Introduction

While previous chapters described motion (Kinematics), this chapter delves into the causes of motion, specifically the role of force.

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Common experience suggests that an external force is required to:

• Start a stationary object moving.
• Stop a moving object.
• Retard motion (slow down an object).

This external force can be applied through direct contact (pushing, pulling, friction) or act from a distance (gravity, magnetic force).

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A fundamental question arises: Is an external force necessary to maintain a body in uniform motion (constant velocity)?

Aristotle’s Fallacy

Ancient Greek philosopher Aristotle believed that an external force was required to keep a body in motion. He thought, for instance, that air pushed an arrow to keep it flying.

This view aligns with everyday observations, where objects in motion (like a toy car) tend to slow down and stop unless a force is continuously applied.

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Aristotle's conclusion was based on observing motion in the real world, where opposing forces like friction are always present. He failed to recognize that these opposing forces are what cause the motion to stop, not that a continuous force is needed for motion itself.

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The error in Aristotle's view is that the external force applied in everyday situations is actually needed to counteract the natural opposing forces (like friction), rather than to maintain the motion itself.

If friction and other opposing forces were absent, an object in motion would continue to move uniformly without any applied force.

The Law Of Inertia

Galileo Galilei challenged Aristotle's ideas through experiments, particularly by studying the motion of objects on inclined planes.

• Objects rolling down an inclined plane accelerate.
• Objects rolling up an inclined plane retard (decelerate).

Galileo reasoned that motion on a horizontal plane is an intermediate case. On an ideal, frictionless horizontal plane, an object should experience neither acceleration nor retardation, meaning it should move with a constant velocity.

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Using a double inclined plane, Galileo observed that a ball rolling down one plane and up another reached nearly the same height, regardless of the slope of the second plane.

He extrapolated that if the second plane were perfectly horizontal (zero slope), the ball, aiming to reach its original height, would continue to move indefinitely with constant velocity in the absence of friction.

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This led Galileo to the concept of Inertia:

• The state of rest and the state of uniform linear motion (constant velocity) are fundamentally equivalent states of motion.
• No net external force is required to maintain a state of uniform motion.
• Inertia is the property of a body that resists changes in its state of motion (either rest or uniform velocity).

A body will maintain its state of rest or uniform motion unless an external force acts upon it to change that state.

Newton’s First Law Of Motion

Building upon Galileo's concept of inertia, Isaac Newton formulated his first law of motion, which essentially restates the law of inertia:

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.

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In simpler terms, Newton's First Law can be stated as:

If the net external force on a body is zero, its acceleration is zero. Conversely, if the acceleration of a body is zero, the net external force on it must be zero.

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The First Law helps in analyzing situations in two ways:

1. If we know the net external force on an object is zero, we can conclude its acceleration is zero (it is either at rest or moving at constant velocity). Example: A spaceship in deep space, far from gravitational influences, with engines off, will move at constant velocity if it was already moving, or remain at rest if it was initially at rest.
2. More commonly, if we observe that an object is not accelerating (it's at rest or in uniform linear motion), we can infer that the net external force acting on it must be zero. Example: A book resting on a table is subject to gravity (weight) pulling it down. Since it is at rest (zero acceleration), the table must exert an equal and opposite upward force (normal force) on the book, resulting in zero net force.

The First Law highlights that acceleration is caused by a net external force. Internal forces within a system do not cause the system as a whole to accelerate.

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Inertia is responsible for phenomena like being thrown backward when a bus starts suddenly or forward when it stops suddenly. Our body resists the change in motion due to inertia, while friction between our feet and the bus floor attempts to change our state of motion along with the bus.

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Answer:

Newton’s Second Law Of Motion

Newton's Second Law addresses situations where a net external force acts on a body, resulting in non-zero acceleration. It quantifies the relationship between force and acceleration.

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The Second Law is based on the concept of momentum.

Momentum ($\mathbf{p}$) of a body is defined as the product of its mass ($m$) and velocity ($\mathbf{v}$):

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

Momentum is a vector quantity, having the same direction as the velocity. Its magnitude is $|\mathbf{p}| = mv$.

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Observations related to force and momentum:

• Greater force is needed to change the speed of a more massive object compared to a lighter one in the same time.
• Greater force is needed to stop an object moving at high speed compared to one moving at low speed in the same time (for the same mass).
• Applying the same force for the same duration results in the same change in momentum for objects of different masses (they acquire different speeds, but the product $mv$ changes by the same amount).
• A force is required to change the direction of momentum, even if the magnitude of momentum (speed) is constant (e.g., circular motion).
• The faster the change in momentum is brought about (shorter time interval), the greater the force required (e.g., catching a ball).

These observations suggest that force is related to the rate of change of momentum.

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Newton's Second Law of Motion (Formal Statement):

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

Mathematically, if $\mathbf{F}$ is the net external force on a body and $\mathbf{p}$ is its momentum, the Second Law is:

$$ \mathbf{F} \propto \frac{d\mathbf{p}}{dt} $$

Introducing a constant of proportionality $k$: $$ \mathbf{F} = k \frac{d\mathbf{p}}{dt} $$

For a body of constant mass $m$, $\frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt} = m \frac{d\mathbf{v}}{dt} = m\mathbf{a}$.

So, $\mathbf{F} = km\mathbf{a}$.

The unit of force is defined such that $k=1$ in the SI system. Thus, the Second Law becomes:

$$ \mathbf{F} = \frac{d\mathbf{p}}{dt} = m\mathbf{a} \quad (\text{for constant mass}) $$

The SI unit of force, the newton (N), is defined using this equation: 1 newton is the force that produces an acceleration of 1 m/s$^2$ in a mass of 1 kg ($1 \text{ N} = 1 \text{ kg} \cdot \text{m/s}^2$).

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Important Points about the Second Law:

1. Consistency with First Law: If $\mathbf{F} = \mathbf{0}$, then $m\mathbf{a} = \mathbf{0}$. For $m \ne 0$, this implies $\mathbf{a} = \mathbf{0}$. Thus, zero net force means zero acceleration, which is consistent with the First Law.
2. Vector Law: The Second Law is a vector equation. It can be broken down into three scalar equations for the components of force and acceleration along perpendicular axes: $F_x = ma_x$, $F_y = ma_y$, $F_z = ma_z$. This means force in one direction affects acceleration only in that direction.
3. Applicable to Systems: The law applies to a single particle. It also applies to a rigid body or a system of particles if $\mathbf{F}$ is the *total external force* acting on the system, and $\mathbf{a}$ is the acceleration of the system's *centre of mass*. Internal forces within the system do not contribute to the acceleration of the centre of mass.
4. Local Relation: The force acting on a particle at a specific location and time determines its acceleration at that same location and time. Acceleration is not determined by the particle's past motion or forces from earlier times. (e.g., Dropping a stone from an accelerating train – the stone's acceleration immediately after release is only due to gravity, not the train's prior acceleration).

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Impulse:

• When a large force acts for a very short duration, the product of force and time is called impulse.
• From the Second Law, $\mathbf{F} = \frac{d\mathbf{p}}{dt}$. Over a small time interval $\Delta t$, $\mathbf{F} \approx \frac{\Delta \mathbf{p}}{\Delta t}$, so $\mathbf{F} \Delta t \approx \Delta \mathbf{p}$.
• Impulse is formally defined as the integral of force over the time interval, but for a constant force or average force over a short time, it's approximately: $$ \text{Impulse} = \mathbf{F}_{avg} \Delta t = \Delta \mathbf{p} = \mathbf{p}_{final} - \mathbf{p}_{initial} $$
• Impulse is equal to the change in momentum of the body.
• The concept of impulse is useful when the force and time interval of a collision or impact are difficult to measure individually, but the change in momentum is known or can be calculated.

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Answer:

Newton’s Third Law Of Motion

Newton's Third Law describes the fundamental nature of forces as interactions between pairs of bodies.

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Newton's Third Law of Motion (Formal Statement):

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

A clearer way to state it:

Forces always occur in pairs. When one body (A) exerts a force on another body (B), the second body (B) simultaneously exerts a force on the first body (A) that is equal in magnitude and opposite in direction.

If $\mathbf{F}_{AB}$ is the force on body A due to body B, and $\mathbf{F}_{BA}$ is the force on body B due to body A, then:

$$ \mathbf{F}_{AB} = - \mathbf{F}_{BA} $$

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Important Points about the Third Law:

1. Action-Reaction Pairs: The terms "action" and "reaction" refer to the two forces in the pair. Either force can be called "action", and the other is then the "reaction". There is no cause-effect relationship or time delay; the forces are simultaneous.
2. Act on Different Bodies: An action force and its corresponding reaction force always act on different bodies. It is a common mistake to think they act on the same body. Because they act on different bodies, they cannot cancel each other out when considering the motion of a single body.
3. Internal Forces Cancel: When considering a system of two bodies as a whole, the action-reaction forces between them are internal forces within the system. These internal forces cancel out in pairs, meaning they do not contribute to the net external force on the system or the acceleration of the system's centre of mass.
4. Universality: The Third Law applies to all types of forces (contact forces, non-contact forces like gravity or electrical forces). The Earth pulls the Moon, and the Moon pulls the Earth with an equal and opposite force.

The statement "forces always occur in pairs" is the most fundamental takeaway from the Third Law.

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Answer:

Conservation Of Momentum

A crucial consequence derived from Newton's Second and Third Laws is the Law of Conservation of Momentum.

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Consider an isolated system of particles (a system on which no net external force acts). Although particles within the system may exert forces on each other (internal forces), the Third Law dictates that these internal forces occur in equal and opposite pairs.

According to the Second Law, the rate of change of momentum of a particle is equal to the net force on it. For a pair of interacting particles (A and B) within an isolated system, the force on A by B ($\mathbf{F}_{AB}$) and the force on B by A ($\mathbf{F}_{BA}$) are an action-reaction pair: $\mathbf{F}_{AB} = - \mathbf{F}_{BA}$.

$$ \frac{d\mathbf{p}_A}{dt} = \mathbf{F}_{AB} \quad \text{and} \quad \frac{d\mathbf{p}_B}{dt} = \mathbf{F}_{BA} $$

Summing the rates of change of momentum for A and B:

$$ \frac{d\mathbf{p}_A}{dt} + \frac{d\mathbf{p}_B}{dt} = \mathbf{F}_{AB} + \mathbf{F}_{BA} = \mathbf{F}_{AB} - \mathbf{F}_{AB} = \mathbf{0} $$ $$ \frac{d(\mathbf{p}_A + \mathbf{p}_B)}{dt} = \mathbf{0} $$

This means that the total momentum of the pair ($\mathbf{p}_A + \mathbf{p}_B$) remains constant over time.

Extending this to a system of many particles, where the total external force is zero, the sum of all internal forces is zero due to the Third Law. Therefore, the rate of change of the total momentum of the system is zero.

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Law of Conservation of Momentum:

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

This means the total momentum of the system remains constant over time if no external forces act on it. $\mathbf{P}_{total} = \sum_i \mathbf{p}_i = \text{constant}$ if $\mathbf{F}_{external, total} = \mathbf{0}$.

This law is particularly useful in analyzing collisions and explosions where external forces may be negligible compared to internal forces during the event.

For a collision between two bodies A and B in an isolated system, the total momentum before collision equals the total momentum after collision:

$$ \mathbf{p}_{A, initial} + \mathbf{p}_{B, initial} = \mathbf{p}_{A, final} + \mathbf{p}_{B, final} $$

This principle holds regardless of whether the collision is elastic or inelastic.

Equilibrium Of A Particle

In mechanics, a particle is said to be in equilibrium when the net external force acting on it is zero.

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According to Newton's First Law, if the net external force on a particle is zero, its acceleration is zero. This means a particle in equilibrium is either:

• At rest ($\mathbf{v} = \mathbf{0}$ and $\mathbf{a} = \mathbf{0}$).
• In uniform motion ($\mathbf{v} = \text{constant}$ and $\mathbf{a} = \mathbf{0}$).

Often, the term "equilibrium" is used to refer to static equilibrium (being at rest).

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If multiple forces $\mathbf{F}_1, \mathbf{F}_2, \dots, \mathbf{F}_n$ act on a particle, the condition for equilibrium is that the vector sum of all these forces is zero:

$$ \sum \mathbf{F}_i = \mathbf{F}_1 + \mathbf{F}_2 + \dots + \mathbf{F}_n = \mathbf{0} $$

Geometrically, if the forces acting on a particle in equilibrium are represented as vectors placed head-to-tail, they form a closed polygon. For three concurrent forces in equilibrium, they form a closed triangle.

In terms of components, the sum of the components of all forces along each of the coordinate axes must be zero:

$$ \sum F_x = 0 \implies F_{1x} + F_{2x} + \dots + F_{nx} = 0 $$ $$ \sum F_y = 0 \implies F_{1y} + F_{2y} + \dots + F_{ny} = 0 $$ $$ \sum F_z = 0 \implies F_{1z} + F_{2z} + \dots + F_{nz} = 0 $$

These conditions allow us to solve for unknown forces or angles in equilibrium problems.

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Answer:

Common Forces In Mechanics

Mechanics problems involve various types of forces. Besides the universal gravitational force, most commonly encountered forces are contact forces.

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Gravitational Force: Acts between any two objects with mass. On Earth, it is the force exerted by the Earth on an object, commonly called the object's weight (W). Weight is given by $W = mg$, where $m$ is the mass and $g$ is the acceleration due to gravity. Gravitational force acts at a distance.

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Contact Forces: These forces arise when objects are in direct contact with each other.

• Normal Reaction (Normal Force, N): The component of the contact force that is perpendicular (normal) to the surfaces in contact. It is a reaction force exerted by a surface to prevent an object from passing through it.
• Friction (Frictional Force, f): The component of the contact force that is parallel to the surfaces in contact. It opposes relative motion or impending relative motion between the surfaces.
• Tension (T): The restoring force in a stretched or compressed string, rope, or cable. For an ideal (massless, inextensible) string, the tension is uniform throughout the string.
• Spring Force: The restoring force exerted by a spring when it is stretched or compressed. For small displacements, it is proportional to the displacement from equilibrium (Hooke's Law: $\mathbf{F} = -k\mathbf{x}$, where $k$ is the spring constant and $\mathbf{x}$ is the displacement).
• Viscous Force / Air Resistance: Opposing force exerted by fluids (liquids or gases) on objects moving through them. These are also contact forces acting on the surface of the object.
• Buoyant Force: The upward force exerted by a fluid on an object immersed in it, equal to the weight of the fluid displaced by the object.

All these contact forces fundamentally arise from the electric forces between the atoms and molecules of the interacting bodies, but they are treated phenomenologically in macroscopic mechanics due to the complexity of their microscopic origins.

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Friction

Friction is a contact force that opposes relative motion or impending relative motion between two surfaces in contact.

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There are two main types of friction between solid surfaces:

1. Static Friction ($f_s$):
2. Kinetic Friction ($f_k$) (also called Sliding Friction):

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Static Friction ($f_s$):

• This force opposes the impending relative motion between surfaces that are at rest relative to each other. "Impending motion" refers to the motion that *would* happen if friction were absent, under the influence of applied forces.
• Static friction is a self-adjusting force. It arises only when an external force attempts to move the object, and its magnitude is equal and opposite to the net applied force parallel to the surface, up to a certain maximum limit.
• The maximum value of static friction ($f_{s, max}$) is found experimentally to be approximately proportional to the normal force ($N$) between the surfaces: $$ f_{s, max} = \mu_s N $$ where $\mu_s$ is the coefficient of static friction, a dimensionless constant that depends on the nature of the surfaces in contact.
• The static friction force satisfies the inequality: $$ f_s \le \mu_s N $$
• If the applied force parallel to the surface is less than or equal to $f_{s, max}$, static friction balances it, and the object remains at rest.
• If the applied force exceeds $f_{s, max}$, the object begins to move.

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Kinetic Friction ($f_k$):

• This force opposes the actual relative motion between surfaces that are sliding or rubbing against each other.
• When relative motion starts, the frictional force decreases from the maximum static value and becomes kinetic friction.
• Kinetic friction is also approximately proportional to the normal force: $$ f_k = \mu_k N $$ where $\mu_k$ is the coefficient of kinetic friction, another dimensionless constant depending on the surfaces.
• Experimentally, $\mu_k$ is usually found to be less than $\mu_s$. This is why it takes more force to start an object moving than to keep it moving at a constant velocity against friction.
• Kinetic friction is nearly independent of the relative velocity between the surfaces and the area of contact.

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Rolling Friction:

• When a body rolls without slipping, ideally there is no relative motion at the single point of contact, so static or kinetic friction should be zero.
• In practice, some resistance to rolling motion exists (rolling friction). This is because the surfaces deform slightly at the contact point, creating a small contact area and resulting in a net force opposing the rolling motion.
• For the same weight, rolling friction is significantly smaller (by orders of magnitude) than static or sliding friction. This makes rolling much easier than sliding, explaining the importance of wheels.

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Importance of Friction:

• Friction is often seen as undesirable (causes wear and tear, dissipates energy as heat in machines). Lubricants and ball bearings are used to reduce kinetic and rolling friction.
• However, friction is essential in many everyday situations: walking (static friction between shoes and ground), acceleration/braking of vehicles (static friction between tires and road), holding objects. Kinetic friction is used in braking systems.

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Circular Motion

As discussed in Chapter 4, an object moving in a circular path undergoes acceleration even if its speed is constant. This acceleration is directed towards the center and is called centripetal acceleration ($a_c$), with magnitude $a_c = v^2/R$, where $v$ is the speed and $R$ is the radius of the circle.

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According to Newton's Second Law ($\mathbf{F} = m\mathbf{a}$), there must be a net force causing this acceleration. This force, directed towards the center of the circle, is called the centripetal force ($\mathbf{f}_c$).

The magnitude of the centripetal force is:

$$ f_c = m a_c = m \frac{v^2}{R} $$

The centripetal force is not a new type of force. It is the name given to the *net* force that provides the necessary inward acceleration. This inward force must be provided by some physical interaction, such as:

• Tension in a string (for a stone whirled in a circle).
• Gravitational force (for a planet orbiting a star).
• Friction (for a car turning on a flat road).
• Normal force or a component of it (for a car on a banked road).

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Circular Motion of a Car:

Let's analyze the forces providing the centripetal force for a car taking a circular turn.

(a) On a Level Road:

Three forces act on the car:

• Weight ($mg$) downwards.
• Normal reaction ($N$) upwards from the road surface.
• Frictional force ($f$) parallel to the road surface.

In the vertical direction, there is no acceleration, so $N - mg = 0 \implies N = mg$.

The centripetal force required for the circular motion is provided by the frictional force ($f$). The direction of this friction is towards the center of the circle (it's static friction opposing the tendency of the car to slide outwards).

$$ f = \frac{mv^2}{R} $$

For the car not to skid, the required static friction must be less than or equal to the maximum possible static friction ($f_s \le \mu_s N$).

$$ \frac{mv^2}{R} \le \mu_s N $$

Substituting $N=mg$:

$$ \frac{mv^2}{R} \le \mu_s mg $$

Assuming $m \ne 0$, divide by $m$:

$$ \frac{v^2}{R} \le \mu_s g $$ $$ v^2 \le \mu_s R g $$

The maximum speed ($v_{max}$) for a car to take a circular turn on a level road without skidding is:

$$ \mathbf{v_{max} = \sqrt{\mu_s R g}} $$

This shows that the maximum safe speed depends on the coefficient of static friction, the radius of the turn, and gravity, but not on the mass of the car.

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(b) On a Banked Road:

Banking a road means tilting it inwards at an angle $\theta$. This helps the normal force to also contribute a component towards the center, reducing the reliance on friction.

Forces on the car on a banked road:

• Weight ($mg$) downwards.
• Normal reaction ($N$) perpendicular to the banked surface.
• Frictional force ($f$) parallel to the banked surface (can be up or down the incline depending on speed).

Resolve forces into horizontal (towards center) and vertical components.

Vertical equilibrium (no vertical acceleration): $N \cos \theta = mg + f \sin \theta$.

Horizontal force (provides centripetal force): $N \sin \theta + f \cos \theta = \frac{mv^2}{R}$.

To find the maximum safe speed ($v_{max}$), friction is at its maximum static value ($f = \mu_s N$) and is directed down the incline (opposing the tendency to slide upwards). Equations become:

$$ N \cos \theta = mg + \mu_s N \sin \theta $$ $$ N \sin \theta + \mu_s N \cos \theta = \frac{mv_{max}^2}{R} $$

Solving these two equations for $v_{max}$ yields:

$$ \mathbf{v_{max} = \sqrt{Rg \left(\frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta}\right)}} $$

This speed is higher than that on a level road (where $\theta=0$, $v_{max} = \sqrt{Rg \mu_s}$).

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Optimum Speed ($v_0$):

On a banked road, there is an optimum speed where the horizontal component of the normal force ($N \sin \theta$) alone provides the necessary centripetal force, so no friction is required ($f=0$).

Vertical equilibrium: $N \cos \theta = mg$.

Horizontal force: $N \sin \theta = \frac{mv_0^2}{R}$.

Divide the second equation by the first:

$$ \frac{N \sin \theta}{N \cos \theta} = \frac{mv_0^2/R}{mg} $$ $$ \tan \theta = \frac{v_0^2}{Rg} $$ $$ \mathbf{v_0 = \sqrt{Rg \tan \theta}} $$

Driving at the optimum speed minimizes wear and tear on the tires as friction is zero.

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Answer:

Given radius of racetrack $R = 300$ m. Angle of banking $\theta = 15^\circ$. Coefficient of friction $\mu_s = 0.2$. Take $g = 9.8$ m/s$^2$.

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(a) Optimum speed ($v_0$) to avoid wear and tear:

The optimum speed is the speed at which no friction is required, and the horizontal component of the normal force provides the entire centripetal force. It is given by the formula $v_0 = \sqrt{Rg \tan \theta}$.

$$ v_0 = \sqrt{(300 \text{ m})(9.8 \text{ m/s}^2)(\tan 15^\circ)} $$

Calculate $\tan 15^\circ \approx 0.2679$.

$$ v_0 = \sqrt{300 \times 9.8 \times 0.2679} = \sqrt{2940 \times 0.2679} = \sqrt{787.566} $$ $$ v_0 \approx 28.06 \text{ m/s} $$

Rounding to three significant figures (based on R and g), the optimum speed is approximately 28.1 m/s.

The optimum speed of the racecar is approximately 28.1 m/s.

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(b) Maximum permissible speed ($v_{max}$) to avoid slipping:

The maximum speed is when the static friction force is at its maximum value $\mu_s N$ and is directed down the incline (to prevent the car from sliding up). It is given by the formula $v_{max} = \sqrt{Rg \left(\frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta}\right)}$.

We have $R=300$ m, $g=9.8$ m/s$^2$, $\theta=15^\circ$ ($\tan 15^\circ \approx 0.2679$), and $\mu_s = 0.2$.

$$ v_{max} = \sqrt{(300)(9.8) \left(\frac{0.2679 + 0.2}{1 - (0.2)(0.2679)}\right)} $$ $$ v_{max} = \sqrt{2940 \left(\frac{0.4679}{1 - 0.05358}\right)} = \sqrt{2940 \left(\frac{0.4679}{0.94642}\right)} $$ $$ v_{max} = \sqrt{2940 \times 0.49442} = \sqrt{1453.96} $$ $$ v_{max} \approx 38.13 \text{ m/s} $$

Rounding to three significant figures, the maximum permissible speed is approximately 38.1 m/s.

The maximum permissible speed is approximately 38.1 m/s.

Solving Problems In Mechanics

Solving problems in mechanics using Newton's Laws often involves identifying all the forces acting on the object or system of interest and applying $\mathbf{F}_{net} = m\mathbf{a}$. A systematic approach is helpful.

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Recommended steps for solving mechanics problems:

1. Draw a schematic diagram: Sketch the physical situation showing the relevant objects, supports, links, etc.
2. Choose a system: Select a specific object or group of objects as the system you will analyze. This is the body (or system) to which you will apply Newton's laws.
3. Draw a free-body diagram (FBD) for the chosen system: This is a diagram showing *only* the system and *all* the external forces acting *on* it. Do not show forces exerted by the system on other objects. Represent each force with an arrow indicating its direction and label it (e.g., weight $mg$, normal force $N$, tension $T$, friction $f$, applied force $F$).
4. Identify knowns and unknowns: In the FBD, note the forces whose magnitudes and directions are given or are clearly determined (like weight, direction of tension). Treat other forces (magnitudes and/or directions) as unknowns.
5. Apply Newton's Laws:
   • If the system is in equilibrium ($\mathbf{a}=\mathbf{0}$), apply the equilibrium condition $\sum \mathbf{F} = \mathbf{0}$, which translates to $\sum F_x = 0, \sum F_y = 0, \sum F_z = 0$ by resolving forces into components along chosen axes.
   • If the system is accelerating ($\mathbf{a} \ne \mathbf{0}$), apply Newton's Second Law $\sum \mathbf{F} = m\mathbf{a}$ (or $\sum F_x = ma_x, \sum F_y = ma_y, \sum F_z = ma_z$).
6. Repeat for other systems (if needed): If the problem involves multiple interacting bodies, draw FBDs for other relevant objects or subsystems and apply Newton's Laws to them. When moving between FBDs of interacting objects, use Newton's Third Law to relate the forces in action-reaction pairs ($\mathbf{F}_{on A by B} = - \mathbf{F}_{on B by A}$).
7. Solve the equations: You will typically get a set of equations from applying Newton's laws to the chosen systems. Solve these equations simultaneously to find the unknown forces, accelerations, or other quantities.

Drawing clear and correct free-body diagrams is often the most critical step in successfully solving mechanics problems.

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Answer:

Given mass of block $m_B = 2$ kg, mass of cylinder $m_C = 25$ kg. Total mass of the system (block + cylinder) $M = m_B + m_C = 2 + 25 = 27$ kg. Take $g = 10$ m/s$^2$. Weight of block $W_B = m_B g = 2 \times 10 = 20$ N. Weight of cylinder $W_C = m_C g = 25 \times 10 = 250$ N. Total weight $W_{total} = Mg = 27 \times 10 = 270$ N.

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(a) Action of the block on the floor before the floor yields:

Before the floor yields, the block (with the cylinder on top, as the problem phrasing implies the cylinder is placed on top before yielding starts) is at rest. Let's assume part (a) refers to the block + cylinder system at rest on the floor, right before it yields.

System: Block + Cylinder. Forces on the system: Total weight $W_{total} = 270$ N downwards, Normal force $N$ from the floor upwards.

Since the system is at rest, acceleration $\mathbf{a} = \mathbf{0}$. By Newton's First Law, net force is zero:

$$ N - W_{total} = 0 \implies N = W_{total} = 270 \text{ N} $$

This is the normal force exerted by the floor on the system. By Newton's Third Law, the action of the system (block + cylinder) on the floor is equal in magnitude and opposite in direction to the force on the system by the floor. So, the force exerted by the system on the floor is 270 N downwards.

If the question intended part (a) to be the block alone on the floor (without the cylinder), then $N = W_B = 20$ N, and the action of the block on the floor would be 20 N downwards.

Given the context of the next part, it's more likely part (a) refers to the block + cylinder system before moving. Let's assume this interpretation.

Action of the system (block + cylinder) on the floor before yielding is 270 N vertically downwards.

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(b) Action of the block on the floor after the floor yields:

After the floor yields, the system (block + cylinder) moves downwards with acceleration $a = 0.1$ m/s$^2$. Take downwards as the positive direction for acceleration.

System: Block + Cylinder. Forces on the system: Total weight $W_{total} = 270$ N downwards, Normal force $N'$ from the floor upwards.

Applying Newton's Second Law ($\sum F = ma$):

$$ W_{total} - N' = M a $$ $$ 270 \text{ N} - N' = (27 \text{ kg})(0.1 \text{ m/s}^2) $$ $$ 270 - N' = 2.7 \text{ N} $$ $$ N' = 270 - 2.7 = 267.3 \text{ N} $$

This is the normal force exerted by the floor on the system. By Newton's Third Law, the action of the system (block + cylinder) on the floor is equal in magnitude and opposite in direction to the force on the system by the floor. So, the force exerted by the system on the floor is 267.3 N downwards.

Action of the system (block + cylinder) on the floor after yielding is 267.3 N vertically downwards.

Note: The action of the *block* on the floor requires considering the forces within the block-cylinder interaction, which wasn't explicitly asked but is implied if 'block on the floor' means force from the block *only*. However, the standard interpretation is the force from the combination touching the floor. Let's stick to the standard interpretation unless further clarity is given.

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Identify the action-reaction pairs in the problem:

Action-reaction pairs always involve mutual forces between two distinct bodies.

Here are some action-reaction pairs:

1. Force of gravity on the system (Block + Cylinder) by the Earth (270 N downwards) $\leftrightarrow$ Force of gravity on the Earth by the system (270 N upwards).
2. Force on the floor by the system (270 N downwards in part a, 267.3 N downwards in part b) $\leftrightarrow$ Force on the system by the floor (Normal force $N$ or $N'$, upwards).
3. Force on the cylinder by the block (contact force) $\leftrightarrow$ Force on the block by the cylinder (equal magnitude, opposite direction - these are internal forces to the system).
4. Force of gravity on the block by the Earth (20 N downwards) $\leftrightarrow$ Force of gravity on the Earth by the block (20 N upwards).
5. Force of gravity on the cylinder by the Earth (250 N downwards) $\leftrightarrow$ Force of gravity on the Earth by the cylinder (250 N upwards).

It is important to remember that the weight of the block ($mg$) and the normal force ($N$) on the block by the floor are NOT an action-reaction pair. They act on the same body (the block or system), and their equality depends on the state of motion (equilibrium or acceleration), not on the Third Law.