Fundamentals of Force — JEE Advanced Physics

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Introduction to Fundamentals of Force

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Introduction to Dynamics

From Kinematics to Dynamics

In our previous explorations, we focused on kinematics, the branch of mechanics dedicated to describing motion - how position, velocity, and acceleration change over time. We learned to quantify these changes using equations and graphs.

Now, we venture into the realm of dynamics, the study that seeks to answer the fundamental question: why does motion change? What are the underlying causes that make objects speed up, slow down, or alter their direction? This chapter is dedicated to understanding the reasons behind acceleration.

The Concept of Inertia

Observe the world around you. An object at rest, like a heavy textbook resting on your desk, tends to stay at rest. An object already in motion, such as a hockey puck gliding across smooth ice, tends to continue moving in a straight line at a constant speed (ignoring friction for a moment).

This inherent property of all matter to resist changes in its state of motion - whether it's at rest or moving uniformly - is called inertia. It is the natural tendency of an object to maintain its current velocity.

Different objects have different resistance to oppose the change in its motion. For example, a truck has more capability to oppose its state of motion in comparison to a bicycle. In other words, truck has more inertia than bicycle.

Mass: The Measure of Inertia

A quantitative measure of an object's inertia, its intrinsic resistance to being accelerated, is defined as its mass, typically 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 its state of motion. Mass is a mathematical measure of inertia of the object.

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

What is Force?

So, if objects naturally resist changes in motion due to their inertia (mass), what is it that causes these changes? What makes the resting textbook start sliding, or the moving ball eventually slow down and stop? The answer lies in the concept of force.

In the simplest terms, a force is an interaction that, when applied to an object, tends to change its state of motion (i.e. cause acceleration). We commonly experience forces as a push or a pull exerted by one object on another, like pushing a door open or pulling a wagon. Forces are the agents of change in dynamics.

If an agent has a tendency to push or pull an object, we say, agent is applying force on the object.

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

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

Fundamental Forces in Nature

The Four Fundamental Forces

While we experience a lot of pushes and pulls in daily life - friction, force by ropes, the force of a spring, the force from a surface - physicists have discovered that these are forms of just a few fundamental interactions governing all phenomena in the universe. At the most basic level, there are four known fundamental forces:

Gravitational Force: An attractive force that exists between any two objects with mass. It's 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's much stronger than gravity and underlies most everyday forces like friction, normal force. 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 Earth. It's 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.

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 pen 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 pen's surface. This microscopic electromagnetic repulsion manifests macroscopically as the upward normal force exerted by the table on the pen, preventing penetration. We are about to learn in detail about such forces and the mathematical equations they follow.

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

Normal Force

What is Normal Force?

One of the most frequently encountered forces in mechanics problems involving objects in contact with surfaces is the Normal Force, denoted by $\vec{N}$ . It's essential to understand its characteristics and how to determine its magnitude.

The normal force is a contact force. It arises only when two objects are physically touching. Fundamentally, as mentioned earlier, it's a manifestation 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 repulsive push-back force exerted by the surface on the object is the normal force. Following diagram shows two such examples:

Normal force applied by ground surface on block

Normal force applied by wall and ground on a person

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's acting upon. The term 'normal' in mathematics means perpendicular, hence the name. It's always a pushing force exerted by the surface.

Crucially. This occurs 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. Imagine trying to interact 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:

Normal force is a push. A surface (like a hand) can push an object, but it cannot pull it via the normal force if contact is broken or separation occurs.

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: $N = mg$ .

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.

Box on surface

FBD of box

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.

FBD for block on horizontal surface. Upward acceleration

a = 2 \, \text{m/s}^2

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.

Block pulled at angle

Resolve the applied force:

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

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

FBD with resolved forces

Vertical ( $a_y = 0$ ):

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

Horizontal:

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

Normal force is 52.5 N (less than weight because of upward pull component!).

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

Tension Force

What is Tension?

Another important force in mechanics problems is the Tension Force, typically denoted by $\vec{T}$ . This force arises when a flexible connector like a string, rope, cable, or chain is pulled taut. Following diagrams illustrate the action of tension force:

Tension force on block pulled horizontally by sting.

Tension force on a block hanging from a string.

Tension is fundamentally a pulling force. Imagine 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's acting upon, pulling on it. A string can only pull; it cannot push as string goes slack if you try to push with it. This is shown in animation below:

String becomes slack, if it attempts to push the block.

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

Ideal Strings

In many introductory physics problems, we make simplifying assumptions about strings and ropes, treating them as ideal strings:

Massless: We assume the string itself has negligible mass compared to the objects it connects.
Inextensible: We assume the string does not stretch or shrink; its length remains constant. This implies that connected objects or parts of the string moving along its length have the same speed and magnitude of acceleration.
Perfectly Flexible: We assume the string can bend easily without resistance.

A crucial consequence of the massless assumption is that, under typical conditions, the tension is uniform throughout the string, even if the string accelerates or passes over ideal (massless and frictionless) pulleys.

Let's demonstrate why tension is often uniform in a massless string. Consider a small segment of a massless string, with tension $T_1$ pulling on one end and $T_2$ pulling on the other end (in opposite directions along the string). Let $F_{along}$ be the sum of any other external forces acting on this segment along its length (e.g. a component of gravity if the string hangs vertically).

Forces on a small segment of a string.

The mass of the segment is $m = 0$ . Applying Newton's Second Law along the string:

$-T_2 + T_1 + F_{along} = ma_{along} = (0)a_{along} \implies -T_2 + T_1 + F_{along} = 0$

If there are no external forces acting along the length of the string segment itself (i.e. $F_{along} = 0$ ), which is common when strings connect objects horizontally or pass over frictionless pulleys connecting vertical motions, then:

$T_2 - T_1 = 0 \implies T_1 = T_2$

This shows that the tension magnitude is the same at both ends of the segment. Since this applies to any segment, the tension is uniform throughout the massless string provided no force acts along its length. Thus we conclude,

For a massless string, tension remains same throughout the string, provided no force acts on string externally along its length.

If a force does act along the length (like the weight of a massive hanging rope), the tension will vary. When a massless string passes over an ideal (massless, frictionless) pulley, the pulley simply changes the direction of the tension force without changing its 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:

Block suspended by a string.

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

FBD of the block:

FBD for block suspended by a string.

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.

Block pulled upward

FBD of pulled block

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.

Setup with hanging block

FBDs:

FBD of block on table

FBD of hanging block

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?

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.

Spring at its natural length:

x = 0

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 $x = 0$ .

Understanding the Restoring Force

Consider a coordinate system where $x = 0$ is the equilibrium position and positive $x$ is to the right:

Elongated State: Spring stretched to the right (

x > 0

). Displacement

\vec{x}

is positive. The spring exerts a restoring force

\vec{F}_{spring}

to the left (negative direction).

Compressed State: Spring compressed to the left (

x < 0

). Displacement

\vec{x}

is negative. The spring exerts a restoring force

\vec{F}_{spring}

to the right (positive direction).

Essentially, stretching the spring right ( $x > 0$ ) causes a leftward pull ( $F < 0$ ). Compressing it left ( $x < 0$ ) causes a rightward push ( $F > 0$ ).

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

Although springs can be combined in a number of possible ways, but following are are some common ways of connecting multiple springs:

Springs in Series

Springs are connected in series when they are attached sequentially, end-to-end, forming a chain. Consider two ideal (massless) springs with constants $k_1$ and $k_2$ connected in series, with one end fixed and an external force $F$ applied to the free end, as shown below.

Two springs k1 and k2 in series, connected at junction P, stretched by external force F.

Let's analyze the forces acting on the junction point P, where the two springs connect. Since the springs (and thus the junction) are assumed massless, the net force on the junction must be zero according to Newton's second law ( $\sum \vec{F}_P = m_P \vec{a} = 0 \times \vec{a} = 0$ ), both in equilibrium and during acceleration (as long as acceleration is finite). Spring 1 exerts a force $\vec{F}_1$ on P (pulling left if stretched). Spring 2 exerts a force $\vec{F}_2$ on P (pulling right if stretched by external force F). Therefore, for the net force on P to be zero, these forces must be equal in magnitude:

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

Since spring 2 is ideal, the force it exerts at the junction ( $F_2$ ) must equal the external force $F$ applied at its other end. Thus, the force transmitted through both springs is the same:

$F_1 = F_2 = F$

Under this common force $F$ , spring 1 stretches by $x_1$ and spring 2 stretches by $x_2$ , given by Hooke's Law:

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

The total elongation of the series combination is $x_{total} = x_1 + x_2$ . We seek an equivalent single spring with constant $k_{eff}$ such that it undergoes the same total elongation $x_{total}$ when subjected to the same force $F$ . For this equivalent spring:

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

Series combination is softer (smaller $k_{eff}$ ) than individual springs.

Springs in Parallel

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

Springs in parallel combination

$k_{eff} = k_1 + k_2$

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

If a uniform spring with constant $k$ and length $L$ is cut into $n$ equal pieces, each piece has:

$k' = nk$

Shorter pieces are stiffer! This is because stiffness is inversely proportional to length.

Example: A spring with $k = 50$ N/m is cut into 2 equal halves:

$k' = 2k = 2 \times 50 = 100 \, \text{N/m}$

Each half has $k' = 100$ N/m.

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:

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

Key Concepts

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Examples

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Introduction to Fundamentals of Force

Welcome to Fundamentals of Force. Content to be added.

Introduction to Dynamics

From Kinematics to Dynamics

The Concept of Inertia

Mass: The Measure of Inertia

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

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

What is Force?

If an agent has a tendency to push or pull an object, we say, agent is applying force on the object.

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

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

Fundamental Forces in Nature

The Four Fundamental Forces

Gravitational Force: An attractive force that exists between any two objects with mass. It's 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's much stronger than gravity and underlies most everyday forces like friction, normal force. 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

Electromagnetic Origins of Contact Forces

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

Normal Force

What is Normal Force?

Normal force applied by ground surface on block

Normal force applied by wall and ground on a person

Key Characteristics of Normal Force

Normal force is a push. A surface (like a hand) can push an object, but it cannot pull it via the normal force if contact is broken or separation occurs.

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.

Box on surface

FBD of box

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

FBD for block on horizontal surface. Upward acceleration

a = 2 \, \text{m/s}^2

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

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

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.

Block pulled at angle

Resolve the applied force:

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

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

FBD with resolved forces

Vertical ( $a_y = 0$ ):

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

Horizontal:

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

Normal force is 52.5 N (less than weight because of upward pull component!).

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

Tension Force

What is Tension?

Tension force on block pulled horizontally by sting.

Tension force on a block hanging from a string.

Key Characteristics of Tension

String becomes slack, if it attempts to push the block.

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

Ideal Strings

In many introductory physics problems, we make simplifying assumptions about strings and ropes, treating them as ideal strings:

Massless: We assume the string itself has negligible mass compared to the objects it connects.
Inextensible: We assume the string does not stretch or shrink; its length remains constant. This implies that connected objects or parts of the string moving along its length have the same speed and magnitude of acceleration.
Perfectly Flexible: We assume the string can bend easily without resistance.

Forces on a small segment of a string.

The mass of the segment is $m = 0$ . Applying Newton's Second Law along the string:

$-T_2 + T_1 + F_{along} = ma_{along} = (0)a_{along} \implies -T_2 + T_1 + F_{along} = 0$

$T_2 - T_1 = 0 \implies T_1 = T_2$

For a massless string, tension remains same throughout the string, provided no force acts on string externally along its length.

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:

Block suspended by a string.

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

FBD of the block:

FBD for block suspended by a string.

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.

Block pulled upward

FBD of pulled block

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.

Setup with hanging block

FBDs:

FBD of block on table

FBD of hanging block

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?

Spring Force and Hooke's Law

Spring Force Basics

Spring at its natural length:

x = 0

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 $x = 0$ .

Understanding the Restoring Force

Consider a coordinate system where $x = 0$ is the equilibrium position and positive $x$ is to the right:

Elongated State: Spring stretched to the right (

x > 0

). Displacement

\vec{x}

is positive. The spring exerts a restoring force

\vec{F}_{spring}

to the left (negative direction).

Compressed State: Spring compressed to the left (

x < 0

). Displacement

\vec{x}

is negative. The spring exerts a restoring force

\vec{F}_{spring}

to the right (positive direction).

Essentially, stretching the spring right ( $x > 0$ ) causes a leftward pull ( $F < 0$ ). Compressing it left ( $x < 0$ ) causes a rightward push ( $F > 0$ ).

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

Although springs can be combined in a number of possible ways, but following are are some common ways of connecting multiple springs:

Springs in Series

Two springs k1 and k2 in series, connected at junction P, stretched by external force F.

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

Since spring 2 is ideal, the force it exerts at the junction ( $F_2$ ) must equal the external force $F$ applied at its other end. Thus, the force transmitted through both springs is the same:

$F_1 = F_2 = F$

Under this common force $F$ , spring 1 stretches by $x_1$ and spring 2 stretches by $x_2$ , given by Hooke's Law:

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

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

Series combination is softer (smaller $k_{eff}$ ) than individual springs.

Springs in Parallel

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

Springs in parallel combination

$k_{eff} = k_1 + k_2$

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

If a uniform spring with constant $k$ and length $L$ is cut into $n$ equal pieces, each piece has:

$k' = nk$

Shorter pieces are stiffer! This is because stiffness is inversely proportional to length.

Example: A spring with $k = 50$ N/m is cut into 2 equal halves:

$k' = 2k = 2 \times 50 = 100 \, \text{N/m}$

Each half has $k' = 100$ N/m.

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:

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