over very short ranges.

Are the force holding planets in orbit around the sun and the force holding electrons within an atom examples of the Gravitational and Electromagnetic forces, respectively?

Forces in Classical Mechanics

For the scope of classical mechanics, particularly the macroscopic world we typically analyze in introductory physics, we are primarily concerned with the effects of the Gravitational Force and the Electromagnetic Force. The nuclear forces operate at scales far smaller than everyday objects and require quantum mechanics for a full description.

The most common form of the gravitational force in mechanics is the weight of an object near a celestial body like the Earth. Weight is the attractive force exerted by the Earth on the object. For example, the weight of a 1 kg mass on the Earth's surface is approximately $W = mg \approx 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N}$ (Newtons), directed towards the center of the Earth.

Weight is the Earth’s pull, towards its centre.

Electromagnetic Origins of Contact Forces

Most other forces we encounter in mechanics, such as the normal force, tension, friction. Spring forces, are ultimately electromagnetic in origin. They arise from the interactions between charged particles (electrons and protons) within the atoms and molecules of the interacting objects.

For instance, when you place a block on a table, it doesn't fall through because the electrons in the atoms of the table's surface repel the electrons in the atoms of the block's surface. This microscopic electromagnetic repulsion manifests macroscopically as the upward normal force exerted by the table on the block, preventing penetration. We will study these forces, and the mathematical equations they follow, in the coming sections.

A block on a table, magnified at the contact.

Is the normal force exerted by a table on a book fundamentally an example of the Electromagnetic force acting at a microscopic level?

Common Forces in Mechanics

Based on our discussion of fundamental forces, we will frequently encounter these types of forces:

Gravitational Force (Weight): $\vec{W} = m\vec{g}$
Normal Force: $\vec{N}$ - perpendicular contact force
Tension Force: $\vec{T}$ - pulling force through strings/ropes
Friction Force: $\vec{f}$ - opposes relative motion between surfaces
Spring Force: $\vec{F}_{spring}$ - restoring force in deformed springs

The strong nuclear force is about 10³⁸ times stronger than gravity. Why is it gravity, not the strong force, that decides how the planets move?

03

Newton's Three Laws of Motion

Newton's Laws: The Foundation

Newton's three laws of motion form the foundation of classical mechanics. Built on the concepts of inertia, mass, and force, they provide a complete framework for analysing the relationship between forces and motion.

Newton's First Law (Law of Inertia)

This law formalizes the concept of inertia we discussed earlier.

A body remains at rest or in uniform straight-line motion unless acted upon by a net external force.

Essentially, the First Law states that an object's velocity remains constant (which includes the case of zero velocity, i.e. being at rest) if, and only if, the net force acting on it is zero. If the net force is zero, the acceleration is zero. This reinforces that force is required to change velocity (i.e. to accelerate), not merely to maintain it. For example, a book of mass 0.5 kg resting on a table ( $\vec{v} = 0$ ) stays at rest unless someone applies a net force to push or lift it. Similarly, a hockey puck of mass 0.1 kg sliding on near-frictionless ice ( $\vec{v} = \text{constant}$ ) will continue sliding at that constant velocity in a straight line until a net force (like friction with rougher ice, collision with a wall, or air resistance) acts upon it.

True or False: For an object moving at constant velocity, Newton's First Law requires the net external force to be zero.

Newton's Second Law (Law of Acceleration)

While the First Law describes motion in the absence of a net force, the Second Law quantifies what happens when a net force is present.

The acceleration ( $\vec{a}$ ) of an object is directly proportional to the net external force ( $\vec{F}$ ) acting on it, and inversely proportional to its mass.

$\vec{F}_{net} = m\vec{a}$

Here $\vec{F}_{net}$ is the vector sum of all external forces on the object, $m$ is its mass, and $\vec{a}$ is its acceleration. The law is a vector statement: $\vec{a}$ always points along $\vec{F}_{net}$ , never along the velocity.

Doubling F doubles a; doubling m halves it.

Reading the law both ways

$\vec{F}_{net} = m\vec{a}$ is used in two directions. Identify which one a problem needs before starting.

Forces known, motion wanted. A net force of 10 N acts horizontally on a 2 kg object at rest. Then $a = 10/2 = 5\ \text{m/s}^2$ in the direction of the force. The kinematics of the previous chapter apply next from there.
Motion known, forces wanted. If you can measure the acceleration, you know the net force, which is how the tension in a string or the normal force from a surface is usually found.

Two consequences deserve careful attention: an object can be moving fast with zero net force (constant velocity), and an object can be momentarily at rest with a large net force (a ball at the top of its flight).

True or False: According to Newton's Second Law, if the same net force is applied to two objects, the object with the larger mass will experience a larger acceleration.

Newton's Third Law (Law of Action-Reaction)

The first two laws are about the motion of one object. The Third Law is about the nature of forces themselves.

If object A exerts a force on object B ( $\vec{F}_{B \leftarrow A}$ ), then B simultaneously exerts a force on A such that $\vec{F}_{A \leftarrow B} = -\vec{F}_{B \leftarrow A}$ .

Forces occur in pairs. A force is always an interaction between two objects; there is no such thing as an isolated force.
Equal in magnitude, opposite in direction. The two act along the same line, pointing opposite ways.
They act on different objects. This point is the most important. Because the two forces act on different bodies, they never cancel in the equation of motion of either.
Simultaneous. There is no delay and no 'first' force.

Why the pair does not cancel

Because Newton's second law is applied to one object at a time. To find the acceleration of a body you add up the forces acting on that body, and nothing else. That sum is its net force.

Newton's third law does something different: it relates a force on one body to a force on another. The two members of a pair therefore land on two different sums, and neither sum ever contains both. Cancellation is a question you can only ask within a single free-body diagram. A pair is never in one.

Worked through the standard examples:

A 60 kg person pushes a wall with 50 N. The wall pushes back on the person with 50 N. The person, on roller skates, accelerates, the wall does not. This occurs because the ground holds it.
The Earth pulls you down with your weight; you pull the Earth up with exactly the same force. You accelerate visibly and the Earth does not. This occurs because the same force divided by the Earth's enormous mass gives a negligible acceleration.
A rocket pushes exhaust gas down; the gas pushes the rocket up. Nothing outside is needed to push against.

Pick a body: the forces shown are the ones acting on it.

When kicking a football, is the action-reaction pair the force exerted by the foot on the ball and the equal and opposite force exerted by the ball on the foot?

Summary of Common Forces

In summary, a Force is an interaction between two objects or between an object and its environment, causing a change in the object's motion (acceleration) unless balanced by other forces. Based on our discussion of fundamental forces and common scenarios in mechanics, we will frequently encounter the following types of forces in this chapter:

Gravitational Force (Weight): The downward pull exerted by a large celestial body (like Earth) on an object near its surface ( $\vec{W} = m\vec{g}$ ).
Normal Force ( $\vec{N}$ ): The perpendicular contact force exerted by a surface on an object pressing against it, preventing penetration.
Tension Force ( $\vec{T}$ ): The pulling force transmitted through a string, rope, cable, or similar object when it is taut.
Friction Force ( $\vec{f}$ ): A contact force parallel to the surface that opposes relative motion or attempted relative motion between surfaces.
Spring Force ( $\vec{F}_{spring}$ ): The restoring force exerted by a deformed spring, opposing the deformation.

A horse pulls a cart forward. The cart pulls back on the horse with exactly the same force. Why does the pair still move off?

04

Free Body Diagrams (FBDs)

What is a Free Body Diagram?

Newton's Second Law, $\vec{F}_{net} = m\vec{a}$ , is the central equation for solving dynamics problems. However, applying it correctly requires us to accurately identify all the external forces acting on the object or system of interest. The essential tool for this is the Free Body Diagram (FBD).

An FBD is a simplified diagram representing a single object (or a system treated as a single entity) isolated from its surroundings. Its purpose is to clearly visualize all the external forces acting on that specific object. By focusing only on the object and the forces exerted on it by other objects or fields, we can systematically apply Newton's Second Law.

True or False: A Free Body Diagram isolates a single body. It represents all external forces acting on that body.

Steps to Construct a Free Body Diagram

Constructing an accurate FBD is usually the most critical step in a dynamics problem.

Identify the system. Decide exactly which object, or group of objects, you are analysing.
Isolate the body. Draw it alone, a dot or a box, with everything else erased.
Draw one arrow per external interaction. Ask what is touching the body, and what fields reach it. Common answers: weight $m\vec{g}$ downwards; the normal force $\vec{N}$ , perpendicular to and away from a contact surface; tension $\vec{T}$ , along a string and away from the body; an applied push or pull $\vec{F}_{app}$ ; friction $\vec{f}$ , along a surface; a spring force $\vec{F}_{spring}$ .
Label every arrow with a symbol you will use in the equations.
Choose axes. Align one axis with the known direction of the acceleration. The equations then simplify: one gives the motion, the other gives the normal force.

Building a free-body diagram in four steps.

The most common mistake

The most common mistake is putting the wrong forces on the diagram: forces the body exerts on other things, or forces internal to a system you have chosen to treat as one object.

For a block resting on a table, the block's diagram carries its weight (Earth on block) and the normal force (table on block). It does not carry the force of the block on the table, that arrow belongs on the table's diagram. Every arrow you draw should be finishable as the sentence "… exerted on this body by ___". If you cannot fill in the blank, the arrow does not belong.

A correct free-body diagram, and a wrong one.

Example: Bottle on Table

Consider a water bottle of mass $m = 0.5$ kg resting in equilibrium on a horizontal table.

We will use $g \approx 10 \, \text{m/s}^2$ for simplicity.

Free Body Diagram (FBD) for the Water Bottle: The system is the water bottle. We isolate it and show the forces acting on it.

Weight ( $\vec{W}_1$ ): Force exerted by the Earth on the bottle, acting vertically downwards. Magnitude $W = mg = 0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N}$ .
Normal Force ( $\vec{N_1}$ ): Force exerted by the table surface on the bottle, acting vertically upwards, perpendicular to the surface.

Since the bottle is in equilibrium ( $\vec{a} = 0$ ), applying Newton's Second Law on the bottle:

$\sum F = N_1 - W_1 = ma = 0$

Therefore, $N_1 = W_1 = 5 \, \text{N}$ . The normal force exerted by the table on the bottle is 5 N upwards.

Identifying Action-Reaction Pairs

Action-Reaction Pairs (Newton's Third Law):

The force of weight $\vec{W}_1$ (Earth on bottle) has a reaction pair: the gravitational force exerted by the bottle on the Earth ( $W_1'$ (Bottle on Earth)). This is equal in magnitude (5 N) and opposite in direction (upwards, towards the bottle).

The gravitational pair: Earth and bottle.

The normal force

N_1

(table on bottle) has a reaction pair: the force exerted by the bottle on the table (

N_1'

(bottle on table)), which is equal in magnitude (5 N) and opposite in direction (downwards).

Note that $W_1$ and $N_1$ are not an action-reaction pair, even though they are equal and opposite in this specific equilibrium case. They both act on the same object (the bottle) and arise from different interactions (gravity and contact).

(Optional) Considering the Table: If we were to draw an FBD for the table, it would include following forces:

The force exerted by the bottle on the table ( $\vec{F}_{table \leftarrow bottle} = N_1'$ ), downwards (5 N).
The weight of the table itself ( $\vec{W}_{table} = W_2$ ), downwards.
The normal forces exerted by the ground on the table legs ( $\vec{N}_{table \leftarrow ground} = N_2$ ), upwards.

In equilibrium, these forces on the table would also sum to zero, because of Newton's second law.

Pick a body: W and N land on the same body, so they are not a pair.

For a block resting on a table, do the downward weight and the upward normal force constitute an action-reaction pair??

Applying Newton's Laws with FBDs

Once the FBD is correctly drawn, we can apply Newton's Second Law: $\sum \vec{F}_{ext} = m\vec{a}$ . If the object is in equilibrium (at rest or moving with constant velocity), then $\vec{a} = 0$ , and the vector sum of forces is zero: $\sum \vec{F}_{ext} = 0$ . If the object is accelerating, the vector sum of forces equals the mass times the acceleration vector. Often, we resolve the forces and acceleration into components along the chosen coordinate axes: $\sum F_x = ma_x$ and $\sum F_y = ma_y$ .

A book rests on a table in a lift that is accelerating downwards. Which pair of forces on the book is still exactly equal and opposite?

A crate is being dragged across a floor at a steady speed. Which of these belongs on the crate's free-body diagram?

05

Normal Force

What is Normal Force?

The Normal Force, denoted by $\vec{N}$ , is the force a surface exerts on an object in contact with it. It is one of the most frequently encountered forces in mechanics problems.

The normal force is a contact force. It arises only when two objects are physically touching. As mentioned earlier, it is a result of the electromagnetic repulsion between the atoms of the surfaces in contact. When an object presses against a surface, the surface deforms slightly (often imperceptibly) and pushes back, preventing the object from penetrating it. This push-back force exerted by the surface on the object is the normal force. The following diagrams show two such examples:

The normal force acts at the contact surface.

Two normal forces, each perpendicular to its surface.

Key Characteristics of Normal Force

The defining characteristic of the normal force is its direction: it always acts perpendicular to the surface of contact and is directed away from the surface, towards the object it is acting upon. The term 'normal' in mathematics means perpendicular, hence the name. It is always a pushing force exerted by the surface.

Because it arises from repulsion at the atomic level preventing interpenetration, the normal force can only push; it can never pull. If the surfaces lose contact or try to move apart, the normal force instantly becomes zero. Consider interacting with a block using only the palm of your hand. You can push the block by moving your hand towards it, and your hand exerts a normal force on the block. However, if you move your hand away from the block, the block does not follow - your hand cannot exert a 'pulling' normal force. This concept is illustrated below:

The magnitude of the normal force depends on the specific situation, determined by applying Newton's Laws, particularly the condition that acceleration perpendicular to the surface is typically zero (assuming the object stays on the surface). There is no general formula like: $N = mg$ .

Four situations, four different normal forces.

Is the normal force always directed perpendicular to the surface of contact?

Example 1: Box on Floor at Rest

A 5 kg box rests motionless on a horizontal floor. Find the normal force exerted by the floor.

Solution: Equilibrium ( $a_y = 0$ ). Vertical forces: Weight $W = mg = 5 \times 10 = 50$ N (down), Normal force $N$ (up).

Applying $\sum F_y = ma_y = 0$ (upwards positive):

$N - W = 0 \implies N = W = 50 \, \text{N}$

The normal force is 50 N upwards.

Example 2: Object Accelerating Upwards

Consider the same block of mass $m = 2$ kg on a horizontal surface (like a table or the floor of an elevator), but now suppose the surface is accelerating upwards at $a = 2 \, \text{m/s}^2$ . We want to find the normal force exerted by the surface on the block.

The surface accelerates upward with the block.

FBD of the block: The forces acting on the block are the same as in Example 1: Weight $W = mg = 2 \times 10 = 20$ N (down), Normal Force $N$ (up). However, the block now has a net upward acceleration $\vec{a}$ .

Applying Newton's Second Law: The block is accelerating upwards, so $\vec{a}$ is non-zero and points up. We apply $\sum F_y = ma_y$ in the vertical direction (taking upwards as positive), where $a_y = +2 \, \text{m/s}^2$ :

$N - W = ma_y \implies N = W + ma_y = mg + ma_y$

$N = 24 \, \text{N}$

In this case, the normal force ( $24 \, \text{N}$ ) is greater than the weight ( $20 \, \text{N}$ ). The surface must not only support the block's weight but also provide the additional upward force required to cause the upward acceleration. This is why you feel heavier in an elevator when it accelerates upwards.

Example 3: Block Pulled at an Angle

A 6 kg block on a smooth surface is pulled by 15 N at 30° above horizontal. Find normal force and acceleration.

Resolve the applied force:

$F_x = F\cos30° = 15 \times 0.866 \approx 13 \, \text{N}$

$F_y = F\sin30° = 15 \times 0.5 = 7.5 \, \text{N}$

Vertical ( $a_y = 0$ ):

$N + F_y - W = 0 \implies N = mg - F_y = 60 - 7.5 = 52.5 \, \text{N}$

Horizontal:

$F_x = ma_x \implies a_x = \dfrac{13}{6} \approx 2.17 \, \text{m/s}^2$

The normal force is 52.5 N. It is less than the weight because the upward component of the pull supports part of the weight.

When a block on a horizontal surface is pulled by an upward-angled force, how does the normal force compare to the weight?

A 2 kg block sits on the floor of a lift. In which case is the normal force on it exactly zero?

Two identical bricks lie on a table, one flat on its large face and one standing on its small end. How do the normal forces on them compare?

A car goes over the top of a humpback bridge fast enough that the passengers feel momentarily weightless. What is the normal force from the seat at that instant?

06

Tension Force

What is Tension?

The Tension Force, typically denoted by $\vec{T}$ , is the pulling force that arises when a flexible connector such as a string, rope, cable, or chain is pulled taut. The following diagrams illustrate the tension force:

Tension is fundamentally a pulling force. Consider pulling on a rope tied to a box; the rope transmits your pull to the box. This transmitted force within the rope is tension. At any point along the rope, the force applied by one part of the rope on the adjacent part across that point (or conceptual intersection) is called the tension at that point. Microscopically, it arises from the electromagnetic forces between adjacent molecules within the material of the string/rope, resisting the tendency to be pulled apart.

Key Characteristics of Tension

The direction of the tension force exerted by a segment of string/rope on an object (or another segment) is always along the line of the string/rope and directed away from the object (or segment) it is acting upon, pulling on it. A string can only pull; it cannot push, because a string goes slack if you try to push with it. This is shown in the figure below:

True or False: The tension force exerted by a string on an object always acts along the string, pulling the object.

Ideal Strings

Most problems treat strings and ropes as ideal:

Massless, negligible mass compared with the bodies it connects.
Inextensible, its length never changes. Consequently, connected bodies move with the same speed and the same magnitude of acceleration. This is the extra equation that makes connected-body problems solvable.
Perfectly flexible, it bends without resistance. Goes slack rather than pushing.

An important consequence is that the tension is the same everywhere along the string, even while it accelerates and even where it runs over an ideal pulley.

Why the tension is uniform

Cut a short segment out of the string. The string on one side pulls it with $T_1$ ; the string on the other side pulls it with $T_2$ ; let $F_{along}$ be any other force acting along its length, such as its own weight if it hangs vertically.

Its mass is zero. Consequently, Newton's Second Law along the string reads

$T_1 - T_2 + F_{along} = m\,a_{along} = 0$

The right-hand side is zero whatever the acceleration is, because the mass is zero. So if no force acts along the segment's length ( $F_{along} = 0$ ):

$T_1 = T_2$

and since the segment was arbitrary, the tension is the same all along.

In a massless string, the tension is the same throughout, provided no external force acts along its length.

If a force does act along the length, the weight of a heavy hanging rope, for instance, the tension varies. It is largest at the top.

A pulley changes direction, not magnitude.

When an ideal string passes over an ideal pulley (pulley can't apply force on string along its length), does the magnitude of the tension in the string remain the same on both sides of the pulley?

Example 1: Suspended Block

A block of mass $m = 1$ kg hangs at rest, suspended from the ceiling by a single light (massless) string as shown:

Find the tension in the string. Use $g \approx 10 \, \text{m/s}^2$ .

FBD of the block:

Weight $\vec{W}$ acting downwards, magnitude $W = mg = 1 \times 10 = 10$ N.
Tension Force $\vec{T}$ exerted by the string, acting upwards along the string.

Applying Newton's Second Law: The block is in equilibrium ( $\vec{a} = 0$ ). Applying $\sum F_y = ma_y$ in the vertical direction (upwards positive):

$T - W = m(0) \implies T = W = mg \implies T = 10 \, \text{N}$

The tension in the string is equal to the weight of the suspended block.

Example 2: Block Pulled Upward

A 4 kg block is pulled upward by a string with 50 N tension. Find its acceleration.

Solution: Weight $W = 40$ N (down), Tension $T = 50$ N (up)

$\sum F_y = ma \implies T - W = ma$

$50 - 40 = 4a \implies a = 2.5 \, \text{m/s}^2 \text{ (upward)}$

Example 3: Two Blocks, One Hanging

A 1 kg block on a table connected via pulley to a hanging 2 kg block. Find acceleration and tension.

A block on a table, pulled by a hanging block.

FBDs:

Free-body diagram of the block on the table.

Solution: $m_1$ accelerates right, $m_2$ down with magnitude $a$ .

For $m_1$ : $T = m_1 a = 1a$ ... (1)

For $m_2$ : $W_2 - T = m_2 a \implies 20 - T = 2a$ ... (2)

Substitute (1) into (2): $20 - a = 2a \implies a = \dfrac{20}{3} \approx 6.67 \, \text{m/s}^2$

$T = \dfrac{20}{3} \approx 6.67 \, \text{N}$

In a system with a hanging mass, is the tension less than the weight of the hanging block when the system accelerates?

In the table-and-pulley example the hanging 2 kg block would fall at 10 m/s² on its own, but in the system it falls at only 6.67 m/s². What is the tension doing?

A heavy chain hangs from the ceiling with a lamp on the bottom. Where in the chain is the tension largest?

07

Spring Force and Hooke's Law

Spring Force Basics

Springs are common elements in mechanical systems. If a spring (assumed massless) is in its natural length (or equilibrium position), it is neither compressed nor stretched and exerts no force on objects attached to it.

When a spring is deformed (stretched or compressed) from its natural length, it exerts a force attempting to return to that equilibrium length. This force is the Spring Force ( $\vec{F}_{spring}$ ).

Hooke's Law

For many springs, within their 'elastic limit', the force exerted is approximately proportional to the displacement from equilibrium. This is Hooke's Law.

$\vec{F}_{spring} = -k\vec{x}$

In this equation:

$\vec{F}_{spring}$ is the force exerted by the spring on the object attached to its end.
$k$ is the positive spring constant (stiffness) in N/m.
$\vec{x}$ is the displacement vector of the spring's end from its equilibrium position ( $x = 0$ ).

The negative sign signifies a restoring force: $\vec{F}_{spring}$ always opposes the displacement $\vec{x}$ , trying to restore the spring to $x = 0$ .

Hooke’s law: a straight line of slope −k.

Understanding the Restoring Force

Take $x = 0$ at the natural length, with positive $x$ to the right. The negative sign in Hooke's law makes the force point back towards $x = 0$ , whichever side the end is displaced to.

Concept Check: with $k = 100$ N/m, which way does the force point when the end is moved 0.1 m to the left. How large is it? Think about the answer first, then study the figure.

Stretching the spring to the right ( $x > 0$ ) produces a leftward pull ( $F < 0$ ); compressing it to the left ( $x < 0$ ) produces a rightward push ( $F > 0$ ). The magnitude is $k|x|$ in both cases, it depends only on how far the spring is deformed, not on which way.

Example: Calculating Spring Force

A spring has a spring constant $k = 100$ N/m. One end is fixed. Calculate the force exerted by the spring if the free end is moved:

(a) To position $x = +0.1$ m (stretched right).

(b) To position $x = -0.1$ m (compressed left).

Solution: Use Hooke's Law, $\vec{F}_{spring} = -k\vec{x}$ . Let $\hat{i}$ be the unit vector to the right.

(a) Stretched to $x = +0.1$ m:

$\vec{x} = +0.1 \, \hat{i} \, \text{m}$

$\vec{F}_{spring} = -(100 \, \text{N/m})(+0.1 \, \text{m}) \, \hat{i} = -10 \, \hat{i} \, \text{N}$

The force is 10 N in the negative x-direction (left).

(b) Compressed to $x = -0.1$ m:

$\vec{x} = -0.1 \, \hat{i} \, \text{m}$

$\vec{F}_{spring} = -(100 \, \text{N/m})(-0.1 \, \text{m}) \, \hat{i} = +10 \, \hat{i} \, \text{N}$

The force is 10 N in the positive x-direction (right). Magnitude is 10 N in both cases, directed towards equilibrium ( $x = 0$ ).

True or False: If you stretch an ideal spring twice as far from equilibrium, the restoring force is four times stronger.

Combination of Springs

Springs can be combined in many ways. The following are some common ways of connecting multiple springs:

Springs in Series

Springs are in series when they are joined end to end in a chain. Take two ideal springs $k_1$ and $k_2$ joined at a point P, with the far end of the first fixed and a force $F$ applied to the free end of the second.

The junction P has no mass. Newton's second law for it therefore reads $\sum \vec{F}_P = 0\times\vec{a} = 0$ . That holds whether the system is in equilibrium or accelerating, as long as the acceleration is finite. So the two springs pull on P with equal and opposite forces:

$|\vec{F}_1| = |\vec{F}_2|$

Spring 2 is also massless. Consequently, the force it exerts at P equals the force $F$ applied at its other end. The same force is therefore carried by both springs:

$F_1 = F_2 = F$

Adding the extensions

Under that common force each spring stretches by its own amount:

$x_1 = \frac{F}{k_1} \quad \text{and} \quad x_2 = \frac{F}{k_2}$

The chain as a whole stretches by $x_{total} = x_1 + x_2$ . An equivalent single spring $k_{eff}$ would stretch by $F/k_{eff}$ under the same force, so

$\frac{F}{k_{eff}} = \frac{F}{k_1} + \frac{F}{k_2} \implies \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$

Because the reciprocals add, $k_{eff}$ is smaller than either $k_1$ or $k_2$ : a series combination is softer than any of its parts. That is the sense in which adding another spring makes the chain easier, not harder, to stretch.

Springs in Parallel

When springs are side-by-side (parallel), both stretch by the same amount:

$k_{eff} = k_1 + k_2$

Why? Both springs stretch by $x$ , exerting forces $F_1 = k_1 x$ and $F_2 = k_2 x$ . Total force: $F_{total} = F_1 + F_2 = (k_1 + k_2)x$

Parallel combination is stiffer (larger $k_{eff}$ ) than individual springs.

Cutting Springs

A uniform spring of stiffness $k$ cut into $n$ equal pieces gives pieces of stiffness

$k' = nk$

Cutting a spring makes each piece stiffer.

Why. Every coil in a stretched spring carries the same force. Each contributes the same share of the total extension. Half the coils therefore stretch half as far under the same force $F$ . Since stiffness is force per unit extension, halving the extension doubles the stiffness:

$k' = \frac{F}{x/2} = \frac{2F}{x} = 2k$

This is the series result read backwards: two identical $2k$ springs in series give $1/k_{eff} = 1/2k + 1/2k = 1/k$ . This is the original spring.

Example: a spring with $k = 50$ N/m cut into two equal halves gives $k' = 2 \times 50 = 100$ N/m for each half.

Is the effective spring constant of springs connected in series always less than the smallest individual spring constant?

Summary Example

Two springs: $k_1 = 100$ N/m, $k_2 = 200$ N/m

In series:

$\frac{1}{k_{eff}} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200} \implies k_{eff} \approx 66.7 \, \text{N/m}$

In parallel:

$k_{eff} = 100 + 200 = 300 \, \text{N/m}$

A spring of stiffness 60 N/m is cut into three equal pieces, and two of those pieces are then joined side by side in parallel. What is the stiffness of that pair?

A spring stretches 4 cm when a 2 kg mass hangs from it. What does it stretch when a 6 kg mass hangs from it instead, assuming it stays within its elastic limit?

A spring is compressed by pushing its free end 5 cm to the left of the natural length. Which way does the spring push on your hand, and how does that compare with stretching it 5 cm to the right?

Two identical springs, each 200 N/m, support a shelf side by side. One breaks. What happens to the sag under the same load?

08

Motion on Inclined Planes

Why Tilted Coordinates?

For objects moving along an inclined plane, it is almost always more convenient to choose a tilted coordinate system:

x-axis: Parallel to the incline
y-axis: Perpendicular to the incline

This choice simplifies the calculation because the acceleration is usually along the incline (x-axis).

Tilted axes: x along the slope, y out of it.

Resolving Weight on an Incline

Weight $\vec{W} = m\vec{g}$ always acts vertically downward. On an incline at angle $\theta$ , we resolve it:

Parallel to incline (down the slope):

$W_{\parallel} = mg\sin\theta$

Perpendicular to incline (into the surface):

$W_{\perp} = mg\cos\theta$

Remember: $mg\sin\theta$ pulls down the slope, $mg\cos\theta$ presses into the surface

Normal Force on Incline

The normal force $N$ acts perpendicular to the incline surface. For a block on an incline with no perpendicular acceleration:

$N = mg\cos\theta$

Note: $N < mg$ (the full weight) because only the perpendicular component of weight is balanced by the normal force.

Example 1: Sliding Down Smooth Incline

A block starts from rest on a smooth (frictionless) incline at $\theta = 45°$ . What is its speed after 2 seconds?

Released from rest on a smooth 45° incline.

Solution: Only force along incline is $mg\sin\theta$

$\sum F_x = ma \implies mg\sin\theta = ma$

$a = g\sin45° = 10 \times \dfrac{1}{\sqrt{2}} = 5\sqrt{2} \, \text{m/s}^2$

Using kinematics: $u = 0$ , $a = 5\sqrt{2}$ , $t = 2$ s

$v = u + at = 0 + (5\sqrt{2})(2) = 10\sqrt{2} \approx 14.14 \, \text{m/s}$

If the incline angle were increased to 60°, would the acceleration be greater or smaller than at 45°?

Example 2: Block Pulled Up Incline

A block on a smooth incline ( $\theta = 30°$ ) is pulled up by a string parallel to the incline with tension $T = 60$ N. Mass = 5 kg. Find acceleration.

Solution: Along incline (up positive):

$\sum F_x = T - mg\sin\theta = ma$

$60 - (5)(10)(0.5) = 5a$

$60 - 25 = 5a \implies a = 7 \, \text{m/s}^2 \text{ (up incline)}$

Example 3: Incline with Friction

A 4 kg block on a rough incline ( $\theta = 37°$ , $\mu_k = 0.25$ ) slides down. Find acceleration.

Solution:

Normal: $N = mg\cos37° = 40 \times 0.8 = 32$ N

Friction (opposes motion, up incline): $f_k = \mu_k N = 0.25 \times 32 = 8$ N

Along incline (down positive):

$mg\sin37° - f_k = ma$

$(40)(0.6) - 8 = 4a$

$24 - 8 = 4a \implies a = 4 \, \text{m/s}^2$

On an inclined plane, does friction reduce the acceleration compared to a smooth (frictionless) incline?

Two blocks, 2 kg and 8 kg, are released from rest together on the same smooth 30° slope. Which reaches the bottom first?

A 10 kg block sits on a smooth 30° slope. What is the normal force on it? (g = 10 m/s²)

09

Connected Systems and Constraint Motion

Systems of Connected Objects

Many problems involve multiple objects connected by strings, in contact, or constrained to move together. Two approaches:

1. System Approach: Treat all objects as one system, find common acceleration

2. Individual FBD Approach: Draw separate FBDs for each object, solve simultaneously

Often, you use both: system approach for acceleration, individual FBDs for internal forces (tension, contact forces).

Example 1: Two Blocks in Contact

Two blocks ( $m_1 = 2$ kg, $m_2 = 4$ kg) on smooth surface. Force $F = 30$ N pushes $m_1$ toward $m_2$ . Find acceleration and contact force.

System approach: Total mass $M = 6$ kg

$F = Ma \implies 30 = 6a \implies a = 5 \, \text{m/s}^2$

Contact force N (using FBD of $m_2$ ):

$N = m_2 a = 4 \times 5 = 20 \, \text{N}$

In the two-block system, is the contact force between blocks less than the applied force?

Example 2: Two Blocks Connected by String

Two blocks ( $m_1 = 5$ kg, $m_2 = 3$ kg) connected by string on smooth surface. Force $F = 40$ N applied to $m_1$ . Find tension.

System: $M = 8$ kg, $a = F/M = 40/8 = 5 \, \text{m/s}^2$

Tension (using FBD of $m_2$ ):

Free-body diagram of the trailing block.

$T = m_2 a = 3 \times 5 = 15 \, \text{N}$

Atwood Machine

The Atwood machine is a classic arrangement: two masses connected by a string passing over a pulley.

Example: $m_1 = 2$ kg, $m_2 = 3$ kg. Find acceleration and tension.

Solution: $m_2$ accelerates down, $m_1$ up, magnitude $a$

For $m_1$ (up positive): $T - 20 = 2a$ ... (1)

For $m_2$ (down positive): $30 - T = 3a$ ... (2)

Add: $10 = 5a \implies a = 2 \, \text{m/s}^2$

From (1): $T = 20 + 4 = 24 \, \text{N}$

In an Atwood machine, does the tension lie between the two weights, allowing the heavier mass to accelerate downwards?

General Atwood Machine Formula

For masses $m_1$ and $m_2$ (where $m_2 > m_1$ ):

Acceleration:

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

Tension:

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

These formulas can be derived by solving the force equations as shown above.

The formula checked against a second pair of masses

Put $m_1 = 5$ kg and $m_2 = 7$ kg into the formulas above and then work the same problem from the two free-body diagrams, to see that they agree.

Masses $m_1 = 5$ kg and $m_2 = 7$ kg on ideal Atwood machine.

FBDs:

Solution: $m_2 > m_1$ . Consequently, $m_2$ goes down, $m_1$ goes up

For $m_1$ : $T - 50 = 5a$ ... (1)

For $m_2$ : $70 - T = 7a$ ... (2)

Add: $20 = 12a \implies a = \dfrac{5}{3} \approx 1.67 \, \text{m/s}^2$

From (1): $T = 50 + 5(5/3) = \dfrac{175}{3} \approx 58.33 \, \text{N}$

Example: Three Blocks in Contact

Three blocks (1 kg, 2 kg, 3 kg) pushed by $F = 24$ N on a smooth surface. Find the acceleration and the contact forces.

System: $M = 6$ kg, $a = 24/6 = 4 \, \text{m/s}^2$

Contact forces:

For $m_3$ : $N_{23} = m_3 a = 3 \times 4 = 12$ N

For $m_2$ : $N_{12} - N_{32} = m_2 a \implies N_{12} = 12 + 8 = 20$ N

Contact forces: $N_{12} = 20$ N, $N_{23} = 12$ N

The same three blocks (1 kg, 2 kg, 3 kg) are turned round so the 24 N push is applied to the 3 kg block instead. What happens to the acceleration and to the largest contact force?

Why can you find the acceleration of two string-connected blocks by treating them as one 8 kg object and ignoring the tension entirely?

10

Non-Inertial Frames and Pseudo Forces

Inertial vs Non-Inertial Frames

Newton's Laws do not hold in all reference frames.

Inertial Frame: A reference frame where Newton's First Law holds - objects with zero net force remain at constant velocity (including rest).

Non-Inertial Frame: An accelerating reference frame where Newton's Laws in standard form do NOT hold.

Example: A block at rest on the floor of an accelerating train:

Ground observer (inertial): Block at rest, forces balanced ✓
Train observer (non-inertial): Block accelerates backward, but forces still balanced? ✗

The Pseudo Force

To use Newton's Laws in a non-inertial frame, we introduce a pseudo force (fictitious force):

$\vec{F}_{pseudo} = -m\vec{a}_{frame}$

where $\vec{a}_{frame}$ is the acceleration of the non-inertial frame relative to an inertial frame.

Key points:

Acts on every object of mass $m$ in the frame
Directed opposite to frame's acceleration
Not a real physical interaction
Does NOT have an action-reaction pair

From the train: a pseudo force is added.

Is the pseudo force always directed opposite to the acceleration of the non-inertial frame?

Modified Newton's Second Law

In a non-inertial frame:

$\vec{F}_{real} + \vec{F}_{pseudo} = m\vec{a}_{relative}$

where $\vec{a}_{relative}$ is acceleration relative to the non-inertial frame.

This allows us to analyze motion within the accelerating frame by treating the pseudo force as one more force.

Example 1: Pendulum in Accelerating Car

A pendulum hangs in a car accelerating at $a_{car}$ . Find the angle $\theta$ from vertical at equilibrium (relative to car).

Solution (car's frame):

Pseudo force: $F_{pseudo} = ma_{car}$ (backward)

Equilibrium in car frame: $a_{relative} = 0$

Forces on bob: Weight $mg$ (down), Tension $T$ (along string), Pseudo force $ma_{car}$ (backward)

Resolve tension: $T_x = T\sin\theta$ , $T_y = T\cos\theta$

Horizontal: $T\sin\theta = ma_{car}$ ... (1)

Vertical: $T\cos\theta = mg$ ... (2)

Divide (1) by (2):

$\tan\theta = \dfrac{a_{car}}{g} \implies \theta = \arctan\left(\dfrac{a_{car}}{g}\right)$

If the car accelerates faster, would the pendulum angle $\theta$ increase or decrease?

Example 2: Apparent Weight in Elevator

A 70 kg person in an elevator accelerating upward at 3 m/s². Find scale reading.

Method 1 (Inertial frame - ground):

$N - W = ma \implies N = mg + ma = 700 + 210 = 910 \, \text{N}$

Method 2 (Non-inertial frame - elevator):

Pseudo force: $F_{pseudo} = 210$ N (downward)

Equilibrium in elevator: $N - W - F_{pseudo} = 0$

$N = W + F_{pseudo} = 700 + 210 = 910 \, \text{N}$

Scale reading: $910/10 = 91$ kg. The person feels heavier than usual.

A physicist in a windowless, smoothly accelerating railway carriage sees a pendulum hanging at 12° from the vertical. What can they conclude?

Which of these is not an inertial frame, to a good approximation?

A 60 kg person stands on a scale in a lift. The scale reads 48 kg. What is the lift doing? (g = 10 m/s²)

11

Problem-Solving Strategies and Practice

Systematic Problem-Solving Approach

Follow these steps for dynamics problems:

Identify the system: What object(s) are you analyzing?
Draw FBDs: For each object, show all external forces
Choose coordinates: Align axes with acceleration when possible
Apply Newton's Second Law: $\sum \vec{F} = m\vec{a}$ in component form
Solve equations: Use algebra or simultaneous equations
Check: Do signs make sense? Are units correct?

Practice 1: Hanging Lamp

A 2 kg lamp hangs motionless from a wire. Find tension.

Solution: Equilibrium ( $a_y = 0$ )

Forces: Weight $W = mg = 20$ N (down), Tension $T$ (up)

$\sum F_y = 0 \implies T - W = 0 \implies T = 20 \, \text{N}$

Practice 2: Block Pushed on Surface

A 10 kg block on smooth horizontal floor pushed with 20 N horizontally. Find normal force and acceleration.

A 10 kg block pushed along a smooth floor.

Solution:

Vertical: Equilibrium → $N = W = 100$ N

Horizontal: $F = ma \implies 20 = 10a \implies a = 2 \, \text{m/s}^2$

Normal force = 100 N, acceleration = 2 m/s². The horizontal push does not affect the vertical forces.

In the pushed block example, does the 20 N horizontal push affect the normal force?

Practice 3: Block Against Wall

A 2 kg block held against smooth vertical wall by 30 N horizontal force and vertical string. Find normal force and tension for equilibrium.

Solution: Equilibrium ( $a_x = 0$ , $a_y = 0$ )

Horizontal: $F - N = 0 \implies N = 30$ N

Vertical: $T - W = 0 \implies T = mg = 20$ N

Practice 4: Weighing Machine in Elevator

An 80 kg person on a scale in an elevator. Find scale reading when elevator:

(a) Moving at constant 2 m/s upward:

$a = 0$ → $N = W = 800$ N → Reading = 80 kg

(b) Accelerating upward at 2 m/s²:

$N - W = ma$ → $N = 800 + 160 = 960$ N → Reading = 96 kg

(c) Accelerating downward at 2 m/s²:

$N - W = m(-2)$ → $N = 800 - 160 = 640$ N → Reading = 64 kg

Does a person feel heavier (higher scale reading) when an elevator accelerates upward?

Practice 5: Force on Pulley Support

An Atwood machine with $m_1 = 4$ kg, $m_2 = 6$ kg has tension $T = 48$ N. What force does the clamp exert on the pulley axle?

Solution: Pulley is massless and stationary (equilibrium)

Forces on pulley:

Tension $T = 48$ N downward (from $m_1$ side)
Tension $T = 48$ N downward (from $m_2$ side)
Support force $F_{clamp}$ upward

$F_{clamp} - T - T = 0 \implies F_{clamp} = 2T = 96 \, \text{N}$

The clamp exerts 96 N upward on the pulley.

Key Problem-Solving Tips

Always start with a clear FBD
Be consistent with sign conventions
For connected systems, identify constraints (same acceleration, etc.)
Check if object is in equilibrium or accelerating
Note that $N \neq mg$ in general
Friction is self-adjusting (static) or constant (kinetic)
Pseudo forces appear only in non-inertial frames

In Practice 5 the two 48 N tensions add to 96 N on the clamp. But the two masses weigh 40 N and 60 N, which is only 100 N. Why is the clamp force not 100 N?

A lift descends at a constant 3 m/s carrying a 50 kg crate. What is the normal force on the crate from the floor? (g = 10 m/s²)