Samacheer Class 11 Physics - Laws of Motion

a) frictional force acting on the vehicle is along the negative x-direction
Normal force formula is the support force exerted upon an object that is in contact with another stable object.

The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:
* When a stationary bus starts to move, the passengers experience a sudden backward push.
* A book lying on the table will remain at rest until it is moved by some external agencies.
2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:
* When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
* An athlete running is a race will continue to run even after reaching the finishing point.
3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:
* When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
* When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Newton second law state that” The force acting on an object is equal to the rate of change of its momentum:
F = \(\frac { dp }{ dt }\)
If m is the mass_of the object, and v its velocity of motion then \(\overline{p}\) = m\(\overline{v}\):
The above equation can be written as
F = \(\frac { dp }{ dt }\)(m\(\overline{v}\) ) = m\(\frac { dv }{ dt }\)
∴ F = ma

If a very large force acts on an object for a very short time, then the force is impulsive force or impulse. If a Force (F) acts on an object in a very short time (∆t), then from Newton’s second law dp = f.dt
Integrating \(\int_{i}^{f} d p=\int_{t_{1}}^{t_{2}} f d t\)
P f – P i =F[t 1 – t 2 ]
P i – initial momentum of the object at time t 1
P f – Final momentum of the object at time t 2
P f – P i = ∆P = > Change is momentum
(t 2 – t 1 ) = ∆t = > Time interval
The above equation can be written as
∴ ∆P = F ∆ t.
ie \(\int_{t_{1}}^{t_{2}} F d t\) = J is called the impulse and t 1 it is equal to change in momentum of the object is ∆P = F ∆ t. When F is kept constant

It is easier to pull on object than to push it.
An object pushed at a angle θ.
Case 1:
When a body is pushed at an arbitrary angle θ (0 to π/2) the applied force. F is resolved into two components
F sin θ = Horizontally – parallel to surface
F cos θ = Vertically – perpendicular to surface
The total downward force = mg + F cos 0 This is equal to normal force (reaction)
N Push = mg + F cos θ … (1)
The static friction is equal to
P s (max) = μS N push = μ s (mg + F cos θ) … (2)
Case 2:
An object pulled at an angle θ
When an object is pulled at an angle 0 the applied force is resolved into two components.
F sin θ – Horizontally – parallel to the surface
F cos θ – Vertically – perpendicular to the surface
The total downward force = mg – F cos θ
This is equal to the normal force (reaction)
N pull = mg – F cos θ … (3)
The static friction is equal to
F s (max) = μ s N pull = μ s (mg – F cos θ) … (4)
Conclusion:
From equations (1) & (3) (or) from (2) and (4) it is clear that normal force or reaction due to pulling is less than that of push. So it is easier to pull an object than push it out.

There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f s lies between
\(0 \leq f_{s} \leq \mu_{s} \mathrm{N}\)
where, µ s – coefficient of static friction
N – Normal force
(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction,
f k – µ k N
where µ k – the coefficient of kinetic friction
N – Normal force exerted by the surface on the object
Methods to reduce friction:
Friction can be reduced
* By using lubricants
* By using Ball bearings
* By polishing
* By streamlining

“A force which does not actually act on particles but appears due to acceleration of the frame is called – Pseudo force”. In order to use Newton I & II law in the rotational frame of reference, one needs to include a “Pseudo force” called “Centrifugal force”. A pseudo force has no origin, It arises due to the non-inertial nature of the frame considered. In order to solve circular motion problems from a rotating frame of reference, pseudo force is necessary.

A frame of reference in which Newton’s I law of motion holds good is called an inertial frame of reference. In such a frame if no force acts on a body it continues to be at rest or in uniform motion. So it is called as an inertial frame.

On a leveled circular road, if the static friction is not able to provide enough centripetal ’force to turn, the vehicle will start to skid \(\mu_{s}<\frac{v^{2}}{r g}\)
III. Long Answer Questions:

Law: If there are no external forces acting on the system, then the total linear momentum of the system ( – tot ) is always a constant vector. In other words, the total linear momentum of the system is conserved in time.
Proof:
By combing Newton’s second and third laws the law of conservation of total linear momentum can be proved. When two particles interact with each other, they exert equal and opposite forces on each other
Consider two particles 1 & 2.
Let F 21 be the force exerted on 2 by 1
Let F 12 be the force exerted on 1 by 2
According to Newtons III law
\(\bar { F }\) 21 = – \(\bar { F }\) 12 → (1)
In terms of momentum, according to Newtons II Law,
F 12 = \(\frac { d }{ dt }\)\(\bar { P }\) 1 → (2)
F 21 = \(\frac { d }{ dt }\)\(\bar { P }\) 2 → (3)
Where is the momentum of p 1 particle P 2 is the momentum of II particle Sub (2) & (3) in (1)
This implies that (P 1 + P 2 ) = constant vector
\(\bar { P }\) 1 + \(\bar { P }\) 2 = \(\bar { P }\) tot – is the total linear momentum of the two particle system.
F 12 & F 21 are internal force and no external force acting on the system form outside. So total linear momentum is conserved.

Concurrent forces:
A collection of forces is said to be concurrent if the lines of forces act at a common point. If the concurrent forces are in same plane they are coplanar also, in additional to concurrent forces.
Lamis is a theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lamis theorem states that ” the magnitude of each force of the system is proportional to sine of the angle between other two forces”.
Proof:
Let F1, F2 and F3 be three coplanar and concurrent forces act at a common point 0 as in figure.
If the point 0 is in equilibrium then according to Lamis theorem.

When blocks are connected by strings and force F is applied vertically or horizontally, it produces Tension (T) in the string which affects acceleration to some extent. Let us discuss vertical and horizontal motion here
Case 1:
1) Vertical motion of connected bodies:
Consider two blocks to masses m 1 and m 2 (m 1 > m 2 ) connected by light and in an extensible string that passes over the pulley.
Let T be the tension in the string and a be the acceleration. When the system is released m 2 move vertically up and m 1 move vertically down with acceleration a. The gravitational force m 1 g on m 1 is used to lift m 2. The upward direction is chosen as y. The free body diagram of both masses can be drawn as
Applying Newton II law for mass m 2
T\(\hat{j}\) – m 2 g\(\hat{j}\) = m 2 a\(\hat{j}\) → (1)
By comparing the components on both sides We get
T – m 2 g = m 2 a
Similarly for mass m 1
If m 1 = m 2 ie both masses are equal, a = 0 This shows that if masses are equal, there is no acceleration, and the system as a whole will be at rest.
Tension on the string:
Substitute the value of ‘a’ from (4) in (1)
Equation (4) gives the magnitude of acceleration for m 1 the acceleration vector is
For m 1 the acceleration vector is
Case 2:
2) Horizontal motion of connected masses:
Let mass m 2 kept on the horizontal surface or table and m 1 is hanging through a small pulley.
Assume there is no friction on the surface. As both the blocks are connected to the unstretchable string, m 1 moves with an acceleration downward then m 2 also moves with the same acceleration ‘a’ horizontally.
The forces acting on m 2 are
* Downward gravitational force (m 2 g)
* Upward normal force (N) exerted by the surface.
* Horizontal Tension (T) exerted by the string.
Free body diagram
i) Downward gravitational force m 1 g
ii) Tension ‘T’ – upwards.
Applying II Law for m 1.
T\(\hat{j}\) – m 1 g\(\hat{j}\) = – m 1 a\(\hat{j}\)
m 1 g\(\hat{j}\) – T\(\hat{j}\) = m 1 a\(\hat{j}\)
By comparing components
m 1 g – T = m 1 a → (1)
Apply Newton’s II Law for m 2
T\(\hat{i}\) = m 2 a\(\hat{i}\) (along x)
Comparing components
N = m 2 → (2)
There is no acceleration along y direction
N = m 2 a → (3)
Substituting (2) in (1)
Tension can be got by substituting (4) in (2)
Conclusion: By comparing motion in both cases it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings.

Frictional is an opposing force exerted by the surface on the object which resists it motion. This force is frictional force. Friction force always opposes the relative motion between two surfaces which are in contact.
“when a force parallel to the surface is applied on the object, the force tries to move the object with respect to the surface. This relative motion is opposed by the surface by exerting a frictional force on the objects in a direction opposite to the applied force. Frictional force always act on the object parallel to surface on which the object is placed”.
To show angle of repose = angle of friction in an inclined plane.
Angle of friction = tan θ = F s /N
Consider an inclined plane on which an object is placed.
Let θ be the angle of inclination of the plane with horizontal.
“Angle of repose is the angle of inclined plane with horizontal such that an object placed on it begins to slide”.
The gravitational force mg is resolved into two components.
mg sin θ – parallel to inclined plane.
mg cos θ – perpendicular to inclined plane.
The component mg sin θ parallel to inclined plane tries to move the object down.
The component mg cos θ perpendicular to inclined plane is balanced by normal reaction N.
N = mg cos θ … (1)
When the object just begins to slide static friction attains its maximum value
tan θ = angle of repose = angle of friction.
Which is equal to tan θ = μ s
Hence proved.

* I law: Every object continues to be in the state of rest or off uniform motion unless there is external force acting on it.
* II law: The force acting on an object is equal to the rate of change of its momentum F = \(\frac{d(\bar{p})}{d t}\)
* III law: For every action there in an equal and an opposite reaction.
Discussion:
1) Newton’s laws are vector laws. The equation \(\bar { F }\) = \(\bar { ma }\) is a vector equation and it is essentially equal to three scalar equations. In Cartesian co-ordinate this equation can be written as
F x \(\hat{i}\) + F y \(\hat{i}\) + F y \(\hat{k}\) = ma x \(\hat{i}\) + ma y \(\hat{j}\) + ma z \(\hat{k}\) From this we can infer Fz cannot affect a y and a z and vice versa.
2) The acceleration experienced at a time ‘t’ depends on the force and the body at that instant does not depend on the force which acted on the body before F(t) = ma(t) In general direction of force may be different form direction of motion.
Case 1: Force and motion in same direction:
Example: when an apple falls towards earth the direction of motion and the force are in same downward direction.
Case 2: Force and motion are not in same direction
Example: The moon experiences a force towards the earth. But if actually moves in an elliptical orbit. In this case direction of motion and force are different.
Case 3: Force and motion in opposite direction:
If an object is thrown vertically upwards direction of motion is upward, but a gravitational force is downward.
Case 4: Zero net force, but there is motion:
Example: When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends the upward drag force increases and cancels the downward force. So the raindrop moves with a constant velocity called terminal velocity till it touches the surface of the earth, Hence the raindrop comes with zero net force and therefore with zero acceleration but with non zero terminal velocity.
Case 5: if multiple forces
\(\bar { F }\) 1, \(\bar { F }\) 2, \(\bar { F }\) 3, …… \(\bar { F }\) n act on the same body, then the total force \(\bar { F }\) net is equal to vector sum of individual forces, which provides the acceleration.
\(\bar { F }\) net = \(\bar { F }\) 1, \(\bar { F }\) 2, \(\bar { F }\) 3, …… \(\bar { F }\) n
\(\bar { F }\)net = m\(\bar { a }\)
Example: Bow & arrow
Case 6: Newtons II Law can be written as
F = m\(\frac { dv }{ dt }\) = m.\(\frac{d^{2} r}{d t^{2}}\)
Newtons II law is basically a second order derivative of position vector, which is not zero, there must be a force acting on it.
Case 7: if no force acts an a body then
m. \(\frac{d \bar{v}}{d t}\) = 0
Which implied \(\bar { V }\) = constant. If is essentially a I law. So Newtons II law is consistent with I law, but cannot be derived form each other. Newton II law is a cause and effect relation. Force is the cause and effect is the acceleration,
(effect) ma = F(cause)
\(\frac{d \bar{p}}{d t}\) = F.

Circular motion can be analyzed from two different frames of reference. One is the Inertial frame where Newton’s laws are obeyed. The other is the rotating frame of reference which is noninertial as it is accelerating. To use Newton’s I and II law in the rotational frame of reference the pseudo force called as centrifugal force is needed.
The centrifugal forces appear to act on objects with respect to rotating frames. To explain consider an example, In the case of a whirling motion of a stone tied to a string, assume the stone has angular velocity ω in an inertial frame. If the motion of the stone is observed from a frame which is also rotating along with the stone with the same angular velocity ω then the stone appears to be at rest.
This implies that in addition to the inward centripetal force (-mrω²) there must be an equal and opposite force that acts on outwards equal to (+ mrω 2 ). So the total force acting on the stone in the rotating frame is equal to zero (- mrω² + mrω² = 0) This outward force acting on the stone + mrω² is called centrifugal force.

When an object moves on a surface essentially it is sliding on it. But the wheels move on the surface through rolling motion. In case of rolling motion when a wheel moves on a surface the point of contact is always at rest. Since the point of contact is at rest, there is no relative motion between the wheel and the surface. Hence the frictional force is very less.
At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object.
Ideally in pure rolling, motion of point of contact with the surface should be at rest, but in practice it is not so. Due to elastic nature of the surface at the point of contact there will be same deformation on the wheel or on the surface. Due to this deformation there will be minimal friction between wheel and surface. It is called rolling friction. In fact rolling friction if much smaller than kinetic friction.

Angle of repose is the minimum angle that an inclined plane makes with horizontal when a body placed on it just begins to slide. Consider a body of mass m placed on an inclined plane. The angle of inclination 0 of the inclined plane is adjusted that the body on the plane begins to slide down. Thus 0 is the angle of repose.
Various forces acting on the body are:
(a) Weight mg of the body acting vertically downwards
(b) The limiting friction fs (max) in upward direction along the inclined plane. It balances the component mg sin θ of the weight mg along the inclined plane.
∴ mg sin θ = f s (max) …. (1)
(c) The normal reaction ‘N’ perpendicular to the inclined plane. It is balanced by the component mg cos θ, perpendicular to the inclined plane.
Where θ is the angle of repose.

In a leveled circular road skidding mainly depends on the co-efficient of static friction n s. The coefficient of static friction depends on the nature of surface which has a maximum limiting value. To avoid this usually “the outer edge of the road is slightly raised compared to inner edge”. This is called banking of roads or tracks. The angle of inclination called banking angle.
Let the surface of the road make angle θ with horizontal surface. Then the normal force makes an angle θ with vertical. When the car takes a turn, two forces are acting on the car.
(a) Gravitational force mg (downwards)
(b) Normal force N (Perpendicular to surface).
Normal force ‘N’ can be resolved into two components N cos θ and N sin θ and balances downward gravitational force.
N sin θ provides necessary centripetal acceleration, According to II law
N cos θ = mg
N sin θ = \(\frac{m v^{2}}{r}\)
Dividing the above equations,
tan θ = \(\frac{v^{2}}{r g}\)
V = \(\sqrt{r g \tan \theta}\)
∴ The banking angle θ and radius of curvature of the road or track determines the safe speed of car at the turning.
If the speed exceeds this safe limit, then it starts to skid outward but the frictional force comes into effect and provides an additional centripetal force to prevent outward skidding. But at the same time if the speed is less than the safe limit it starts to skid inward and again frictional force come into effect which reduces centripetal force to prevent inward skidding
However if the speed of the vehicle is sufficiently greater than the correct speed the frictional force cannot stop the car from skidding. So to avoid skidding in circular road or tracks they are banked.

The moon orbits the earth once in 27.3 days in an almost circular orbit.
Radius of earth = 6.4 x 10 6 m.
The centripetal accelerations given by a = \(\frac{ v^{2}}{r}\)
This can be related with moon a m = R m ω²
ω → angular velocity
a m → centripetal acceleration of the moon due to earths gravity to angular velocity
R m → Distance between the center of earth to moon. Which is 60 times the radius of the earth.
IV. Conceptual Questions:

While trying to push a car from outside, he pushes the ground backward at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

When surfaces are highly polished the area of contact between them increases. As result of this a large number of atoms and molecules lying on both surfaces start exerting strong attractive forces on each other. Therefore the frictional force increases.

No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

When a parachute is descending down in the atmosphere an equal and opposite force to motion of parachute is acting due to air resistance. As the area of the parachute is large the air resistance resists the motion and so it descends slowly.

When a person is walking on ice he presses the ice downward with his feet and in turn the ice pushes the person with an equal force. Since ice slippery the person is not able to press it hardly. So the action legs and so the reaction which is also less. So by making small steps, with larger normal force one can walk without slipping.

No, it cannot be more than 1 for normal plane surfaces. But when surfaces are so irregular that they have sharp minute projections and cavities on them. Then the coefficient of friction may be more than one.

Yes. The direction of motion is always opposite to the force of kinetic friction. By using the principle of equilibrium, the direction of force of static friction can be determined. When the object is in equilibrium, the frictional force must point in the direction which results as a net force is zero.

∵ Forces on both sides are equal the reading in the spring balance is zero.
(ii)
The spring is pulled by a force along the inclined plane so.
F = mg sin θ
= 2 x 9.8 x sin 30
= 9.8 x 2 x 1/2
F = 9.8 N

Force on book A
1) Downward gravitational force exerted by earth (m A g)
2) Upward normal force (N B ) exerted by book B (N B ).
Force on book B
i) Downward gravitational force exerted by earth (m B g)
ii) Downward force exerted by books A (N A )
iii) Upward normal force exerted by book C (Nc).
Force on book C
i) Downward gravitational force exerted by earth (m C g)
ii) Downward force exerted by books B (N B )
iii) Upward normal force exerted by book D (N D ).
Force on book D
i) Downward gravitational force exerted by earth (m D g)
ii) Downward force exerted by books C (N c )
iii) Upward force exerted by the N Table.

(i) Gravitational force(mg) acting downwards.
(ii) Tension (T) exerted by the string on the bob whose position determines the direction of T.
The bob is moving is a circular arc and so it has centripetal acceleration. At points A and C bob comes to rest momentarily and then its velocity increases when move towards B.
Hence tangential acceleration is along the arc.
The gravitational force can be resolved into two components.
mg cos θ along the string
mg sin θ perpendicular to the string
At point A & C T=mg cos θ and at all other points T is greater than mg cos θ.
∴ Centripetal force = T – mg cos θ
∴ ma c = T- mg cos θ
Centripetal acceleration
a c = \(\frac { T – mg cos θ}{ 2 }\)m

Case 1:
Resultant force =
forward force – Frictional force
= 500 – 400
F R = 100N
F R = ma = 100N
a = \(\frac { 100 }{ 25 }\)
a = 4 ms -2
Case 2:
Resultant force = forward force – Frictional force = 400 – 400 = 0 N
a = 0.

m = 0.8 kg
v = 12 ms -1
t = \(\frac { 1 }{ 60 }\)S
F = ?
Change in momentum = impulse
P f – P i = F t
mv – 0 = Ft

m = 2 kg
I = 1 m = r
T = F = 200 N
V = ?
During whirling motion, the force acting on the stone is a centripetal force which provides the necessary Tension in the string.
T = F = \(\frac{m v^{2}}{r}\)
200 = \(\frac{2 \times v^{2}}{1}\)
V max = 10 ms -1

:
V = 50 ms -1
r = 10m
centrifugal reaction = ?
m = 60kg
F = \(\frac{m v^{2}}{r}\)
= \(\frac { 60×50×50 }{ 10 }\)
= 3 x 5 x 10 3
F = 15,000 N

Force on the mass = μ k N = μ k mg
F = 0.7 x 0.5 x 9.8
F = 3.43 N
But F = ma
a = \(\frac { 3.43 }{ 0.5 }\)
= 6.86ms -2
m = 0.5kg
W.K.T
v² – u² = 2as
|u²| = 2as
u = \(\sqrt{2 \times 6.86 \times 10}\)
= 11.71 ms -2
Velocity with which the mass is thrown = 11.71 ms -1
11th Physics Guide Laws of Motion Additional Important Questions and Answers
I. Multiple choice questions: