The story of the Laws of Motion is a fascinating journey through time, beginning with ancient philosophers and culminating in the work of Sir Isaac Newton. The earliest ideas about motion date back to philosophers like Aristotle, who believed that the natural state of objects was to be at rest and that they required a force to be kept in motion. This view held for many centuries until the Renaissance period.
In the 17th century, Galileo Galilei challenged Aristotle’s views through his experiments. He observed that objects tend to keep moving unless a force—like friction—stops them. This idea was revolutionary because it suggested that motion, not rest, could be a natural state.
Isaac Newton, building on the work of Galileo and others, formulated the three Laws of Motion. In 1687, he published the “Philosophiæ Naturalis Principia Mathematica” (Mathematical Principles of Natural Philosophy), often referred to as the “Principia.” This work laid the foundation for classical mechanics and changed our understanding of the universe.
Newton’s Laws were a turning point. They not only explained earthly motion but also celestial movements, providing a universal framework that could predict the motion of planets and stars. Today, Newton’s Laws are fundamental in various fields, from engineering to space exploration. They help us design vehicles, understand athletic performance, and even navigate spacecraft to distant planets.
Newton’s First Law (Law of Inertia)
Newton’s First Law of Motion, also known as the Law of Inertia, states that
An object at rest will remain at rest, and an object in motion will continue in motion with the same speed and in the same direction unless acted upon by an external force.
Inertia: Inertia is the property of massive objects to resist changes in their state of motion. Think of it as an object’s natural stubbornness—it wants to keep doing what it’s doing.
Rest or Uniform Motion: Whether an object is at rest or moving uniformly (at a constant speed in a straight line), it will continue in that state unless a net external force acts upon it.
Velocity and Acceleration: If an object is at rest, its velocity (v) and acceleration (a) are zero. It will stay at rest. If an object is in motion (v ≠ 0), but acceleration is zero (a = 0), it will continue moving with constant velocity and in the same direction.
In other words, objects tend to maintain their state of motion unless an external force compels them to change. This law highlights the concept of inertia, which is the tendency of an object to resist changes in its state of motion.
Let’s consider a football resting on the ground. According to the Law of Inertia, the football will remain at rest unless an external force acts upon it. This is where the action of kicking comes into play. When a player kicks the football, they are applying an external force to it. This force causes the football to move from its state of rest and fly through the air.
However, the football doesn’t continue moving indefinitely. This is because other forces are acting on the football as it moves. These forces include gravity, which pulls the football towards the Earth, and air resistance, which opposes the motion of the football. These forces eventually cause the football to slow down, stop, and fall to the ground.
So, in the context of a football game, Newton’s First Law of Motion explains why a football at rest needs a kick (an external force) to start moving, and why it eventually stops moving due to other forces like gravity and air resistance.
Example: Car and Passenger – Imagine you are sitting in a car that is at rest. According to Newton’s first law, you (and everything else inside the car) are also at rest, and you will remain at rest unless a force is applied.
When the driver starts the car and accelerates, you might feel pushed back into your seat. This sensation is due to your body’s initial state of rest. As the car accelerates, your body remains at rest due to inertia. The car seat, which is now moving forward, exerts a force on you, accelerating you forward. This force is what you perceive as being pushed into your seat.
On the other hand, if the driver suddenly applies the brakes, your body wants to keep moving forward due to its inertia. In this case, the force applied by the seatbelt or the friction between you and the seat brings you to a stop. Without these forces, you would continue moving forward even as the car slows down.
Newton’s Second Law
Newton’s second law of motion, a cornerstone of classical mechanics, provides a fundamental insight into the relationship between force, mass, and acceleration in the dynamics of physical systems.
“Newton’s second law of motion states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass.”
Mathematically, this law is expressed as:
\(\displaystyle\vec{F} = m\vec{a}\)
Where, F represents the net force acting on the object; m is the mass of the object; a is the acceleration of the object.
This formula indicates that the force applied to an object is equal to the mass of the object multiplied by its resulting acceleration. In simple terms, the more force applied to an object, the greater its acceleration will be, assuming its mass remains constant. Conversely, a larger mass requires more force to achieve the same acceleration. Newton’s second law provides a quantitative relationship between force, mass, and acceleration, forming a fundamental principle in classical mechanics.
The direction of the total acceleration vector points in the same direction as the net force vector.
Derivation
Newton’s Second Law states that the net force acting on an object is equal to the rate of change of its momentum, which is the product of its mass and velocity. Mathematically, this can be written as:
\(\displaystyle\vec{F} = \frac{d\vec{p}}{dt}\)
where \(\displaystyle\vec{F}\) is the net force vector, \(\displaystyle\vec{p}\) is the momentum vector, and $t$ is the time.
If the mass of the object is constant, then the momentum vector can be expressed as:
\(\displaystyle\vec{p} = m\vec{v}\)
where m is the mass and \(\displaystyle\vec{v}\) is the velocity vector.
Substituting this into the first equation, we get:
\(\displaystyle\vec{F} = \frac{d(m\vec{v})}{dt}\)
Using the product rule of differentiation, we can simplify this as:
\(\displaystyle\vec{F} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}\)
If the mass of the object does not change with time, then the second term on the right-hand side becomes zero. Therefore, we have:
\(\displaystyle\vec{F} = m\frac{d\vec{v}}{dt}\)
The term \(\displaystyle\frac{d\vec{v}}{dt}\) is the definition of acceleration, which is the rate of change of velocity. Hence, we can write:
\(\displaystyle\vec{F} = m\vec{a}\)
where \(\displaystyle\vec{a}\) is the acceleration vector.
This is the final equation for Newton’s Second Law of Motion for a constant mass. It shows that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. The direction of the acceleration is the same as the direction of the net force.
Newton’s Third Law
Newton’s Third Law of Motion stands as a fundamental principle in classical mechanics, revealing the reciprocal nature of forces in the physical world. Proposed by Sir Isaac Newton in the 17th century, this law articulates a crucial concept that governs the interactions between objects. At its core, Newton’s Third Law states that
For every action, there is an equal and opposite reaction.
One way to write Newton’s third law mathematically is:
\(\displaystyle\vec{F}_{AB} = -\vec{F}_{BA}\)
This equation means that the force exerted by object A on object B is equal in magnitude but opposite in direction to the force exerted by object B on object A.
Newton’s Third Law is like a secret rule in the universe. It says that for every action, there’s an equal and opposite reaction. Sounds mysterious, right? Well, it’s not so mysterious when you see it in action, especially during a rocket launch.
Imagine you have a toy rocket in your hand, and you blow up a balloon attached to it. As you let go of the balloon, air rushes out in one direction (action), and your rocket zooms off in the opposite direction (reaction). That’s Newton’s Third Law at play!
Now, picture a real rocket taking off. Instead of a balloon, rockets use powerful engines that shoot out gases at super high speeds. This creates an action force pushing the rocket upward, and in return, the rocket experiences a reaction force pushing it in the opposite direction – that’s how rockets lift off the ground.
As the rocket keeps going up, it faces another force – gravity, trying to pull it back to Earth. But guess what? Newton’s Third Law helps the rocket overcome gravity. The engine keeps pushing down (action), and the rocket keeps going up (reaction). It’s like a cosmic balancing act in space!
Even after the rocket reaches space, Newton’s Third Law is still in action. The engines can be adjusted to make the rocket go left, right, or even spin around. Every move the rocket makes has an equal and opposite reaction.
Derivation by Conservation of Momentum
Newton’s Third Law states that whenever one object exerts a force on another object, the second object exerts an equal and opposite force on the first object. Mathematically, this can be written as:
\(\displaystyle\vec{F}_{AB} = -\vec{F}_{BA}\)
where \(\displaystyle\vec{F}_{AB}\) is the force vector exerted by object A on object B, and \(\displaystyle\vec{F}_{BA}\) is the force vector exerted by object B on object A. The negative sign indicates that the forces are in opposite directions.
To derive this equation, we can use the law of conservation of momentum, which is a consequence of Newton’s Second Law of Motion. The law of conservation of momentum states that the total momentum of an isolated system of objects remains constant. Mathematically, this can be written as:
\(\displaystyle\vec{p}_{total} = \vec{p}_{1} + \vec{p}_{2} + … + \vec{p}_{n} = constant\)
where
- \(\displaystyle\vec{p}_{total}\) is the total momentum vector of the system,
- \(\displaystyle\vec{p}_{i}\) is the momentum vector of the ith object in the system.
Now, consider a system of two objects A and B that interact with each other by exerting forces on each other. The system is isolated, meaning that no external forces are acting on it. Therefore, the total momentum of the system is conserved. This means that the initial total momentum of the system is equal to the final total momentum of the system. Mathematically, this can be written as:
\(\displaystyle\vec{p}_{total(initial)} = \vec{p}_{total(final)}\)
Expanding the terms, we get:
\(\displaystyle\vec{p}_{A(initial)} + \vec{p}_{B(initial)} = \vec{p}_{A(final)} + \vec{p}_{B(final)}\)
where
- \(\displaystyle\vec{p}_{A(initial)}\) and \(\displaystyle\vec{p}_{B(initial)}\) are the initial momentum vectors of objects A and B,
- \(\displaystyle\vec{p}_{A(final)}\) and \(\displaystyle\vec{p}_{B(final)}\) are the final momentum vectors of objects A and B.
Rearranging the terms, we get:
\(\displaystyle\vec{p}_{A(final)} – \vec{p}_{A(initial)} = \vec{p}_{B(initial)} – \vec{p}_{B(final)}\)
The change in momentum of an object is equal to the impulse applied to it, which is the product of the force and the time interval. Therefore, we can write:
\(\displaystyle\vec{F}_{AB} \Delta t = \vec{p}_{A(final)} – \vec{p}_{A(initial)}\)
and
\(\displaystyle\vec{F}_{BA} \Delta t = \vec{p}_{B(final)} – \vec{p}_{B(initial)}\)
where
- \(\displaystyle\vec{F}_{AB}\) is the force exerted by object A on object B,
- \(\displaystyle\vec{F}_{BA}\) is the force exerted by object B on object A,
- \(\displaystyle\Delta t\) is the time interval of the interaction.
Substituting these expressions into the previous equation, we get:
\(\displaystyle\vec{F}_{AB} \Delta t = -\vec{F}_{BA} \Delta t\)
Dividing both sides by \(\displaystyle\Delta t\), we get:
\(\displaystyle\vec{F}_{AB} = -\vec{F}_{BA}\)
This is the equation for Newton’s Third Law of Motion for a system of two objects.
Some examples of Newton’s third law in action are:
- When you jump, you push the ground with your feet, and the ground pushes you back with the same force, making you go up.
- When a rocket launches, it pushes the exhaust gases backward, and the gases push the rocket forward with the same force, making it accelerate.
- When a bird flies, it flaps its wings downward, pushing the air downward, and the air pushes the bird upward with the same force, making it lift.
Also Read: Kepler’s Law of Planetary Motion
Solved Examples
Example 1: A hockey puck is sliding on a frictionless ice surface with a constant velocity of 5 m/s. Suddenly, an external force of 10 N is applied perpendicular to its direction of motion for 2 seconds. What will the puck’s velocity be after the force is removed?
Solution: According to Newton’s First Law, an object in motion will stay in motion with a constant velocity unless acted upon by an external force. Here, the force is applied perpendicular to the initial velocity, so we need to consider the change in velocity due to this force.
First, let’s find the acceleration due to the force:
\(\displaystyle F = ma \)
\(\displaystyle a = \frac{F}{m} \)
Let’s assume the mass of the puck ( m = 1 ) kg (as it’s not provided, we use 1 kg for simplicity):
\(\displaystyle a = \frac{10 \text{ N}}{1 \text{ kg}} = 10 \text{ m/s}^2 \)
Now, let’s find the change in velocity due to this acceleration over 2 seconds:
\(\displaystyle \Delta v = a \cdot t \)
\(\displaystyle \Delta v = 10 \text{ m/s}^2 \times 2 \text{ s} = 20 \text{ m/s} \)
Since this change in velocity is perpendicular to the initial velocity, we use vector addition to find the resultant velocity. The initial velocity in the x-direction (let’s say (vx)) is 5 m/s, and the change in velocity in the y-direction (let’s say (vy)) is 20 m/s.
The magnitude of the resultant velocity ( v ) is:
\(\displaystyle v = \sqrt{v_x^2 + v_y^2} \)
\(\displaystyle v = \sqrt{(5 \text{ m/s})^2 + (20 \text{ m/s})^2} \)
\(\displaystyle v = \sqrt{25 + 400} \)
\(\displaystyle v = \sqrt{425} \)
\(\displaystyle v \approx 20.62 \text{ m/s}\)
The puck’s velocity after the force is removed will be approximately 20.62 m/s.
Example 2: A 2 kg block is initially at rest on a frictionless horizontal surface. A horizontal force of 10 N is applied to the block. What is the velocity of the block after 5 seconds?
Solution: According to Newton’s Second Law, the acceleration ( a ) is given by:
\(\displaystyle F = ma \)
\(\displaystyle a = \frac{F}{m} \)
\(\displaystyle a = \frac{10 \text{ N}}{2 \text{ kg}} = 5 \text{ m/s}^2 \)
Now, we use the kinematic equation to find the velocity after 5 seconds:
\(\displaystyle v = u + at \)
\(\displaystyle v = 0 + (5 \text{ m/s}^2 \times 5 \text{ s}) \)
\(\displaystyle v = 25 \text{ m/s} \)
The velocity of the block after 5 seconds is 25 m/s.
Example 3: Two skaters, each of mass 50 kg, are standing on ice facing each other. They push off each other with a force of 100 N. What is the acceleration of each skater?
Solution: According to Newton’s Third Law, for every action, there is an equal and opposite reaction. When the skaters push off each other, they exert equal and opposite forces on each other.
The force exerted on each skater is 100 N. Using Newton’s Second Law:
\(\displaystyle F = ma \)
\(\displaystyle a = \frac{F}{m} \)
\(\displaystyle a = \frac{100 \text{ N}}{50 \text{ kg}} = 2 \text{ m/s}^2 \)
Each skater will accelerate at 2 m/s² in opposite directions.
Example 4: A 10 kg block is placed on a frictionless inclined plane that makes an angle of 30° with the horizontal. Calculate the acceleration of the block down the incline and the normal force acting on the block.
Solution: First, resolve the weight of the block into components parallel and perpendicular to the inclined plane. The weight of the block is ( mg ), where ( m = 10 ) kg and ( g = 9.8 ) m/s².
The parallel component of the weight:
\(\displaystyle F_{\parallel} = mg \sin \theta \)
\(\displaystyle F_{\parallel} = 10 \times 9.8 \times \sin 30^\circ \)
\(\displaystyle F_{\parallel} = 98 \times 0.5 \)
\(\displaystyle F_{\parallel} = 49 \text{ N} \)
The perpendicular component of the weight:
\(\displaystyle F_{\perp} = mg \cos \theta \)
\(\displaystyle F_{\perp} = 10 \times 9.8 \times \cos 30^\circ \)
\(\displaystyle F_{\perp} = 98 \times \frac{\sqrt{3}}{2} \)
\(\displaystyle F_{\perp} \approx 84.87 \text{ N} \)
The acceleration of the block down the incline is given by:
\(\displaystyle a = \frac{F_{\parallel}}{m} \)
\(\displaystyle a = \frac{49 \text{ N}}{10 \text{ kg}} \)
\(\displaystyle a = 4.9 \text{ m/s}^2 \)
The normal force acting on the block is ( \(\displaystyle F_{\perp} \)).
The acceleration of the block down the incline is 4.9 m/s², and the normal force acting on the block is approximately 84.87 N.
Example 5: Two masses, ( m1 = 5 ) kg and ( m2 = 3 ) kg, are connected by a light, inextensible string over a frictionless pulley. Calculate the acceleration of the system and the tension in the string.
Solution: First, let’s set up the equations of motion for both masses.
For ( m1 ) (heavier mass moving down):
\(\displaystyle m_1 g – T = m_1 a \)
\(\displaystyle 5g – T = 5a \)
For ( m2 ) (lighter mass moving up):
\(\displaystyle T – m_2 g = m_2 a \)
\(\displaystyle T – 3g = 3a \)
Adding these two equations to eliminate ( T ):
\(\displaystyle 5g – T + T – 3g = 5a + 3a \)
\(\displaystyle 2g = 8a \)
\(\displaystyle a = \frac{2g}{8} = \frac{g}{4} = \frac{9.8}{4} = 2.45 \text{ m/s}^2 \)
Now, substitute ( a ) back into one of the original equations to find ( T ):
\(\displaystyle T – 3g = 3a \)
\(\displaystyle T – 3 \times 9.8 = 3 \times 2.45\)
\(\displaystyle T – 29.4 = 7.35 \)
\(\displaystyle T = 36.75 \text{ N} \)
The acceleration of the system is 2.45 m/s², and the tension in the string is 36.75 N.
FAQs
What does Newton’s First Law imply about objects in space?
Newton’s First Law states that an object at rest will stay at rest and an object in motion will stay in motion at a constant velocity unless acted upon by an external force. This implies that in the vacuum of space, where there is very little to no friction or resistance, an object can keep moving indefinitely in the same direction and at the same speed unless it encounters another force.
How does Newton’s Second Law explain the acceleration of a car?
Newton’s Second Law explains that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. For a car, this means that the greater the force applied by the engine (via the wheels on the road), the greater the car’s acceleration, assuming the mass of the car remains constant.
Can you provide an example of Newton’s Third Law in everyday life?
A common example of Newton’s Third Law is the action of walking. When you push backward on the ground with your feet, the ground pushes forward with an equal and opposite force, propelling you forward. This is the essence of “for every action, there is an equal and opposite reaction.”
Why don’t we feel the effects of inertia when sitting still?
Inertia is the tendency of an object to resist changes in its state of motion. When you are sitting still, you do not feel the effects of inertia because there are no forces acting on you to change your state of rest. You only feel inertia when there is a change in motion, such as when a car suddenly stops or accelerates.
How does the concept of inertia apply to safety in vehicles?
The concept of inertia is critical in vehicle safety. Seat belts and airbags are designed to counteract the effects of inertia during a sudden stop or collision. Without these safety devices, passengers would continue moving forward at the vehicle’s previous speed due to their inertia, potentially causing injury.
In what way does Newton’s Second Law relate to the force experienced in a collision?
Newton’s Second Law relates to the force experienced in a collision through the relationship between force, mass, and acceleration (or deceleration). In a collision, a rapid deceleration occurs, and the force experienced by the objects involved is dependent on their masses and the rate of deceleration. This is why more massive vehicles or those with stronger structures can influence the severity of the collision forces.
How can Newton’s Third Law be observed in the use of rockets?
Newton’s Third Law is fundamental in rocketry. When a rocket expels gas out of its engines, the action of expelling the gas produces an equal and opposite reaction that propels the rocket forward. This principle allows rockets to launch into space and maneuver in the vacuum of space where there is no atmosphere to push against.