Laws of Motion

Newton’s Laws of Motion are fundamental ideas in classical mechanics that explain how forces influence the way things move.

The First Law, also known as the Law of Inertia, tells us that an object will either stay at rest or keep moving at a constant speed in a straight line unless something (an external force) makes it change.

The Second Law connects force with mass and acceleration using the equation F = ma, which helps us calculate how objects move when forces are applied.

The Third Law says that whenever you push or pull on something, it pushes or pulls back with the same amount of force but in the opposite direction. These laws are crucial for understanding everything from the simplest objects to complex systems, including things like momentum, friction, and tension.

1.0Force and its Effects

Force is any push or pull that alters an object's state of rest or uniform motion. It can have the following effects on a body:

1. Change its speed
2. Change its direction
3. Change both Example: Pushing and direction
4. Alter its size or shape.

Force is a vector quantity, with magnitude and direction.

Absolute units: SI Units-Newton(N)         C.G.S Units-Dyne

Conversion: $1N=10^5$ Dyne

Push: A force that moves an object away, changing its state of rest.

Example: Pushing a door to open it.

Pull: A force that draws an object closer, changing its direction.

Example: Pulling a rope or drawing a cart towards you.

2.0Balanced And Unbalanced Forces

1. Balanced Forces: It occurs when the sum of all forces acting on an object equals zero, keeping its motion unchanged.
2. Unbalanced Forces: It occurs when forces acting on an object change its state of rest or motion. For example, when you kick a football, unbalanced forces make it move from one place to another.

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3.0Newton’s Laws of Motion

Newton’s First Law of Motion

It states that an object at rest stays at rest, and an object moving with constant velocity continues to move unless acted upon by an external force.

Inertia: It is the property of a body that resists any change in its state of rest or motion. It is directly proportional to mass, meaning a body with greater mass has greater inertia.

• Inertia of Rest: An object resists changing its state of rest.
  Example: Passengers fall backwards when a bus starts suddenly.
• Inertia of Motion: An object resists changes to its uniform motion.
  Example: Passengers lean forward when a bus stops suddenly.
• Inertia of Direction: An object resists changing its direction.
  Example: A person inside a turning car feels a force pushing outward.

Linear Momentum

• The total amount of motion possessed by a moving body is known as the momentum of the body.
• It is the product of the mass and velocity of a body.

$\vec{p}=mv=\text{Momentum}=\text{Mass}\times \text{Velocity}$

• SI unit-kg-m/s, N-s
• Dimension - $[ML{T}^{-1}]$

Change in Momentum

Change=Final-Initial

$\Delta \vec{p}={\vec{p}}_f-{\vec{p}}_i=m{\vec{v}}_f-m{\vec{v}}_i=m\Delta \vec{v}$

Newton’s Second Law of Motion

According to this law, "rate of change of momentum of any system is directly proportional to the applied external force".

${\vec{F}}_{\text{net}}=\frac{d\vec{p}}{dt}$

$\vec{F}=\frac{d\vec{p}}{dt}=\frac{d(m\vec{v})}{dt}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}$

$\frac{dm}{dt}=0\text{if }m\text{ is constant}$

$\vec{F}=m\frac{d\vec{v}}{dt}=m\vec{a}$

$F=\frac{dp}{dt}=\tan \theta =\text{slope}$

Note: $\overrightarrow{F_{net}}=\vec{v}\frac{dm}{dt}$ (variable mass system)

Impulse and Average Force

Impulse: When a large force is applied on a body for a very short interval of time, then the product of force and time interval is known as impulse.

$\overrightarrow{dI}=\vec{F}dt=\overrightarrow{dP}$ $[\therefore \frac{\overrightarrow{dp}}{dt}=\vec{F}]$

• Unit of impulse = N-s or kg-m/s.
• Dimension- $[M^1{L}^1{T}^{-1}]$

Average Force: Average force acting on a body in a given time interval can be calculated by,

$\frac{\overrightarrow{dp}}{dt}=\vec{F}\Rightarrow \overrightarrow{F_{avg}}=\frac{\overrightarrow{\Delta p}}{\Delta t}$

$F_{avg}=\frac{\overrightarrow{\Delta p}}{\Delta t}$ Therefore, for a certain momentum change if the time interval is increased, then the average force exerted on the body will decrease.

Impulse Momentum Theorem

This theorem states that the impulse exerted on an object is equivalent to the change in its momentum. This theorem can be derived from Newton's Second Law of Motion. Since the rate of change of momentum is instantaneous, the impulse is effectively equal to the change in momentum.

$\vec{I}=\vec{F}t=(\overrightarrow{p_2}-\overrightarrow{p_1})$

The impulse applied to an object is equal to the resulting change in its momentum.

Conservation of Linear Momentum

When a system of interacting particles experiences no extrinsic forces, the total linear momentum of the system remains conserved. This total linear momentum is calculated as the vector sum of the individual linear momenta of all the particles within the system.

${\vec{F}}_{\text{ext}}=\frac{d\vec{p}}{dt}$

$If{\vec{F}}_{\text{ext}}=0\Rightarrow \frac{d\vec{p}}{dt}=0\Rightarrow \vec{p}=\text{constant}\Rightarrow {\vec{p}}_{\text{initial}}={\vec{p}}_{\text{final}}$

For two particle system ${\vec{p}}_1+{\vec{p}}_2=\text{constant}$

$\Delta {\vec{p}}_1+\Delta {\vec{p}}_2=\text{constant}$

$\Delta {\vec{p}}_1=-\Delta {\vec{p}}_2$

Change in momentum of 1st particle=Change in momentum of 2`nd particle

Newton's Third Law of Motion

According to Newton's third law, to every action, there is always an equal (in magnitude) and opposite (in direction) reaction. This law is also known as action-reaction law.

$\overrightarrow{F_{12}}=-\overrightarrow{F_{21}}$

4.0Rocket Propulsion(Variable Mass Problems)

m=Mass of rocket

$v_{rel}$=Velocity of exhaust gases w.r.t rocket

$\frac{dm}{dt}$=Rate of burning of fuel

Case-1 : If rocket is accelerating upwards, then $v_{rel}|\frac{dm}{dt}|-mg=ma$

acceleration of the rocket, $a=\frac{v_{rel}}{m}|\frac{dm}{dt}|-mg$

Case-2 : If rocket is moving with constant velocity, then a = 0

$v_{rel}|\frac{dm}{dt}|=mg$

5.0Common Forces in Mechanics

6.0Free Body Diagram

• A Free Body Diagram (F.B.D.) shows all external forces acting on an object.
• To create a Free Body Diagram, choose a body, identify all forces acting on it, and draw them on the object.
• Important Point: In Newton's Third Law, action and reaction forces never appear in the same Free Body Diagram. This is because these forces act on different objects.

7.0Normal Reaction Force

The contact force by which two surfaces in contact push each other perpendicular to the contacting surfaces, is known as normal reaction.

Problems on Normal Reaction

Two bodies in contact $a=\frac{F}{m_1+m_2}$ or $a=\frac{F_{net}}{m_{total}}$ $N=\frac{m_{2}F}{m_1+m_2}$
Three bodies in contact $a=\frac{F}{m_1+m_2+m_3}$ $N_2=\frac{F{m}_3}{m_1+m_2+m_3}$ $N_1=\frac{(m_2+m_3)F}{m_1+m_2+m_3}$

8.0Tension in String

Tension is the intermolecular force in a stretched string, acting away from the point of contact or tied end.

Important Points about Tension:

• Nature: Tension is electromagnetic and a pull-type force.
• Slack String: Tension in a slack string is always zero.
• Minimum Value: Tension can never be negative; its minimum value is zero.
• Direction: Tension acts away from the point of contact or tied end of the string.
• Inextensibility: The magnitude of accelerations of blocks tied to the string is the same, assuming the string is inextensible.
• Massless and Frictionless String: Tension is constant throughout the string.
• Massless but Frictional String: Tension varies at every contact point due to friction.
• Massive String: Tension changes at different points depending on the acceleration.
• Applied Force: If a force is applied directly to the string, the tension equals the applied force, regardless of motion.
• Massless String Assumption: Tension is uniform and equal to the applied force unless stated otherwise.
• Stretching Forces: To produce tension, two equal and opposite stretching forces must be applied, making tension equal to either applied force.
• Breaking Strength: A string can bear only a maximum tension before breaking, known as its "breaking strength," which depends on its material and dimensions.

9.0System of Masses Tied by Strings

Two Connected Bodies  $a=\frac{F}{m_1+m_2}$ Or $a=\frac{F_{net}}{m_{total}}$ $T=\frac{m_{2}F}{m_1+m_2}$
Three Connected Bodies  $a=\frac{F}{m_1+m_2+m_3}$ or $a=\frac{F_{net}}{m_{total}}$ $T_2=\frac{m_{3}F}{m_1+m_2+m_3}$ $T_1=\frac{(m_2+m_3)F}{m_1+m_2+m_3}$
Bodies Hanging Vertically $T_1=m_{1}g$ $T_2=(m_1+m_2)g$ $T_3=(m_1+m_2+m_3)g$
Bodies Accelerating Vertically Upwards $T_1=m_1(g+a)$ $T_2=(m_1+m_2)(g+a)$ $T_3=(m_1+m_2+m_3)(g+a)$
Bodies Accelerating Vertically Downwards $T_1=m_1(g-a)$ $T_2=(m_1+m_2)(g-a)$ $T_3=(m_1+m_2+m_3)(g-a)$
Tension in rod and heavy strings $T=F(1-\frac{x}{L})$

10.0Pulley Block Systems

• Ideal Pulley: Massless and frictionless.
• Ideal String: Massless and inextensible.
• Pulley Function: Changes the direction of the force but does not affect the tension in the string.
• No Friction: A frictionless pulley does not retard rotation; its only role is to change the direction of force through the connecting cord.

Some Cases of Pulley

1. $m_1=m_2=m$ $T=mg$ $a=0$ Reaction at the point of suspension $R=2T=2mg.$
2. $m_1>m_2$ $a=\frac{(m_1-m_2)}{(m_1+m_2)}g$ and $T=\frac{2m_1{m}_2}{(m_1+m_2)}g$ and  $g=\frac{W_1{W}_2}{W_1+W_2}$ and $R=\frac{4W_1{W}_2}{W_1+W_2}$
3. $a=\frac{m_{1}g}{(m_1+m_2)}$ and $T=\frac{m_1{m}_2}{(m_1+m_2)}g$ And  $R=\sqrt{2}T$
4. $(m_1>m_2)$ $a=\frac{(m_1-m_2)}{(m_1+m_2+M)}g$
5. $a=\frac{(m_1-m_2\sin \theta )}{(m_1+m_2)}g$ $T=\frac{m_1{m}_2(1+\sin \theta )}{(m_1+m_2)}g$
6. $m_1\sin \alpha >m_2\sin \beta$ $a=\frac{(m_1\sin \alpha -m_2\sin \beta )}{(m_1+m_2)}g$ $T=\frac{m_1{m}_2\left(m_1\sin \alpha +m_2\sin \beta \right)}{(m_1+m_2)}g$
7. $a=\frac{(m_2+m_3-m_1)}{m_1+m_2+m_3)}g$

11.0Translational Equilibrium

• A body is in translational equilibrium when the net force acting on it is zero (vector sum of all forces = 0).
• If multiple forces $F_1,F_2,......F_n$ act on the body, body and the body is in translational  equilibrium, their resultant must be zero $\sum F=0$

• In Cartesian components, $\sum F_x=0,\sum F_y=0,\sum F_z=0$
• A body cannot be in equilibrium under a single external force.
• With two forces, they must be equal and opposite for equilibrium.
• For three forces, their resultant must be zero.

12.0Spring Force

• A spring is typically a helical metallic wire.
• When stretched or compressed, the spring's length changes.
• The spring resists this change and tries to return to its original length.
• The force opposing the change in length is called the restoring force.
• The restoring force is electromagnetic in nature.

Hooke’s Law: Under small extension/compression, The opposing force is directly proportional to the change in length and opposite to direction of pull/push.

$F_{spring}=-kx$

For a pulley – spring system (at steady state) 

$a=\frac{m_{2}g}{m_1+m_2}$

$T=\frac{m_1{m}_2}{(m_1+m_2)}g$

$T=kx\Rightarrow x=\frac{T}{k}=\frac{m_1{m}_2}{k(m_1+m_2)}g$

13.0Frame of Reference and Pseudo Force

Frame of Reference: A system with respect to which position or motion of a particle is described is known as frame of reference.

Pseudo Force: To apply Newton’s law of motion in a non-inertial reference frame we need to apply pseudo force. It is an imaginary force which is used to explain the motion of objects from non-inertial reference frames.

$F_{pseudo}=\text{mass of body}\times \text{acceleration f non-inertial frame w.r.t observer}$

Note: Opposite to the direction of acceleration of non-inertial frame(observer)

$\overrightarrow{F_{pseudo}}=-m\times \overrightarrow{a_{obs}}$

Pseudo force does not follow action reaction law.

14.0Weighing Machine

• Weighing machine measures the Normal Reaction applied on it
• Reading of weighing machine is also called the Apparent weight.

$W_{app}$== Normal Reaction on machine

• The Reading is expressed in two ways

1. SI unit of Force is Newton(N)
2. Practical/Gravitational unit of Force is kg-f or kg-wt

$\text{Reading (kg-wt)}=\frac{NormalReactioninNewton}{g}$

$g=9.8m/s^2\approx 10m/s^2$

15.0Effective or Apparent weight of a man in lift

Case-1 If the lift is at stationary or moving uniformly (a = 0), then $N=mg$ $W_{app}=W_{actual}$ $W_{app}=N$ $W_{actual}=mg$
Case-2 If the lift is accelerated upwards, $N=m(g+a)$ $W_{app}>W_{actual}$
Case-3 If the lift is accelerated downwards, $N=m(g-a)$ $W_{app}<W_{actual}$

Note: If the lift is under free fall, it implies that its acceleration is equal to the

acceleration due to gravity.

$a=g$

$W_{app}=0$

It means that person in lift will feel weightless

The apparent weight of any body falling freely is zero

16.0Friction and its Types

Friction: It is the force that resists the relative motion between two surfaces in contact, not the motion itself. Rather it opposes relative motion between two bodies.

Cause of Friction

Old view: Friction occurs when irregularities on two surfaces interlock, resisting relative motion.

Modern view: Friction is caused by intermolecular forces between surfaces at the point of contact.

Friction depends on the following factors:

1. Friction depends on the actual contact area, not the apparent contact area.
2. Except for static friction (fs)friction is proportional to the normal Reaction (N).

Graph between applied force and force of friction

Contact Force: Let f be the force of friction and N the normal reaction, then the net contact force by the surface on the object is

$F_{Surface}=\sqrt{N^2+f^2}$

Minimum Value:(when f = 0) is N

Maximum value:(when $f=\mu N$) is $N\sqrt{1+{\mu }^2}$

Therefore

$N\le F_{Surface}\le N\sqrt{1+{\mu }^2}$

Angle of Friction ($\lambda$): The angle which the resultant of the force of limiting friction $f_L$ and normal force N makes with the direction of normal reaction N

$tan\lambda ={\mu }_s$

Angle of Repose or Angle of Sliding($\theta$): It is defined as the smallest angle of inclination of a plane with the horizontal at which a body just starts to slide, or equivalently, the greatest angle at which the body remains stationary without sliding.

$s=\frac{f_l}{N}=\frac{mgsin\theta }{mgcos\theta }=tan\theta \Rightarrow \theta =ta{n}^{-1}{\mu }_s$

Angle of Repose$(\theta )$=Angle of Friction ($\lambda$)

Downward sliding on rough incline plane

If angle of inclination is greater than the angle of repose, then the body accelerates down the incline.

$a=g[sin\theta -{\mu }_{k}cos\theta ](a<g)$

Acceleration of a body down a rough inclined plane is always less than 'g'.

For sliding down with constant velocity: a = 0

$sin\theta ={\mu }_{k}gcos\theta$

$tan\theta ={\mu }_k$

Upward sliding on rough inclined Plane

$a=g[sin\theta +{\mu }_{k}gcos\theta ]$

Note:

(1) If we want to prevent the downward slipping of body then minimum upward force required is $mgsin\theta -{\mu }_{k}mgcos\theta$

(2) If a body is projected in upward direction along the inclined plane then retardation of body is $a=[gsin\theta +{\mu }_{k}gcos\theta ]$

Note: Retardation of a body up a rough inclined plane may be greater than 'g'

Pulley with friction between block and surface

$\text{Acceleration, }a=\left[\frac{m_1-m_2(\sin \theta +\mu \cos \theta )}{m_1+m_2}\right]g$

$\text{Tension, }T=\frac{m_1{m}_2(1+\sin \theta +\mu \cos \theta )g}{m_1+m_2}$

17.0Two Block System in Friction

Step-1. Draw the FBD of the combined system:

1. Include limiting static friction.
2. If the applied force is greater than limiting friction, motion is possible; otherwise, no motion occurs.
3. Determine if the blocks move together or separately based on frictional forces.

Step-2. Assume combined motion

1. Find the common acceleration ac
2. Draw the FBD of the block without the applied force and calculate the frictional force needed to move it together.
3. Compare the frictional force with limiting friction:

• If   $f\le f_L$ both blocks move together with acceleration ac
• If $f\ge f_L$ They move separately.

Step-3.

1. Draw individual FBDs for both blocks, including kinetic friction.
2. Use Newton’s second law to find the accelerations of each block.

18.0Sample Questions on Laws of Motion

Q-1. A pendulum of mass m is hanging from the top of a train moving with an acceleration 'a' as shown in figure. Find the angle in which pendulum is in equilibrium w.r.t. train. Also calculate tension in the string.

Solution:

With respect to train, the bob is in equilibrium

$\sum F_y=0\Rightarrow Tcos\theta =mg$…….(1)

$\sum F_x=0\Rightarrow Tsin\theta =ma$……..(2)

Dividing equation (2) by (1)

$tan\theta =\frac{a}{g}\Rightarrow \theta =ta{n}^{-1}(\frac{a}{g})$

Squaring and adding equation (1) and (2)

$T=m\sqrt{a^2+g^2}$

Q-2. What horizontal acceleration should be provided to the wedge so that the block of mass m kept on wedge remains at rest w.r.t. wedge?

Solution:

For equilibrium along wedge

(In reference of wedge frame)

$macos\theta =mgsin\theta$

$a=g\frac{sin\theta }{cos\theta }$

$a=gtan\theta$

Q-3.A 12 kg monkey climbs a light rope as shown in fig. The rope passes over a pulley and is attached to a 16 kg bunch of bananas. Mass and friction in the pulley are negligible so that the effect of pulley is only to reverse the direction of force of the rope. What maximum acceleration can the monkey have without lifting the bananas?

Solution:

For monkey

$T-120=12\times a$……..(1)

For Bananas

$160-T=0\Rightarrow T=160$…..(2)

From equations (1) and (2)

$160-120=12\times a\Rightarrow a=3.33m/s^2$

Q-4.A block of mass M is dragged along a horizontal unresisting surface by a rope of mass m as shown in fig. A horizontal force F is applied to one end of the rope. Find the (i) Acceleration of the rope and the block (ii) Force that the rope exerts on the block.  (iii) Tension in the rope at its mid-point.

Solution:

1. Acceleration $a=\frac{F}{(m+M)}$
2. Force exerted by the rope on the block IS

$T=Ma=\frac{MF}{(m+M)}$

(3)

$T_1=(\frac{m}{2}+M)a=(\frac{m+2M}{2})(\frac{F}{(m+M)})$

Tension in rope at midpoint is $T_1=\frac{(m+2M)F}{2(m+M)}$

Q-5.A 800 kg rocket is set for vertical firing. If the exhaust speed of gases is 2000 m/s w.r.t rocket, calculate the rate at which gas must be ejected (in terms of mass per second) to generate enough thrust to counteract the weight of the rocket.

Solution:

Force required to overcome the weight of rocket $F=mg$

Thrust Needed= $v_{rel}\frac{dm}{dt}$

$v_{rel}\frac{dm}{dt}=mg\Rightarrow \frac{dm}{dt}=\frac{mg}{v_{rel}}=\frac{800\times 9.8}{2000}=3.92kg/s$

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