Newton’s law of Motions

Newton's laws of motion

Motion occurs when a body covers distance with time. The quantity of motion is the product of mass of a body and the speed at which it moves. Momentum represents the quantity of motion, and it is the time rate of change of momentum that determines which keeps the body moving. Through ages, man has observed motions of various bodies both in space and on smooth and rough planes. It was Isaac Newton who formulated the three laws we now call Newton's laws of motion. By applying these laws with certain conditions, motion problems can be solved.

First law of motion

Statement: "A body remains in a state of rest or uniform motion in a straight line unless acted upon by an external force."

This law is also referred to as the law of inertia. Inertia means reluctance of a body to be set into motion or to stop if already moving. A body with less mass has small inertia and vice versa. Mass is the measure of inertia of a given body. The greater the mass of the body, the less the acceleration when an external force is applied.

Formula: No specific equation, but can be stated as:

$\sum F = 0 \quad \text{(if object is at rest or moving with constant velocity)}$

This implies that if the net external force acting on a body is zero, the body remains at rest or continues with constant velocity.

Newton's second law of motion

The law states that: "The rate of change of linear momentum is directly proportional to the net externally applied force and the change takes place in the direction of the net force."

Let a body of mass $m$ be moving with velocity $v$. Then the linear momentum $P$ is given by:

$P = mv$

According to Newton's second law:

$F_{\text{ext}} = \frac{dP}{dt}$

Since $P = mv$, and assuming $m$ is constant, we have:

$F_{\text{ext}} = \frac{d(mv)}{dt} = m\frac{dv}{dt}$

Therefore:

$F_{\text{ext}} = ma$

Relationship to Newton's first law of motion

If $F_{\text{ext}} = 0$, then from the equation:

$F_{\text{ext}} = ma$

We get:

$0 = ma$

Since $m \neq 0$, it follows that:

$a = 0$

This means:

$\frac{v - u}{t} = 0 \Rightarrow v = u$

Hence, the body either remains at rest or continues to move with constant velocity — this is Newton's first law.

Newton's third law of motion

The law states that: "To every action, there is an equal and opposite reaction."

Let object A exert a force on object B:

$F_{AB}$

Then object B exerts a force on object A:

$F_{BA}$

According to the third law:

$F_{AB} = -F_{BA}$

Important note: The two forces:

• Are equal in magnitude
• Opposite in direction
• Act on different objects

Therefore, they never cancel each other.

Implications of Newton's laws of motion

The laws we have seen above can be used to explain some of the occurrences that we encounter in daily life, such as:

1. Friction
2. Impulse
3. Weightlessness
4. Collisions
5. Motion in fluids

Equilibrant forces on a body

Equilibrant forces are those forces that produce zero acceleration on an object on which they act, thereby establishing equilibrium for that object.

One of the simplest cases of a body in equilibrium is a book resting on a table. The forces acting on the book are:

Its weight, $W$, acting downwards

The normal reaction, $R_N$, that the table exerts upward on the book

Since the forces are balanced, the book remains at rest, and the net force on the book is zero, meaning the book is in equilibrium.

Newton's first law of motion

"A body remains in a state of rest or uniform motion in a straight line unless acted upon by an external force."

This law is also called the law of inertia.

Example: A book on a table stays at rest until a force is applied to push it.

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Newton's second law of motion

"The rate of change of momentum of a body is directly proportional to the applied force and occurs in the direction of the force."

Formula: $F = ma$

Example: Calculate the force needed to accelerate a 5 kg object at $3\,\text{ms}^{-2}$

$F = ma = 5 \times 3 = 15\,\text{N}$

Final answer: 15 N

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Newton's third law of motion

"For every action, there is an equal and opposite reaction."

Example: A book exerts a downward force (its weight) on a table, and the table exerts an equal upward force on the book.

Equilibrium condition:

$F_{\text{ext}} = R_N + (-mg) = 0 \Rightarrow R_N = mg$

Example: If the mass of the book is 2 kg, then:

$R_N = 2 \times 9.8 = 19.6\,\text{N}$

Final answer: 19.6 N

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Friction and normal reaction

Formula: $\mu = \frac{f}{R_N}$

Example: If friction force is 4 N and normal reaction is 20 N, then:

$\mu = \frac{4}{20} = 0.2$

Final answer: Coefficient of friction = 0.2

Frictional force: Friction is a force that opposes the relative motion or tendency of such motion of two surfaces in contact.

Normal reaction: Normal reaction (R) is the perpendicular force exerted by a surface on an object resting on it.

Relationship between friction and normal reaction: Frictional force is directly proportional to the normal reaction. Mathematically,

$f \propto R \Rightarrow f = \mu R$

Where:

• $f$ is the frictional force
• $R$ is the normal reaction
• $\mu$ is the coefficient of friction

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Example 1: Block on horizontal plane connected to hanging mass

Given: Mass of block on table: $m_1 = 2\, \text{kg}$ Coefficient of static friction: $\mu_s = 0.3$ Acceleration due to gravity: $g = 9.8\, \text{m/s}^2$

Step 1: Calculate normal reaction Since the surface is horizontal:

$R = m_1 g = 2 \times 9.8 = 19.6\, \text{N}$

Step 2: Calculate maximum static friction

$f_s = \mu_s R = 0.3 \times 19.6 = 5.88\, \text{N}$

Step 3: Calculate the mass $m_2$ that will just move the block At the point of sliding, tension in the string equals frictional force:

$T = m_2 g = f_s \Rightarrow m_2 = \frac{f_s}{g} = \frac{5.88}{9.8} = 0.6\, \text{kg}$

Final answer:

$\text{The block starts sliding when } m_2 = \boxed{0.6\, \text{kg}}$

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Apparent weight in a lift

Apparent weight is the normal force exerted by the floor of the lift on a person or object. It varies depending on the lift's acceleration.

Formulas:

• When ascending: $R = m(g + a)$
• When descending: $R = m(g - a)$
• In free fall: $a = g \Rightarrow R = 0$

Example 2: Lift motion

Given: Mass $m = 50\, \text{kg}$ Acceleration $a = 2\, \text{m/s}^2$ Gravity $g = 9.8\, \text{m/s}^2$

Step 1: Apparent weight when lift ascends

$R = m(g + a) = 50(9.8 + 2) = 50 \times 11.8 = 590\, \text{N}$

Step 2: Apparent weight when lift descends

$R = m(g - a) = 50(9.8 - 2) = 50 \times 7.8 = 390\, \text{N}$

Step 3: Apparent weight in free fall

$R = m(g - g) = 50 \times 0 = 0\, \text{N}$

Final answer: Ascending lift: $\boxed{590\, \text{N}}$ Descending lift: $\boxed{390\, \text{N}}$ Free fall: $\boxed{0\, \text{N}}$

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Collision

A collision is an event where two or more bodies come into contact with each other over a short time, resulting in an exchange of momentum and possibly energy.

Types of collision:

1. Elastic collision: Both momentum and kinetic energy are conserved.
2. Inelastic collision: Only momentum is conserved; kinetic energy is lost.

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Example 1: Elastic collision

Two balls of mass $m_1 = 2\, \text{kg}$ and $m_2 = 3\, \text{kg}$ are moving towards each other with velocities $u_1 = 5\, \text{m/s}$ and $u_2 = -2\, \text{m/s}$. Find their velocities after an elastic collision.

Step 1: Use conservation of momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

$(2)(5) + (3)(-2) = 2v_1 + 3v_2$

$10 - 6 = 2v_1 + 3v_2 \Rightarrow 4 = 2v_1 + 3v_2 \quad \text{(Equation 1)}$

Step 2: Use conservation of kinetic energy:

$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

$\frac{1}{2}(2)(5)^2 + \frac{1}{2}(3)(-2)^2 = \frac{1}{2}(2)v_1^2 + \frac{1}{2}(3)v_2^2$

$25 + 6 = v_1^2 + \frac{3}{2}v_2^2 \Rightarrow 31 = v_1^2 + \frac{3}{2}v_2^2 \quad \text{(Equation 2)}$

(Solving these equations gives values of $v_1$ and $v_2$)

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Impulse

The product of force and the time interval during which the force acts.

$\text{Impulse} = F \cdot \Delta t$

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Example 2: Impulse calculation

A force of $F = 10\, \text{N}$ acts on a body for $t = 4\, \text{s}$. Calculate the impulse.

Step:

$\text{Impulse} = 10 \times 4 = 40\, \text{Ns}$

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Coefficient of restitution (e)

The ratio of relative speed after to that before collision.

$e = \frac{v_2 - v_1}{u_1 - u_2}$

Example 3: Coefficient of restitution

Two bodies approach each other with $u_1 = 6\, \text{m/s}$, $u_2 = 2\, \text{m/s}$, and after collision move with $v_1 = 3\, \text{m/s}$, $v_2 = 4\, \text{m/s}$.

Step:

$e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{4 - 3}{6 - 2} = \frac{1}{4} = 0.25$

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Example 4: Reaction from a gun (recoil)

Gun mass $M_g = 5\, \text{kg}$, bullet mass $m_b = 0.1\, \text{kg}$, bullet speed after firing $v_b = 400\, \text{m/s}$. Find recoil velocity of the gun.

Step: Using conservation of momentum:

$0 = M_g v_g + m_b v_b \Rightarrow v_g = -\frac{m_b v_b}{M_g}$

$v_g = -\frac{0.1 \times 400}{5} = -8\, \text{m/s}$

The gun recoils at 8 m/s in the opposite direction.

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Example 5: Water jet on wall

Density $\rho = 1000\, \text{kg/m}^3$, velocity $u = 10\, \text{m/s}$, pipe area $A = 0.01\, \text{m}^2$. Find force on wall.

Step: Mass per second hitting wall is:

$\dot{m} = \rho A u = 1000 \times 0.01 \times 10 = 100\, \text{kg/s}$

Force = rate of change of momentum:

$F = \dot{m} \cdot u = 100 \cdot 10 = 1000\, \text{N}$

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Equilibrant forces definition

Forces that bring an object to equilibrium. Their vector sum must be zero.

$\sum F_x = 0,\quad \sum F_y = 0$

Applications of Newton's laws of motion

1. When turning a corner in a moving car, your body tends to continue in a straight line due to Newton's first law (inertia), while the car turns — causing you to feel a force pushing outward.
2. In a moving bus that stops suddenly, your body continues moving forward due to inertia (first law), hence you feel a forward jerk.
3. Seat belts are advised for safety because they prevent you from continuing in motion during a sudden stop (first law).
4. When beating a dusty carpet, the coat moves but the dust remains in place due to inertia and gets dislodged (first law).
5. Jumping from a boat causes the boat to move in the opposite direction — Newton's third law (action-reaction pair).
6. Walking involves pushing the ground backward, and the ground pushes you forward (third law).
7. Kicking a ball in the air makes it rise and fall due to air resistance and gravity — Newton's second law.
8. Rocket propulsion is based on Newton's third law: the rocket expels gas (action), and gas pushes the rocket forward (reaction), and the net momentum change causes acceleration.