Chapter Notes

Introduction to Laws of Motion

So far, we've learned how to describe motion using concepts like velocity and acceleration. But what causes motion? Why do objects start moving, stop, or change direction? This chapter explores the fundamental question of what governs the motion of bodies.

Our everyday experience tells us that a force—a push or a pull—is needed to change an object's state of motion.

• To move a football, you must kick it.
• To stop a rolling ball, you must apply a force against its motion.

This force is provided by an external agency, like your hand, the wind, or a river's current. Sometimes this agency is in direct contact with the object (contact forces), but not always.

• The Earth's gravity pulls a falling stone without touching it (non-contact force).
• A magnet can pull an iron nail from a distance (non-contact force).

In short, a force is required to start or stop motion. But what about an object that is already moving at a constant speed in a straight line (uniform motion)? Does it need a force to keep going? This simple question puzzled thinkers for centuries.

Aristotle's Fallacy

The ancient Greek philosopher Aristotle believed that a continuous external force is required to keep an object in motion. For example, he thought that an arrow keeps flying because the air behind it is constantly pushing it forward. This idea, known as the Aristotelian law of motion, can be summarized as:

An external force is required to keep a body in motion.

This seems logical based on everyday observations. If you stop pushing a toy car, it quickly comes to rest. All objects, when left to themselves, seem to eventually stop moving.

So, what was wrong with Aristotle's idea? The flaw is that he didn't account for the invisible force of friction.

• A toy car stops because of the friction between its wheels and the floor.
• To keep the car moving at a constant speed, the child must apply a pulling force that is exactly equal and opposite to the force of friction.

When these two forces cancel each other out, the net external force on the car is zero. The key insight is:

• Aristotle's View (Incorrect): Force is needed to maintain motion.
• Modern View (Correct): Force is needed to overcome opposing forces like friction. If there were no friction, no force would be needed to keep an object in uniform motion.

To understand the true laws of nature, we must imagine a world without friction, which is precisely what Galileo did.

The Law of Inertia

Galileo Galilei was the first to correctly describe the relationship between force and motion by studying objects on inclined planes. He observed that:

1. An object moving down an inclined plane accelerates.
2. An object moving up an inclined plane retards (decelerates).
3. Therefore, an object moving on a perfectly smooth, frictionless horizontal plane should neither accelerate nor retard. It should move with a constant velocity.

He also conducted a thought experiment with a double inclined plane. A ball released from a certain height on one plane rolls down and climbs up the other plane to nearly the same height.

• If the planes are perfectly smooth (no friction), the ball will reach the exact same height.
• If the slope of the second plane is made gentler, the ball travels a longer distance to reach the same height.
• If the second plane is made perfectly horizontal, the ball will never reach its original height and, in an ideal frictionless world, would travel forever.

This led Galileo to a revolutionary conclusion: the state of rest and the state of uniform motion in a straight line are equivalent. In both states, there is no net external force acting on the body.

This property of a body to resist a change in its state of rest or uniform motion is called inertia. The word inertia means 'resistance to change'. An object will not change its velocity unless an external force makes it do so.

Newton's First Law of Motion

Isaac Newton built upon Galileo's ideas and formulated them into his three laws of motion, which form the foundation of classical mechanics. Newton's first law is essentially a restatement of Galileo's law of inertia.

Newton's First Law of Motion states:

Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.

In simpler terms, if the net external force on a body is zero, its acceleration is zero. An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction.

• A book on a table: The book is at rest. The force of gravity pulls it down, but the table exerts an equal and opposite upward force (the normal force). Since the net force is zero, the book remains at rest.
• Standing in a bus: When a bus suddenly starts, your feet move forward with the bus, but your upper body, due to inertia, tends to stay at rest. This makes you feel thrown backward. When the bus suddenly stops, your feet stop, but your upper body continues to move forward due to inertia, throwing you forward.

Given

• The astronaut is in interstellar space, far from any gravitational sources.
• The spaceship no longer exerts a force on him.

To Find

The acceleration of the astronaut.

Concept

According to Newton's First Law, if the net external force on a body is zero, its acceleration is zero.

Solution

Since there are no nearby stars or other objects to exert a gravitational force on the astronaut, and he is no longer in contact with the accelerating spaceship, the net force acting on him is zero.

Final Answer By the first law of motion, the acceleration of the astronaut is zero.

Newton's Second Law of Motion

Newton's first law describes what happens when the net force is zero. The second law describes what happens when there is a net force. It connects force, mass, and acceleration.

Momentum

To understand the second law, we first need to define momentum. The momentum ($\mathbf{p}$) of a body is the product of its mass ($m$) and velocity ($\mathbf{v}$).

Formula: $\mathbf{p} = m\mathbf{v}$

Momentum is a vector quantity, meaning it has both magnitude and direction. It's a measure of the "quantity of motion" an object has.

• A heavy truck and a small car moving at the same speed have different momenta; the truck's is much larger. It takes a much greater force to stop the truck.
• A fast-moving bullet has a large momentum even with a small mass, making it very destructive.

Experience shows that the force required to change an object's motion depends not just on the change in momentum, but also on how quickly that change happens.

Statement of the Second Law

Newton's Second Law of Motion states:

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

Mathematically, if a force $\mathbf{F}$ acts for a time interval $\Delta t$ and causes a change in momentum $\Delta\mathbf{p}$, then: $\mathbf{F} \propto \frac{\Delta\mathbf{p}}{\Delta t}$ We can write this as an equation with a constant of proportionality, $k$: $\mathbf{F} = k \frac{\Delta\mathbf{p}}{\Delta t}$ In calculus terms, this is: $\mathbf{F} = k \frac{d\mathbf{p}}{dt}$ Since $\mathbf{p} = m\mathbf{v}$, for a body with constant mass $m$, we have: $\mathbf{F} = k \frac{d(m\mathbf{v})}{dt} = km \frac{d\mathbf{v}}{dt}$ And since the rate of change of velocity ($\frac{d\mathbf{v}}{dt}$) is acceleration ($\mathbf{a}$), the law becomes: $\mathbf{F} = km\mathbf{a}$ For simplicity, in the SI system, the constant $k$ is chosen to be 1. This gives us the famous equation:

Formula: $\mathbf{F} = m\mathbf{a}$

The SI unit of force is the newton (N). One newton is the force required to give a mass of $1 \text{ kg}$ an acceleration of $1 \text{ m s}^{-2}$. $1 \text{ N} = 1 \text{ kg m s}^{-2}$

Important Points about the Second Law

1. Consistency with the First Law: If $\mathbf{F} = 0$, then $\mathbf{a} = 0$, which is exactly what the first law states.
2. Vector Law: The equation $\mathbf{F} = m\mathbf{a}$ is a vector equation. This means it can be broken down into components for each axis: $F_x = ma_x$, $F_y = ma_y$, and $F_z = ma_z$. A force in one direction only affects the acceleration in that same direction.
3. Net External Force: The force $\mathbf{F}$ in the equation is the net external force—the vector sum of all forces acting on the body from the outside. Internal forces within the body do not count.
4. Local Relation: The acceleration of a body at a specific instant is determined by the net force acting on it at that exact instant. The body has no "memory" of past forces.

Impulse

Sometimes, a very large force acts for a very short time, like a bat hitting a baseball. This is called an impulsive force. In such cases, it's hard to measure the force and time separately, but their product is measurable.

Impulse is defined as the product of force and the time duration over which it acts. It is equal to the change in momentum of the body.

Formula: Impulse = Force $\times$ time duration = Change in momentum

Given

• Mass of bullet, $m = 0.04 \text{ kg}$
• Initial speed, $u = 90 \text{ m s}^{-1}$
• Final speed, $v = 0 \text{ m s}^{-1}$
• Distance, $s = 60 \text{ cm} = 0.6 \text{ m}$

To Find

The average resistive force, $F$.

Formula

First, find the acceleration using the equation of motion: $v^2 = u^2 + 2as$. Then, use Newton's Second Law: $F = ma$.

Solution

First, we find the acceleration 'a'. $a = \frac{v^2 - u^2}{2s} = \frac{0^2 - (90)^2}{2 \times 0.6} = \frac{-8100}{1.2} = -6750 \text{ m s}^{-2}$ The negative sign indicates retardation (deceleration).

Now, we calculate the force using the second law. $F = ma = (0.04 \text{ kg}) \times (-6750 \text{ m s}^{-2}) = -270 \text{ N}$ The resistive force exerted by the block is 270 N, opposing the motion of the bullet.

Final Answer The average resistive force is $270 \text{ N}$.

Given

• Mass of ball, $m = 0.15 \text{ kg}$
• Initial speed = $12 \text{ m s}^{-1}$. Let's take the bowler's direction as positive. So, initial velocity $v_i = -12 \text{ m s}^{-1}$.
• Final speed = $12 \text{ m s}^{-1}$. The ball is hit back, so final velocity $v_f = +12 \text{ m s}^{-1}$.

To Find

The impulse imparted to the ball.

Formula

Impulse = Change in momentum = Final momentum - Initial momentum $\text{Impulse} = m v_f - m v_i$

Solution

$\text{Impulse} = (0.15 \text{ kg})(12 \text{ m s}^{-1}) - (0.15 \text{ kg})(-12 \text{ m s}^{-1})$ $\text{Impulse} = 1.8 - (-1.8) = 1.8 + 1.8 = 3.6 \text{ N s}$ The positive sign indicates the impulse is in the direction from the batsman to the bowler.

Final Answer The impulse imparted to the ball is $3.6 \text{ N s}$.

Newton's Third Law of Motion

The second law explains how a force affects a body. But where do forces come from? Newton's third law states that forces are always the result of an interaction between two bodies. You can't have a single, isolated force.

Newton's Third Law of Motion states:

To every action, there is always an equal and opposite reaction.

This means that forces always occur in pairs. If body A exerts a force on body B (the "action"), then body B simultaneously exerts an equal and opposite force on body A (the "reaction").

Formula: $\mathbf{F}_{AB} = -\mathbf{F}_{BA}$ (The force on A by B is equal and opposite to the force on B by A).

Important Points about the Third Law

1. Action and Reaction are Forces: The terms "action" and "reaction" are just names for the pair of forces.
2. Simultaneous Action: Action and reaction happen at the exact same time. One does not cause the other.
3. Forces Act on Different Bodies: This is the most crucial point. The action force acts on one body, and the reaction force acts on the other body. Because they act on different bodies, they do not cancel each other out.

• Action: Earth pulls on the book (gravity). Reaction: The book pulls on the Earth.
• Action: The book pushes down on the table. Reaction: The table pushes up on the book (normal force).

Given

• Two identical balls, mass $m$, speed $u$.
• Case (a): Ball strikes normally (perpendicularly).
• Case (b): Ball strikes at an angle of $30^\circ$ to the normal.

To Find

(i) The direction of force on the wall in each case. (ii) The ratio of the magnitudes of impulse in case (a) to case (b).

Solution

We will find the impulse (change in momentum) on the ball first. The force on the ball is in the direction of the impulse. By Newton's third law, the force on the wall is in the opposite direction. Let's set up a coordinate system with the x-axis normal to the wall.

(i) Direction of the Force

Case (a): Normal collision

• Initial momentum ($p_x$): $mu$
• Final momentum ($p_x$): $-mu$
• Change in momentum (Impulse on ball): $\Delta p_x = p_{final} - p_{initial} = -mu - mu = -2mu$ The impulse on the ball is in the negative x-direction (away from the wall). Therefore, the force on the ball is normal to the wall, pointing away from it. By the third law, the force on the wall is normal to the wall, pointing into it (positive x-direction).

Case (b): Angled collision

• Initial x-momentum: $mu \cos 30^\circ$
• Final x-momentum: $-mu \cos 30^\circ$
• Initial y-momentum: $-mu \sin 30^\circ$
• Final y-momentum: $-mu \sin 30^\circ$ The y-component of momentum does not change. The change in momentum is only in the x-direction.
• Change in x-momentum (Impulse on ball): $\Delta p_x = -mu \cos 30^\circ - mu \cos 30^\circ = -2mu \cos 30^\circ$ Again, the impulse on the ball is in the negative x-direction. So, the force on the wall is normal to the wall, pointing into it (positive x-direction).

Answer for part (i): In both cases, the force on the wall is normal to the wall.

(ii) Ratio of Impulses

• Magnitude of impulse in case (a): $|I_a| = |-2mu| = 2mu$
• Magnitude of impulse in case (b): $|I_b| = |-2mu \cos 30^\circ| = 2mu \cos 30^\circ = 2mu (\frac{\sqrt{3}}{2}) = \sqrt{3}mu$

The ratio of the magnitudes of the impulses is: $\frac{|I_a|}{|I_b|} = \frac{2mu}{2mu \cos 30^\circ} = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} \approx 1.2$

Answer for part (ii): The ratio of the magnitudes of the impulses is $\frac{2}{\sqrt{3}}$, which is approximately 1.2.

Conservation of Momentum

A very important consequence of Newton's second and third laws is the law of conservation of momentum.

Consider an isolated system, which is a system with no net external force acting on it. Inside this system, particles can interact and exert forces on each other (internal forces).

• For example, a gun firing a bullet. The exploding gunpowder creates a force on the bullet ($\mathbf{F}$) and an equal and opposite force on the gun ($-\mathbf{F}$).
• These forces act for the same short time, $\Delta t$.
• Change in bullet's momentum = $\mathbf{F}\Delta t$.
• Change in gun's momentum = $-\mathbf{F}\Delta t$.

The total change in momentum of the (gun + bullet) system is $(\mathbf{F}\Delta t) + (-\mathbf{F}\Delta t) = 0$. This means the total momentum of the system does not change. If the gun and bullet were initially at rest (zero momentum), their total momentum after firing is also zero. The bullet's forward momentum is perfectly balanced by the gun's backward recoil momentum.

The Law of Conservation of Momentum states:

The total momentum of an isolated system of interacting particles is conserved.

This law is fundamental and applies to many situations, especially collisions. If two bodies A and B collide, the total momentum before the collision is equal to the total momentum after the collision. $\mathbf{p}_A + \mathbf{p}_B = \mathbf{p}'_A + \mathbf{p}'_B$ (Initial total momentum = Final total momentum)

Equilibrium of a Particle

In mechanics, a particle is in equilibrium when the net external force acting on it is zero. $\sum \mathbf{F} = 0$ According to the first law, this means the particle is either at rest or moving with a constant velocity (zero acceleration).

• Equilibrium with two forces: If two forces, $\mathbf{F}_1$ and $\mathbf{F}_2$, act on a particle, it is in equilibrium if $\mathbf{F}_1 = -\mathbf{F}_2$. The forces must be equal in magnitude and opposite in direction.
• Equilibrium with three forces: If three forces, $\mathbf{F}_1$, $\mathbf{F}_2$, and $\mathbf{F}_3$, act on a particle, it is in equilibrium if their vector sum is zero: $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 0$. This means the three force vectors can form a closed triangle when placed head-to-tail.

Given

• Mass, $m = 6 \text{ kg}$
• Weight, $W = mg = 6 \times 10 = 60 \text{ N}$
• Horizontal force, $F_H = 50 \text{ N}$

To Find

The angle $\theta$ the rope makes with the vertical.

Concept

The system is in equilibrium, so the net force at point P is zero. We can resolve the forces into horizontal (x) and vertical (y) components and set the sum of components in each direction to zero.

Solution

Let's analyze the forces at point P using a free-body diagram.

• The weight of the mass pulls down, so the tension in the lower rope, $T_2$, is equal to the weight: $T_2 = 60 \text{ N}$.
• The horizontal force is $F_H = 50 \text{ N}$.
• The tension in the upper rope is $T_1$, acting at an angle $\theta$ to the vertical.

For equilibrium at point P, the net forces in the x and y directions must be zero. Vertical forces ($\sum F_y = 0$): The upward vertical component of $T_1$ must balance the downward tension $T_2$. $T_1 \cos\theta = T_2$ $T_1 \cos\theta = 60 \text{ N}$

Horizontal forces ($\sum F_x = 0$): The horizontal component of $T_1$ must balance the applied horizontal force. $T_1 \sin\theta = 50 \text{ N}$

Now we have two equations. If we divide the second equation by the first: $\frac{T_1 \sin\theta}{T_1 \cos\theta} = \frac{50}{60}$ $\tan\theta = \frac{5}{6}$ $\theta = \tan^{-1}\left(\frac{5}{6}\right) \approx 40^\circ$

Final Answer The angle the rope makes with the vertical is approximately $40^\circ$.

Common Forces in Mechanics

In mechanics, we deal with several types of forces. They can be broadly classified as non-contact and contact forces.

• Gravitational Force: A non-contact force that acts between any two objects with mass. It's pervasive and governs everything from falling apples to planetary orbits.
• Contact Forces: These forces arise when objects are in physical contact. At the microscopic level, they are due to electrical forces between the atoms of the objects. Common examples include:
  • Normal Reaction: The component of the contact force perpendicular to the surfaces in contact. It prevents objects from passing through each other.
  • Friction: The component of the contact force parallel to the surfaces. It opposes relative motion.
  • Tension: The restoring force in a stretched string or rope.
  • Spring Force: The restoring force in a compressed or extended spring, given by $F = -kx$ (Hooke's Law).

Friction

Friction is the force that opposes impending (about to happen) or actual relative motion between surfaces in contact.

Static Friction

Static friction ($f_s$) is the force that prevents an object from moving when an external force is applied.

• It is a self-adjusting force. It increases as the applied force increases, always remaining equal and opposite to it.
• It has a maximum possible value, called the limiting friction, $\left(f_s\right)_{max}$.
• This maximum value is proportional to the normal reaction force ($N$).

Formula: $f_s \le \mu_s N$ where $\mu_s$ is the coefficient of static friction, a constant that depends on the nature of the two surfaces.

Kinetic Friction

Kinetic friction ($f_k$) is the force that opposes the relative motion of surfaces that are already sliding against each other.

• It is generally less than the maximum static friction. This is why it's often harder to get an object moving than to keep it moving.
• It is approximately constant and independent of the relative speed.
• It is also proportional to the normal reaction force.

Formula: $f_k = \mu_k N$ where $\mu_k$ is the coefficient of kinetic friction. Typically, $\mu_k < \mu_s$.

Rolling Friction

When an object like a wheel or a sphere rolls over a surface, the resistance it experiences is called rolling friction.

• Rolling friction is much, much smaller than static or kinetic friction.
• This is because during rolling, the surfaces get slightly deformed, but there is no large-scale sliding.
• The invention of the wheel was a major milestone because it replaced sliding friction with the much weaker rolling friction.

Given

• Coefficient of static friction, $\mu_s = 0.15$

To Find

The maximum acceleration, $a_{max}$.

Concept

The box accelerates along with the train because of the static friction force between the box and the floor. This force has a maximum limit.

Formula

The force causing acceleration is the static friction: $F = f_s = ma$. The maximum static friction is: $f_{s,max} = \mu_s N$. The normal force $N$ on a horizontal surface is equal to the weight, $mg$.

Solution

The accelerating force on the box is static friction, so $ma = f_s$. The box will remain stationary as long as this required force is less than or equal to the maximum available static friction. $ma \le f_{s,max}$ $ma \le \mu_s N$ Since $N = mg$: $ma \le \mu_s mg$ We can cancel the mass $m$ from both sides: $a \le \mu_s g$ The maximum acceleration occurs when the equality holds: $a_{max} = \mu_s g$ Using $g = 10 \text{ m s}^{-2}$: $a_{max} = 0.15 \times 10 \text{ m s}^{-2} = 1.5 \text{ m s}^{-2}$

Final Answer The maximum acceleration of the train is $1.5 \text{ m s}^{-2}$.

Circular Motion

When an object moves in a circle at a constant speed, its direction of velocity is continuously changing. A change in velocity means there is an acceleration. This acceleration is directed towards the center of the circle and is called centripetal acceleration ($a_c = v^2/R$).

According to Newton's second law, if there is an acceleration, there must be a net force causing it. The force that provides the centripetal acceleration is called the centripetal force ($f_c$). It is also directed towards the center of the circle.

Formula: $f_c = m a_c = \frac{mv^2}{R}$

• Tension in a string for a whirling stone.
• Gravity for a planet orbiting the sun.
• Friction for a car turning on a level road.

Motion of a Car on a Level Road

When a car takes a circular turn on a flat road, the centripetal force required is provided entirely by the static friction between the tires and the road.

• The vertical forces (gravity down, normal force up) are balanced: $N = mg$.
• The required centripetal force is $f_c = \frac{mv^2}{R}$.
• This force is provided by static friction, $f_s$. For the car not to slip, the required force cannot exceed the maximum static friction: $\frac{mv^2}{R} \le f_{s,max}$ $\frac{mv^2}{R} \le \mu_s N = \mu_s mg$ This gives the condition for the maximum safe speed, $v_{max}$: $v_{max} = \sqrt{\mu_s R g}$

Motion of a Car on a Banked Road

To allow for higher speeds on turns, roads are often banked (tilted inwards). On a banked road, the centripetal force is provided by two sources:

1. The horizontal component of the normal force ($N \sin\theta$).
2. The horizontal component of the friction force ($f \cos\theta$).

By analyzing the forces, we can find the maximum permissible speed on a banked road: $v_{max} = \sqrt{R g \left(\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}\right)}$ This speed is greater than on a flat road.

There is also an optimum speed ($v_o$) at which the horizontal component of the normal force alone is sufficient to provide the centripetal force, and no friction is needed. This is the ideal speed to take the turn to avoid wear and tear on the tires. $v_o = \sqrt{R g \tan\theta}$

Given

• Speed, $v = 18 \text{ km/h} = 18 \times \frac{1000}{3600} \text{ m/s} = 5 \text{ m s}^{-1}$
• Radius of turn, $R = 3 \text{ m}$
• Coefficient of static friction, $\mu_s = 0.1$
• Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$

To Find

Whether the cyclist will slip.

Formula

The condition for not slipping is that the required centripetal force must be less than or equal to the maximum static friction. This translates to: $v^2 \le \mu_s R g$

Solution

Let's calculate the maximum possible value for the square of the speed ($v^2_{max}$): $v^2_{max} = \mu_s R g = (0.1) \times (3 \text{ m}) \times (9.8 \text{ m s}^{-2}) = 2.94 \text{ m}^2\text{s}^{-2}$ Now let's find the actual square of the cyclist's speed: $v^2 = (5 \text{ m s}^{-1})^2 = 25 \text{ m}^2\text{s}^{-2}$ Comparing the two values: $25 \text{ m}^2\text{s}^{-2} > 2.94 \text{ m}^2\text{s}^{-2}$ The cyclist's speed is much higher than the maximum safe speed for this turn.

Final Answer The condition $v^2 \le \mu_s R g$ is not obeyed. The cyclist will slip while taking the circular turn.

Solving Problems in Mechanics

Solving problems in mechanics often involves multiple bodies interacting with each other. A systematic approach is crucial.

1. Draw a Diagram: Sketch the entire setup, showing all bodies, links, and supports.
2. Choose a System: Isolate a convenient part of the assembly to analyze. This is your "system." Everything else is the "environment."
3. Draw a Free-Body Diagram (FBD): This is the most important step. Draw a separate diagram showing only your chosen system. Draw arrows representing all the external forces acting on the system. Do not include forces that the system exerts on the environment.
4. Apply Newton's Laws: Apply $\sum \mathbf{F} = m\mathbf{a}$ (or $\sum \mathbf{F} = 0$ for equilibrium) to the FBD. Resolve forces into components if necessary.
5. Repeat if Necessary: If you have too many unknowns, choose another part of the assembly as a new system, draw its FBD, and apply the laws. Use Newton's third law to relate forces between the systems (if system A exerts force $\mathbf{F}$ on B, then B exerts force $-\mathbf{F}$ on A).

Given

• Mass of block, $m_b = 2 \text{ kg}$
• Mass of cylinder, $m_c = 25 \text{ kg}$
• Acceleration (after yielding), $a = 0.1 \text{ m s}^{-2}$ (downwards)
• $g = 10 \text{ m s}^{-2}$

To Find

(a) The action (force) of the block on the floor before yielding. (b) The action (force) of the system on the floor after yielding. And identify action-reaction pairs.

Solution

(a) Before the floor yields

The block is at rest (in equilibrium). Let's draw a free-body diagram for the block.

• Downward force: Weight of the block, $W_b = m_b g = 2 \times 10 = 20 \text{ N}$.
• Upward force: Normal reaction from the floor, $R$. Since the block is in equilibrium, the net force is zero. $R - W_b = 0 \implies R = W_b = 20 \text{ N}$ The force on the block by the floor is 20 N upwards (this is the reaction). By Newton's third law, the action of the block on the floor is equal and opposite.

Answer for part (a): The action of the block on the floor is $20 \text{ N}$ directed vertically downwards.

(b) After the floor yields

The system (block + cylinder) is accelerating downwards. Let's draw a free-body diagram for the combined system.

• Total mass, $M = m_b + m_c = 2 + 25 = 27 \text{ kg}$.
• Downward force: Total weight, $W_{total} = M g = 27 \times 10 = 270 \text{ N}$.
• Upward force: Normal reaction from the floor, $R'$. Now, apply Newton's second law ($\sum F = ma$). We'll take the downward direction as positive. $W_{total} - R' = M a$ $270 - R' = (27) \times (0.1) = 2.7$ $R' = 270 - 2.7 = 267.3 \text{ N}$ The force on the system by the floor is 267.3 N upwards (reaction). The action of the system on the floor is equal and opposite.

Answer for part (b): The action of the system on the floor is $267.3 \text{ N}$ directed vertically downwards.

Action-Reaction Pairs

• For (a):
  • Pair 1: Force of gravity on the block by the Earth (action) & Force of gravity on the Earth by the block (reaction).
  • Pair 2: Force on the floor by the block (action) & Force on the block by the floor (reaction).
• For (b):
  • Pair 1: Force of gravity on the system by the Earth (action) & Force of gravity on the Earth by the system (reaction).
  • Pair 2: Force on the floor by the system (action) & Force on the system by the floor (reaction).
  • Pair 3 (Internal): Force on the block by the cylinder (action) & Force on the cylinder by the block (reaction).