Newton’s Laws of motionTypes of Forces

Force"Force" is an external or internal agent present to "influence" the natural state of motion of an object. So this is an influence (force) needed to change the natural state of body; that is of rest or of uniform motion.Classification of ForcesThere are different types of forces in our universe. Based on the nature of the interaction between two bodies, forces may be broadly classified as follows.(a)  Contact ForcesForces that act between the bodies in contact with each other. For example, Normal Reaction, Friction etc.(b)  Field Forces (Non-Contact Forces)Forces that act between bodies separated by a distance without any actual contact. For example, Tension, Spring, Weight etc.

Tension (T)When a string, thread or wire is held taut, the ends of the string or thread (or wire) pull on whatever bodies are attached to them in the direction of the string. This force is known as Tension.If the string is massless, then the tension T has the same magnitude at all points throughout the string.The direction of tension is always from the point of attachment to the body. In the given figure, two segments of tension act at O towards A and B.

SpringForce in an extended (or compressed) spring is proportional to the magnitude of extension (or compression).

i.e. F α x, in magnitude, but opposite in direction.So, F = -kx, where k is a positive constant, also known as the spring constant of the spring; and x is the compression or elongation from the natural length.

Normal reactionWhen a body is in contact with another body, there exists a force that prevents them from penetrating each other. Each one "stops" the other by applying a force away from itself.

The forces , shown in the diagram acting on bodies A and B, respectively, act away from the surface of contact, and prevent the two bodies from "occupying the same space".If , is the action  is reaction: they are equal in magnitude but opposite in direction. Further and,  are both perpendicular to the surfaces in contact and note that they act on two different bodies.

Examples:

Friction

It is a force that acts between bodies in contact with each other along the surface of contact and it opposes relative motion (or tendency of relative motion) between the two bodies. The direction of friction force on A is opposite to that of force on B and magnitude is same for both.

Weight (W)

It is a field force, the force with which a body is pulled towards the centre of the earth due to its gravity. It has the magnitude mg, where m is the mass of the body and g is the acceleration due to gravity.

Newton’s First Law of MotionIt states that every object persists in its natural state of motion i.e. continues to be at rest or moves in a straight line with uniform (constant) velocity! (This is what is meant by natural state of motion); in the absence of a net external force acting (impressed) on it

Mathematically, it is equivalent to say that for producing acceleration (that is for changing velocity) in a body, we need to have a net external force. (By net external force we mean vector sum of all the external forces acting on it).

It can be easily deduced from the statement of change in the state of motion. It is directly related to a frame of reference about which we have discussed earlier. To mark the point here, we can discover that by viewing objects from different frame of references the natural state of motion as perceived by different observers will be obviously different (can only be same if the frames are truly equivalent). Therefore, the change in state will also depend on the choice of reference frame. Finally, the amount of acceleration produced in a

body (or change in velocity) will depend on our choice of reference frames.

Inertia

The first law of motion 'describes a property of matter: No body (dead or living) can change its velocity by itself, without an external force acting on it. This property is known as inertia. When a car suddenly starts moving, the passengers in the car are thrown back due to inertia of rest. Similarly, when the car stops suddenly, the passengers stoop forward due to inertia (of motion).

Momentum

Momentum is the measure of motion contained in a body. It is measured by the product of mass and velocity of the body. Its direction is same as that of the velocity of the body.

= m , where  = momentum of the  body, m= mass of the body,  = velocity of the body

Illustration:

A mass 'M' is lying (figure shown below) on a table which is at rest (w.r.t. the table on which it is kept). Explain its state with the help of Newton's First Law of motion.

Since 'M' is lying on a table, there is no external force acting on it (forget about gravity just for the immediate discussion). As per Newton's first law of motion it will keep on lying at rest with respect to table for infinite time.

Here, comes out a very important, intrinsic (that is inherent) property of a body which is that it retains its state of motionlessness (as well as of motion, if it is in motion) which is termed as INERTIA of an object. This is present in all materialistic bodies in this universe.Newton’s Second Law of Motion

It states that rate of change of momentum of a body is equal to the force applied on it, in terms of the magnitude as well as in the sense of direction. Here

the momentum is defined as the product f mass and velocity i.e. m .

Therefore we can write mathematically.

F = d(mv)/dt, if 'm' remains constant then

=> F = m dv/dt

but d /dt = 

                       = m           which is mathematically Newton's second law of motion.

Here, if F = 0 then we find a = 0. This reminds us of first law of motion. That is, if net external force is absent, then there will be no change in state of motion, that means its acceleration is zero.

Further we can extend second law of motion, (in fact its decomposition) to three mutually perpendicular directions as per our coordinate system.

If components in x, y and z direction are Fx, Fy & Fz respectively, the three acceleration produced when Fx, Fy & Fz act simultaneously) in the body are, Now,

If we add three forces then resultant is called net external force.

Similarly

is called net acceleration produced in the body.

Impulse:

A large force acting for a short time to produce a finite change in momentum which is called impulse of this force and the force acted is called impulsive force or force of impulse.

Mathematically it is described as the product of force and time.

.·. Impulse (J) = F Δt (For constant force)

.·. Impulse (J) = mv - mu and since force is variable, hence J = 

The area under F - t curve gives the magnitude of impulse.

Impulse is a vector quantity and its direction is same as the direction of    .

Unit of Impulse : The unit in S.I. system is kgm/sec or newton -second.

Dimension         : MLT1

Illustration:

In Figure shown below, let us have M = 10 kg and a new net external force in the direction as shown in figure below is 150N. Find its acceleration.

= 15m/sec2

(in the direction of force)

.·.   = 15 m/sec2 (X-direction)

Newton’s Third Law of Motion

Newton's third law of motion was discovered and formulated, during the investigation of the fact that in all experiments it appeared that "whenever a body exerts a force on a second appeared that "whenever a body exerts a force on a second body, the second body always exerts a force on the first one".

Let us visualize and understand this phenomena with an experiment:

Suppose, we throw a stone on a surface of good strength; and the surface is made of glass, one finds it broken (the surface). From here one concludes that a force was exerted by stone on the surface and consequently it was broken.

Now, the question is, did that surface also exert a force on the stone. Just to know about it let us change our throwing object from stone to an egg of almost equal mass. Now, one throws this egg on the same surface of good strength with the same throwing force which he used for the stone. What happens? Obviously with your daily experience you know that the egg will be broken (And the damage to the surface will not be visible due to egg's spoiling the observation).

This is only possible if there was a force acting on the egg at the time it hit the surface. In fact we can now conclude that there is mutual force acting on the contact point of the surface and the object thrown. The breaking of either one (or may be both) depends on their ability to absorb forces without getting

damaged (that is their strength) so in precise words:-

To every action there is always opposite and equal reaction, it is equivalent to say that mutual actions of two bodies upon each other are always equal and directed to contrary parts.

Note: The most important fact to notice here is that these oppositely directed equal action and reaction can never balance or cancel each other because they always act, on two different point (broadly on two different objects) For balancing any two forces the first requirement is that they should act one one and the same object. (or point, if object can be treated as a point mass, which is a common practice).

Illustration of Newton's Third Law:

Some of the examples of Newton's third law of motion are given below:

1.      Book kept on a table: A book lying on a table exerts a force on the table which is equal to the weight of the book. This is the force of action. The table supports the book, by exerting an equal force on the book. This is the force of reaction, as shown in the figure. As the system is at rest, net force on it is zero. Therefore, forces of action and reaction must be equal and opposite.

2.     Walking: while walking a person presses the ground in the backward direction (action) by his feet. The ground pushes the person in forward direction with an equal force (reaction). The component of reaction in the horizontal direction makes the person move forward.

3.     Swimming: A swimmer pushes the water backwards (action). The water pushed the swimmer forward (reaction) with the same force. Hence the swimmer swims.

4.     Firing from a gun: When a gun is fired, the bullet moves forward (action). The gun recoils backwards (reaction).

5.     Fight of jet planes and rockets: The burnt fuel which appears in the form of hot and highly compressed gases escapes through the nozzle (action) in

the backward direction. The escaping gases push the jet plane or rocket forward (reaction) with the same force, hence, the jet or rocket moves.

6.     Rebounding of a rubber ball: When a rubber ball is struck against a wall or floor it exerts a force on a wall (action). The ball rebounds with an equal force (reaction) exerted by the wall or floor on the ball.

7.     It is difficult to walk on sand or ice: This is because on pushing, sand gets displaced and reaction from sandy ground is very little. In case of ice, force of reaction is again small because friction between feet and ice is very small.

8.     Driving a nail into a wooden block without holding the block is difficult:This is because when the wooden block is not resting against a support, the block and nails both move forward on being hit with a hammer. However, when the block is held firmly against a support, and the nail is hit, an equal reaction of the support drives the nail into the block.

Illustration:

Suppose in figure shown above we put one more block of 5 kg mass adjacent to 10 kg and a force of 150 N acts as shown in the figure below, then find the forces acting on the interface.

The combined acceleration of the two bodies when treated as one is                             a = F/((10+5))=150/15=10/sec2

So each one moves with a = 10m/sec2 keeping their contact established.

Here you can feel that due to 150N force the body of 5 kg feels as if it is being pushed by the 10 kg mass. There is force acting on 5kg called R1, to oppose it by third law this body exerts a force R2 on 10kg. The interface is as shown in Figure given below.

Also, third law tells us that R1 = R2 in magnitude and is opposite in direction.

.·. R1 = R2 = R

Here since 150 N force acts on the 10kg mass and only r acts on the 5kg mass. For motion in 5kg only R is responsible. We can write the initial equation as:

F = 150 = (10 + 5) a

150 = 10a + 5a

Here 10a is force experienced by 10kg mass. And 5a is experienced by 5kg mass.

.·. R = 5a   a = 10m/sec2

.·. R = 50N

.·. Net force experienced by 10kg block is (150-R) = 10a 150-R = 1010 = 100 N

.·. R = 50

Therefore we get R = 50N for both blocks. Hence we find "action and reaction are equal and opposite". Now net force on the body of 10kg mass is 100N & Net force on the body of 5kg mass is 50N and on the interface action and reaction are both equal and also are equal to force experienced by second body.

Friction

Whenever the surface of a body slides over another, each body experiences a contact force which always opposes the relative motion between the surfaces. This contact force is called frictional force.

Intermolecular interaction arising due to elastic properties of matter is the cause of frictional force.

Cause of sliding friction

Old view

Earlier it was thought that roughness of the two surfaces causes friction in the figure because it can be easily seen that smoother the surfaces, lesser is the friction. Interlocking of irregularities of the two surfaces causes hindrance to sliding. This, however, is not the current view

Current view

The current view is a slight deviation form the old view. Earlier we thought that interlocking of irregularities of surfaces was causing friction. Now, it is though that due to irregularities, the common surface area which is in actual contact of the two surfaces, is much less than the total overall area in contact. In one experiment, it came out to be 1/10,000th of the apparent area.

Thus, while the total interactive (action and reaction) forces between the two surfaces remain the same, the pressures at the points of contact are extremely high and cause the humps to flatten out (undergoing plastic deformation) until the increased area of contact enables the upper solid to be supported. It is thought that at the points of contact, small, cold-welded joints are formed by the strong adhesive forces between molecules which are very close together. These have to be broken away before one surface can move over the other.

Types of frictions

(a)    Static friction: it is the friction force that acts between surfaces at rest. It is almost independent of the area of contact (Although it is dependent on micro area of contact). It is proportional to normal force. This static friction force can vary from zero to a certain maximum value.

Maximum static friction force is the smallest applied force necessary to start the motion of the body.

The ratio of the magnitude of the maximum force of static friction tot eh magnitude of normal force is called the coefficient of static frictional (μs). If the magnitude of the applied force is less than the maximum frictional force, then we can sayfS = Applied force < μs N (Maximum frictional force)

(b)    Kinetic Friction: The kinetic friction is present between surfaces which have relative motion. The ratio of the magnitude of maximum kinetic frictional force to the magnitude of the applied force during motion is called coefficient of kinetic friction.

Friction

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Cause and direction of rolling friction

A wheel of radius R rolling without sliding on a flat surface will experience a resistance due to the very small local deformation that takes place, which is sometimes elastic sometimes inelastic, i.e., a kind of ridge is formed in front of the wheel as shown exaggerated in the figure. This gives rise to the force FR, whose line of action passes through the centre C of the wheel and P the horizontal force necessary to force the wheel to topple over the point M, the total clockwise torque acting on the wheel about M must be more than or marginally more than the total anticlockwise torque about M.

F = Frictional force

.·.     P × R cos θ > mg × R sin θ

P > mg tan θ

The value of "tan θ" is called the coefficient of rolling friction (μg). This value does not depend upon R. If the two surfaces are absolutely rigid, then no ridge will be formed and q will be zero i.e., coefficient of rolling friction will be zero.

Typical values are μR = 0.006 for steel and 0.02 - 0.04 for rubber tyres on concrete surfaces. Rolling friction is very small compared to the sliding friction.

In case of pure rolling μR = 0.

Laws of Friction

1.      Laws of Static Friction

Static friction force is always equal and opposite to the net external force acting on the body.

2.      Laws of limiting friction

(i)     Limiting frictional force is independent of the apparent area of contact till the value of the normal reaction remains same.

(ii)    The direction of limiting frictional force is opposite to the direction in which one body is on the verge of starting its motion.

(iii)    The limiting frictional force depends upon the nature of surfaces in contact.

(iv)   Quantitatively, the magnitude of the force of limiting friction (F) on one of the two bodies in contact, is directly proportional to the normal reaction (R) on this body due to the other.

F ∞ R

Or F = μR

Where m is the constant of proportionally called coefficient of friction.

Co-efficient of Friction

According to the law of limiting friction,

F µ R

Or F = μR           ............ (3)

where μ is a constant of proportionality and is called the coefficient of limiting friction between the two surfaces in contact.

From (3),

μ = F/R              ............ (4)

Hence coefficient of limiting friction between any two surfaces in contact is defined as the ratio of the force of limiting friction and normal reaction between

them. The value of μ depends on

(i)     nature of the surfaces in contact i.e., whether dry or wet; rough or smooth; polished or not polished.

(ii)    material of the surfaces in contact.

For example, when two polished metal surfaces are in contact, μ ≈ 0.2, when these surfaces are lubricated, μ ≈ 0.1. Between two smooth wooden surfaces, μ varies between 0.2 and 0.5. Obviously,μ has no units.

When a body is actually moving over the surface of another body, we place F by Fx, the kinetic friction, and μ and μk.

Therefore,

μk = Fk/r

μk is then called the coefficient of kinetic or dynamic friction. As Fk < F, therefore, μk is always less than μ i.e. coefficient of kinetic or dynamic friction is always less than the coefficient of limiting friction.

Table gives the values of coefficient of limiting/kinetic friction between some pairs of materials:

S.No.Surface in contact Coefficient of

limiting frictionCoefficient of kinetic friction

1. Wood on wood 0.70 0.40

2. Wood on leather 0.50 0.40

3. Steel on Steel (mild) 0.74 0.57

4. Steel on Steel (hard) 0.78 0.42

5. Steel on Steel (greased)

0.10 0.05

Angle of friction

It is the angle between normal Reaction and the resultant of the normal reaction and limiting frictional force. i.e. we have a body of mass m which is placed on a table and we say that the body and the surface of the table have a coefficient of friction μ between them.

If we apply a small force F, the body will not move. Let us gradually increase the force until the body starts moving. At one stage the applied force will be equal to the frictional force. The coefficient of static friction μs = F/N where F is the applied force and N = mg is the normal force.

Angle of repose

It is the minimum inclination of an inclined plane at which a body just starts moving under its own weight.

Let us take a body of mass m placed on a table. The body and the surface of the table have a coefficient of friction μ between them. If we gradually increase the inclination of the table at a certain stage the body will start moving under gravity. The angle made by the table to horizontal is called the angle of repose.

Illustration:

A body of mass 100kg is placed on a table. The coefficients of static & kinetic frictions are 0.4 & 0.35 respectively. A force equal to the maximum static frictional force is applied and the body is slightly disturbed.

Analyse the motion (a) find acceleration (b) a relation between angle of repose & angle of friction.

F = μs N and N = mg

so, for the motion of the body

F - Fk = ma

or a = (Fs-Fk)/m = (μsmg - μkmg)/m = (μs-μk)g

= (0.4 - 0.5) × 10 = 0.5m/s2

So, only a slight disturbance will move the body with an acceleration a = 0.5 m/s. though the force which is reducing the acceleration was enough to balance the body in static friction.

Let us now take the same body placed on the table again. Now limiting frictional force is applied without disturbing it.

R cos θ = N                                        R sin θ = μ N

=> m = tan θ                                        θ = tan-1 μ

we can say that the resultant of friction force and the normal force is making an angle θ with the normal force. θ is called limiting angle of friction :

The same body is placed on an inclined plane and the angle of inclination is increased slowly. In the limiting case when the body starts moving. We draw the

force diagram as shown in figure above.

mg sin α = m mg cos α

tan α = m

where α is called angle of repose

also since m = tan α

α = θ         i.e. angle of repose = limiting angle of friction

(c)    Rolling Friction:

(In pure rolling only the static friction acts which is necessary for pure rolling). This will be dealt with in detail in module 2 (with Rigid body dynamics)

After knowing three laws of motion and friction let us see that how can we use these laws for solving problems. But just to tell you, to actually apply the Laws related to force and acceleration, we always need to know all the forces acting on the system on which we want to apply Newton's Law's because without knowing all the forces we can't know the net external force and for that we have to convert our system in a free body diagram.

Free body diagram

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It is to be noticed here that during previous examples, we were using a concept called FBD implicitly which can now be brought to you conscious attention.

It is like this. Whenever one attempts a problem involving forces and acceleration (say of dynamics or statics) one must show all forces and acceleration (possible acceleration may be unknown also) on each part of the system treating that part separately (it is called dividing system into possible subsystems). By all forces we mean external as well as internal forces (internal forces refer to mutual reactions). Now each system is ready to get treatment of laws of motion e.g. acceleration, velocity.

Coming to actual situation one should first identify all the component involved in the system, say mass m1, mass m2 .........., pulley 1, pulley 2, etc. Now separate them from others by cutting the string contacts (sort of imaginary separation) In effect, make them free (that's where comes the name free body) from other components and at the same time show all the forces acting on it, external as well as internal, arising due to separation from other parts. These are called mutual interaction forces (one of this type we have seen in the illustration of IIIrd law) and show all possible acceleration. The diagram thus obtained is called a free body diagram.

Illustration :

Let us draw FBD for various given systems

[Here we are assuming that all the surfaces strings and pulleys are ideal that is we are neglecting their masses & any friction present]

FBD (1) :  Block of mass M is resting on a frictionless rigid surface

There are only two forces in the system in figure above. mg which is the weight of the block, and is acting on the surface. R2 = mg through its centre (since the body is symmetric). So, here itself from FBD we can see that net external force

on the block is zero. That is why, it is stationary on the surface.

FBD (2) :  Draw the free body diagram of the block shown in figure 1.12.

Where R is reaction from the surface (R = Mg)

FBD (3) :  Draw free body diagrams of both the blocks (figure 1.13 a), Assuming a reaction of magnitude 'R' is present at the interface.

Now a question comes Can 'R' have a direction opposite to what is shown here? Answer is, of course it can have,

[before we proceed further let me point out that once you make calculations you will find that values of R will automatically come negative, which tells us that earlier direction were the correct ones].

FBD (4) :  Suppose situations is as shown in figure above that a light inextensible string pulls a block of mass M on a frictionless rigid surface.

Here string is acting as a force transmitting element

It will experience a tension T in it. Let us cut it (imagine) at then the situation is as shown in figure shown below.

(We join it again then net force should become zero at that point. Since t, and T are oppositely directed their sum will come out to be zero).

So tension t is responsible for dragging mass of block.

Therefore free body representation is

Caution :   What happens if rope is not massless?

FBD (5) :  Draw free body in case of system shown (figure shown below).

Suppose T is tension in the string, then by cutting it at 1-1' and 2-2' we can draw FBD's as shown in figure shown below.

R1 and R2 are reactions from plane, and from Newton's law they are equal to the weights of respective blocks.

FBD (6) :  Draw FBD where one of the block is resting on an inclined plane and rope goes over a frictionless massless pulley (Figure given below).

FBD becomes

Here the noticeable fact is that R2 is perpendicular to the inclined plane and mg is perpendicular to horizontal plane. Since Mg is always directed downwards. R2 is due to the component of Mg, which is acting, perpendicular to inclined plane. (Here, the rule is that reaction force is always normal to the surface, which provides the reaction).

Here for "ease" we can resolve force Mg in two components. One parallel to inclined plane and other on perpendicular to inclined plane. As shown in figure below.

Free body diagram

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.·. free Body, of the block on the inclined plane can be represented by figure shown below.

Finally at the end of this FBD, I want to point out that T2 will always be equal to T2for a continuous homogeneous massless inextensible string passing over a massless frictionless pulley.

FBD (7) :  Suppose both the block are on inclined planes as shown in figure given below.

For FBD cut the strings at 1-1' and 2-2' and separate the two blocks and pulleys as subsystems.

FBD (8) :  If pulley is hanging from a rigid support say roof and masses are connected as shown in figure given below.

FBD's become.

Once we are comfortable in drawing FBD then we can proceed to write equation of motion, using Newton's II Law. But prior to that, it is necessary to briefly introduce you with the concept of equilibrium of bodies.

Equilibrium:

An object is said to be in equilibrium if the vector sum of all the forces acting on the body (externally applied + forces arising due to mutual interaction;) is zero.

That is  1 +  2 +......+  N = 0

But since it is impractical to apply it, as it is, to problems, therefore we resolve all forces in the three directions X, Y & Z equilibrium and if some of the forces is zero in each direction then the body is said to be in equilibrium. (to be more

specific - translator equilibrium).

F1x + F2x + ............ + Fnx = 0

F1y + F2y + ............ + Fny = 0

F1z + F2z + ............ + Fnz = 0

Enquiry : How to apply equations of motion to any problem?

Writing Down Equations of Motion: Once we have made a free body diagram then we can write equations of motion for each part of the system, for which we have drawn FBD. To write down equations of motion for a sub-system for which we have already drawn the FBD the requirement is to choose two direction (if required we will need three directions) in which we shall work to reduce the

complexity say, direction X and Y   which in most of the cases is natural. In other cases, where we are dealing with inclined planes we can fix our coordinate axis x-y in any desired orientation which reduces our trouble of finding out components of the forces which are acting in various directions and by experience one knows that it is better to choose X axis parallel to the inclined

plane and Y axis perpendicular to it. For inclined plane    is the right choice of coordinate axis, tilted as per inclination. Now once can go for writing F = ma equations in two directions for any ith sub-system.

∑Fx = miax

∑Fy = miay

Let us write down equation of motion for the FBD's we have already drawn.

Refer to FBD (2):      Here figure is given below.

∑Fy = R - Mg

It will be zero since we know that on the surface the body can't move upwards or downwards.

.·. R - mg = 0             => R = Mg and ∑Fx = F

And this alone should accelerate it

∑ F = Ma,   => a = F/m

Refer to FBD (3): Here Figure is given below.

for the smaller block

∑Fy = R1 - mg = 0             .·. R1 = mg

∑Fx = F-R                        .·. F-R = m a1

For the Large Block          ∑Fy = R2 - Mg = 0 .·. R2 = Mg

∑Fx = R                           .·. R = Ma2

Since both Blocks move as on system, their accelerations will be equal

a1 = a2                           .·. F - R = ma

& R = Ma                    from here we can solve for R & a.

Refer to FBD (4): Here figure is given below.

For block

∑Fy = R - Mg = 0

.·. R = mg

∑Fx = T = Ma

.·. a = T/m

Refer to FBD (5): Here Figure is given below.

Here also, since system is fully connected and we assume that string is not loose, then the acceleration in different parts will be equal. That is, Blocks m and M as well as the string will move with the same acceleration 'a'.

Equation for smaller Block

∑Fy = R1 - mg = 0

.·. R1 = mg

∑Fx = T = ma ............ (1)

For Block - 'M'            ∑Fy = R2 - Mg = 0

.·. R2 = Mg

∑Fx = F - T = Ma ......... (ii)

solving (i) & (ii) we get T & a.

Refer to FBD (6): Here Figure is given below.

Since, system is fully connected by light inextensible string therefore acceleration in all the parts will be equal.

For Block 'm'

R1 - mg = 0 .·. R1 = mg

T1 = ma.

For Block 'M' (Refer to the diagram made with mg cos α & mg sin α components.

R2 - Mg cos α = 0 (since it is not moving to the plane)

R2 = Mg cos α and is parallel to the plane

Mg sin α - T2 = Ma                       ............ (ii)

It is already explained that T1 = T2 = T

.·. solving (i) & (ii) we get 'T' & 'a' for the system.

Refer to FBD (7) : Here Figure is given below.

T1 = T2 = T (similar to the analysis done in FBD (6))

For block m1 perpendicular to inclined plane T1 - m1g cos α = 0 R1 = m1g cos α.

Parallel to the plane.

T-m1g sin α = m1a ............. (1)

For block m2 perpendicular to plane

R2 - m2g cos β = 0 R2 = m2g cos β and m2 g sin β - T = m2 a ...... (ii)

Solving (i) & (ii) we get 'a' & 'T.

Refer to FBD (8) : Here Figure is given below.

In Y-direction :

For block m1 T - m1g = m1a

For Block m2 m2g - T = m2 a

From (i) & (ii) we can solve for 'T' & 'a'.

On pulley 2T force is acting downwards. Therefore the pulley experiences a force F = 2T. Now you are well conversant with making FBD's and writing down equations of motion for the given system. Therefore, now we can write down some definite steps to follow for any given problem, which can considerably reduce your diversion or confusion or say the possibility of getting trapped into complexity of a problem. (of course for that you have to use your brain all the time with these steps).

(i)      Draw the fully connected clear diagram.

(ii)     Define subsystems, that is parts of the system on which you will work to get your answers.

(iii)    Draw free body diagrams for all possible subsystems.

(iv)    Resolve forces as well as accelerations in x y & z direction. (or perpendicular and || to the plane whenever require).

(v)     Write, ∑Fx, ∑Fy, ∑Fz equations with the physical constraints appearing in the problem.

(vi)    Eliminate some variables and get the required one(s).

Centripetal Force

If a body is moving with a constant speed in a circle, as seen from an inertial frame, it is continuously towards the centre of rotation with magnitude v r/r (known as centripetal acceleration), where v is the speed of the particle and r is the radius of the circular path.

According to Newton's second law, this body will experience net force directed towards the centre called the centripetal force.

Therefore, net force acting on the body towards the centre = mv2/r, where m is mass of body.

Centrifugal force is a pseudo force acting on the body from a rotating frame.

Illustration:

A bob of mass m is suspended form a inextensible, massless describe a horizontal uniform circular motion as shown in figure 1.42. about a vertical. Analyze the dynamics of this system.

Solution:

The path of the system described above is shown in figure 1.42. Let the radius of circular path of bob is r equal to l sin θ and tension in string is T. The string makes an angle θ with the vertical. Consider the bob is at A. We can draw the free body diagram of bob at a as shown in figure 1.43. The force acting on the

bob is it's weight mg and tension T of the string. Tenstion T is resolved in two components T cos θ and T sin θ as shown in figure 1.43. we can write the equation of motion

T cos θ = mg                      T sin θ = mv2/r

Perhaps you have noticed that when a road is straight, it is horizontal too. However, when a sharp turn comes, the surface of the road does not remain horizontal. This is called banking of the roads.

Purpose of banking

Banking is done

1.     To contribute in providing necessary centripetal force.

2.     To reduce frictional wear and tear of tyres.

3.     To avoid skidding

4.     To avoid overturning of vehicles.

Case I: μ = 0

What we really wish is that even if there is no friction between the tyres and the road, yet we should be able to take a round turn. In the given figure Vertical N cos θ component of the normal reaction N will be equal to mg and the horizontal

N sin θ component will provide for the necessary centripetal force. [Please note that as we are assuming μ to be zero here, the total reaction of the road will be the normal reaction.] Frictional forces will not act in such a case.

.·.      N cos θ = mg                      .......... (i)

N sin θ = mv2/r                       .......... (ii)

Dividing equation (ii) by (i), we get

tan θ = v2/rg

where θ is the angle of banking.

Case - II : When μ ≠ 0

In the figure shows a section of the banked road and the view of the vehicle form the rear end.

The total forces acting are

N1 and N2 = normal reactions

F1 and F2 = frictional forces

mg = weight

r = radius

θ = angle of banking

Let N = Resultant of N1 and N2.

F = Resultant of F1 and F2.

Let us resolve all the forces horizontally and vertically. As the vehicle has

Equilibrium in vertical direction.

.·. N cos θ + F sin θ = mg                        ............ (i)

The resultant of horizontal components i.e., (F cos θ + N sin θ), however, this becomes the net external force acting on the vehicle in the radially inward direction of the round-turn. This thus provides for the necessary centripetal force (mv2/r).

.·. F cos θ + N sin θ = mv2/r                     ............ (ii)

Further, if μ is the coefficient of friction, we have

F = μN                                                  ............ (iii)

These are the three basic equations from which, we can find out whatever we want to find out.'

Putting (iii) in (i) gives

N cos θ = μN sin θ + mg

=>    N(cos θ - μsin θ) = mg

=>    N = mg/cos θ - μsin θ                ...............(iv)

Putting (iii) and (iv) in (ii) gives

μ × mgcos θ/(cos θ - μsin θ) + mgsinθ/(cos θ - μsin θ) = mv2/r

=>    μ mgr cos θ + mgr sin θ = mv2 cos θ - μmv2 sin θ

.·.     tan θ = (v2 - μrg)/(rg + μv2)               ............... (A)

or     v2 = rg(μ + tanθ)/(1-μtanθ)                ................ (B)

(A)    Gives the angle of banking for the maximum velocity v and (B) gives the value of the maximum velocity which the vehicles should be allowed on a road banked at an angle θ.

Notes : 1. The value of μ is the minimum value of μ required. The value of v is the maximum allowable velocity.

.·.      The best angle of banking θ so that there is absolutely nil wear and tear due to frictional force for the given values of v and r can be determined by putting μ = 0 in this formula.

If we put μ = 0 in formula (A), we get

tan θ = v2/rg

Further, for zero frictional wear and tear, the velocity for the given values of θ and r will be v = √rgtanθ.

Skidding

What is skidding?

Let us consider the situation in the figure. You are cycling fast on road I. You then want to take a turn to go to road II. However, due to big leak of mobile oil from some truck, the portions of the roads within the area ACBD have become slippery. You do not know about it. You are cycling fast. When you reach the line AC, you turn the handle mounted on the front wheel towards road II. What will happen? Will you be able to take the turn? No, you won't be. Although your front wheel is aligned to go towards road II, you still continue to go straight to road III. This is called skidding. You will skid.

The tendency to slip transverse (transverse means across) to the intended line of run, is called skidding. Thus as soon as you turn the handle of your cycle (or the driving wheel in a motor car), skidding will try to occur. If there is enough friction, this start of skidding brings into action a frictional force between the

road and the bottom surface of the wheels of the vehicle. This frictional force then provides for the necessary centripetal force required to negotiate the turn. If friction is not enough, skidding will start which will not let you take the turn in a normal way. Skidding will also cause additional friction wear and tear of the tyres of your vehicle.

How to avoid skidding?

Let us consider the situation given in the figure. Let r is the radius of turn which you have to take. N1 and N2 are normal reactions mg the weight and F1, F2 the frictional forces on the inside and outside wheels.

.·.      (N1 + N2) = mg                          ......... (i)

.·.      F1 + F2 = μ(N1 + N2) = μgm         ......... (ii)

This must be greater than or equal to the centripetal force required.

.·.      μmg > mv2/r

or, v < √μrg

or,. Vmax = √μrg

If the velocity of the vehicle is more than √μrg, it will skid.

Overturning

You may have seen overturned trucks lying on the road. Such heavily loaded trucks! Who could have overturned them!! Overturning occurs on the roads when the trucks try to change directions, take sharp turns. Overturning occurs, more after, in case of vehicles which have greater height or whose centre of gravity are much high up from the surface of roads.

Let us first consider why over turning would take place at all. Suppose a heavily loaded truck is going straight. Suddenly it takes a sharp turn towards its left.

Now what actually happens is that while the upper portion of the truck still tends to go straight because of its inertia (Newton's first law of motion), the lower portion starts going towards left because you have turned the driving wheel accordingly.

Thus, if the inertial forces on the upper portion are much height, they provide so much torque on the truck at its centre of gravity that overturning takes place.

Thus overturning always takes place by lifting off the inner wheels from the ground on the curved path.

Limiting case when a four wheeler just begins to overturn on a plain horizontal road

Let

Mg = weight

N = Total normal reaction = N1 + N2

F = Total friction force

v = velocity

r = radius of the round

2a = distance between inner and outer wheels.

G = centre of gravity

h = height of centre of gravity from Earth. Frictional force F will provide for the necessary centripetal force

.·.      F = mv2/r                               ............... (i)

When the vehicle just begins to overturn, the inner wheels will just begin to lift

off from the ground. Their pressure on ground will become zero, so the reaction N1 on the inner wheels will become zero.

.·.      N1 = 0

N1 + N2 = N2 = mg

Let us take moments about G

N2 × a = F × h

Putting (i) and (iii) in (iv) gives

mg × a = mv2/r × h, or v

i.e., vmax = √rga/h

If speed goes beyond it, the vehicle will overturn.

.·.     This topic will be discussed in rotation.

Minimum μ required to prevent overturning

We know frictional force is μN.

=>     F = μN = μmg                                     ............ (v)

Putting equation (v) in (i) gives

μmg = mv2/r,

Or     μ = μmin = v2/rg

F = mv2/r = mω2r

Centripetal force is not a new kind of force. It is the radial component of the net force acting on the particle moving along a circle. Centrifugal force is a type of pseudo force used by an observer moving in a circle. Numerically, it is equal to the centripetal force but is oppositely directed if observer and the body both are moving on same circle as a single unit.

Illustration:

A 1200 kg automobile rounds a level curve of radius 200 m, on a unbanked road with a velocity of 72 km/hr. What is the minimum co-efficient of friction between the tyres and road in order that the automobile may not skid. (g = 10 m/s2)

Solution:

In a unbanked road, the centripetal force is provided by the frictional force.

.·.      ffriction = mv2/r But flimiting friction > ffriction

or μmg  ffriction or μmg  mv2/r

.·.      μmin = v2/gr = (20×20)/(10×200) = 0.2.

Solved examples

Example 1:

Find acceleration of masses m1 and m2 in the set up shown, pulleys and string are massless. Neglect any friction.

Solution:

First we notice that the acceleration a1 = 1/2 a2. This can be seen as follows. Suppose m2 descends by Δx. Then the reduction in the length of string between the point of attachment on the wall and pulley attached to the end of incline also decrease by Δx. The pulley attached to m1 then moves by Δx/2.

If T is the tension in the string then the free body diagram for m1 is as shown in figure.

x component : 2T - m1gsinθ = m1a1                               (I)

y component : N - m1gcosθ = 0

The free body diagram for m2 is                                     (II)

m2g - T = m2a2                                                          (III)

Solving equations (I) and (II)

a2 = (4m2g - 2m1gsinθ)/(m1+4m2)

Example 2:

Three masses m1, m2 and m3 are being pulled by force F (Figure given below). Strings connecting m1 to m2 and m2 to m3 are massless. The coefficient of kinetic friction between masses and table is 0.1. If m1 = 20 kg, m2 = 30 kg and m3 = 50 kg and F-1500N then

(a)    What is the acceleration of books?

(b)    What are the tensions in strings tied between m1 and m2 and m2 and m3?

Solution:

Let T1 = Tension in the string tying m1 to m2.

and T2 = Tension in the string tying m2 to m3.

Free body diagram for the masses and corresponding equations are

T1-μkm1g = M1a

T2-T1-μkm2g = m2a

F-T2-μkm3g = m3a

Solving these equations we get

(a)    m3a = F-μkg(m1+m2+m3)-m1a-m2a

a = F-(μk g)(m1 + m2 + m3)/(m1 + m2 + m3) = (1500-0.95×100)/100

= 14.02 m/s2

(b)    T1 = m1 a + m1 μk g

T1 = 20 (14 + 0.98) = 299.60 N

T2 = 299.60 + 30 × 14.98 = 749 N

Example 3:

The total mass of a lift and passengers is 1000 kg. The variation of velocity with me is given in the graph with velocity being positive in the upward direction.

(a)    What is the tension in the rope pulling the left at t = 1 sec.

(b)    What is the tension at t = 3 sec.

(c)    What is the tension at t = 9 sec.

Solution:

General equation is T = mg + m

(a)    t = 1 sec            a = 1 m/s2

T = 1000(10.8) = 10800 N (t - 1 sec.)

(b)    t = 3 sec            a = 0

T = 98000 N

(c)    t = 3 sec            a = -1m/s2

T = 8800N

Example 4:

A man pulls himself, and the cradle he is sitting in, by pulling down on the rope which is tied to the cradle and would over a fixed pulley. If the man weights 60 kg and the cradle weighs 25 kg then how much force must the man exert on the rope to pull himself up at an acceleration of l m/s2.

Solution:

M = mass of man = 60 kg

m = mass of cradle 25 kg

|F| = force exerted by the man on cradle

= force on man by cradle

Free body diagram for cradle

T-F-mg = ma

Mass T = (a+g)(M+m)/2       T = (10.8×85)/2 N

Example 5:

Two masses m1 and m2 are placed touching each other on an inclined plane forming an angle θ with the horizontal the coefficient of friction between the inclined plane and m1 is μ1 and that of between plane and m2 is μ2 where μ1 > μ2.

(a)    Find the interaction force between the two masses.

(b)    What is the minimum angle θ at which masses start sliding.

Solution:

(a)    Let R be the interaction force and F1 and F2 be the frictional force on m1 and m2, respectively.

m1a = m1g sin θ + R - μ1N1            m2a = m2g sinθ - R - μ2N2

N1 - m1gcosθ = 0                        N2 - N2 - μ2 cosθ = 0

Solving the above equations R = m1m2/(m1 + m2) (μ1-μ2)g cosθ

(b)    The two masses begin to slide just when the acceleration of m1 is not zero.

The limiting case is given by

m1g sinθmin + R - μ1m1g cosθmin = 0

Using R from part (a)

tanθmin = (m1u1 + m2u2)/m1+m2

Example 6:

A small block of mass m is placed on a wedge lying on a horizontal table. The wedge has mass M and face of the wedge is inclined at angle θ with the horizontal. Neglecting any friction, find the acceleration of two blocks.

Solution:

This Problem can be done in two ways.

Method 1.

Take the free body diagram of two masses.

From this ax = g sin θ                                           ...... (1)

and, m ay = mg cos θ - N                                      ...... (2)

From this Nsinθ = ma                                            ...... (3)

The Mass m is in contact with M

Hence ay = a sin θ                                               ...... (4)

From eqn (1) (2) (3) & (4)

Ax = g sin θ ay = m cos θ . sin2θ/(msin2θ + M)

& a = mgcosθsinθ/(msin2θ+M)

Example 7:

A mass m1 is placed at an inclined plane which makes a angle θ with the horizontal. A string is attached to the mass. The other end of the string supports mass m2 hanging freely. The string passes over a frictionless pulley. Calculate acceleration of blocks.

Solution:

Three forces acting on the m1 are the force of gravity normal reaction by the inclined plane and the tension in the string. The free body diagram for this mass is shown in figure.

Taking the x-axis along incline and y-axis perpendicular it, we have for the x and y components.

x component              m1 ax = T - m1g sinθ                    (I)

y component              0 = N - m1g cos θ                         (II)

The only forces acting on m2 are the force of gravity and tension in the string. Free body diagram for mass m2 is shown in the figure. Taking vertically downward to be the positive y direction we get

m2 a = m2 g - T                                                           (III)

(Note : If we choose positive y to be vertically upwards then the acceleration of m2 will be opposite of m1)

Solving I and III

a = g(m2 - m1sinθ)/(m1+m3)

If m2 < m1 sin θ then m1 will slide downward.

Example 8:

Masses M and m are tied together by a string that goes over three pulleys as shown in figure. If m = 3kg, M = 10kg an acceleration of M is 1m/s2, then what is the frictional force on M.

Solution:

Free body diagram for m and M are

If a is the acceleration of M, then acceleration of m is a/2.

Equations for m and M are

T - F = Ma and m a/2 = mg - 2T

Ff = (2mg-4Ma-ma)/4

Example 9:

Two blocks of 2kg each are placed as shown on smooth, frictionless surface. The coefficient of friction between the two blocks is 0.2. A force of 5N acts on the upper blocks. What is the acceleration of two blocks? Take g = 10 m/s2.

Solution:

There are only two possibilities

1.     The upper block moves relative to the lower block in the direction of the applied force.

2.     The two blocks move together

Case I:

Free body for two blocks are

Ff = 2 × g × m = 4N

Block 1:

a1 = (5-Ff)/2 = 0.5 m/s2

a2 = Ff/2 = 2m/s2

This is against the assumption of Case I.

Thus case I is contradictory. It is not allowed.

Case II:

a = a1 = a2                         a = 5/4 = 1.25 m/s2

Example 10:

A balloon of mass M is descending vertically with acceleration a. How much weight must be thrown from the case attached to the balloon so that it begins to ascend vertically with acceleration a?

Assume that the upward lift of the air in the balloon does not change.

Solution:

Take upward direction to be the positive y-direction. If m is the mass thrown out, then the equation of motion for before and after the mass m is thrown out are

Ma = F - Mg

(M-m)a = F - (M-m)g

Solving the two equations m = 2Ma/(a+g).

Example 11:

An elevator consists of the elevator cage M and counterweight m, driving mechanism C and cables and pulleys as shown.

M = 1000 Kg, m = 1400 kg

The elevator is accelerating upward at 2.30 m/s2.

Find (a) T1, (b) T2

and the force exerted by the driving mechanism C on the cable.

Solution:

From the free body diagrams

Ma = T1 - Mg                              ma = mg - T2

(a)    T1 = M (a + g)                    (b)    T2 = m(g - a)

= 12.1 kN                                  = 10.5 kN

Assume the cable is massless. Mass even though it is accelerating no net force acts on it.

This means T2 + C2 = T1

C2 = 1.6 kN towards the counter weight.

Example 13:

In turntable shown in figure, block A has mass 0.9 kg and mass of 1.7 kg and the blocks are 13 cm from the axis of rotation. The coefficient of static friction between all the surface is 0.1. Consider the friction and mass of pulley negligible. Find angular speed of rotation of turntable for which the blocks begin

to slide.

Solution:

If we make free body diagram of B and A and assume that B moves radially outwards while A moves radially inward then, if friction force between A & B is f and between B and surface is f'

For slipping

∑FB = T+f+f' = mBrωw,  ∑FA = T+f+f' = mArωw

Or, 2 f + f' = (mB - mA)rω2

Now, ω increases until f and f' attain their maximum values.

f' = μ(mA + mB)g

f = μmAg.

So, 2μmAg + μ(mA + mB)g = (mB - mA)rω2max

ωmax = [(0.1×9.8(2.7+1.7))/0.13(1.7-0.9) ] = 6.4rad/sec

Example 14:

In the diagram shown, the blocks, A, B and C weigh 3kg, 4kg and 8kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a massless rigid rod fixed to the wall, while B and C are connected by a light flexible cord passing around a fixed frictionless pulley. Find the force F necessary to drag C along the horizontal surface to the left at the constant speed.

Solution:

First we analyze the motion, of A is fixed and C moves to left, B moves to right. If we draw the free body diagram of B,

T = μ (weight of A) + μ (weight of B & A)

= 0.25 × 3 + 0.25 × (2 + 4) - 2.5 kg F.

F = μ (A + B) + T + μ (A + B + C)

= 2.5 + 0.25 (3 + 4) + 0.25 (3 + 4 + 8)

= 2.5 + 1.75 + 3.75            = 8 kg F

= 8 × 9.8 N                        = 78.4 N

Example 15:

Two cubes of masses m1 and m2 lie on two frictionless slopes of the block A which rests on a horizontal table. The cubes are connected by a string which passes over a pulley as shown in the Figure. To what horizontal acceleration 'a' the whole system (the block and the cubes) be subjected so that cubes don't

slide down the planes what is the tension of string?

Solution:

Let us take free body diagram of m2, where (m2a) gives the pseudo acceleration a.

T = m2 g sin β = m2 a cosβ                  (1)

T - m1 g sin α = -m1 a Cosα                 (2)

from (1) and (2)

T = m1m2gsin(α-β)/(m1cosα + m2cosβ)

Example 16:

A man pulls one end of a rigid rod attached to a block of mass 10 kg on a smooth horizontal surface with a force of F1 = 24N horizontally. The rod has a mass of 2 kg. Find the contact force between rod and the block.

Solution:

Since total mass of the system is 12 kg, so acceleration of system as whole

24/12 = 2 m/s2.

Since the system is rigid the rod as well as block will have this acceleration.

Now, if we take free body diagram of rod and block.

Where F1 in the internal force on the contact force between rod and block, and we order block

F1 = 10 × 2 = 20 N

So, contact force is 20 N.

Example 17:

In the system shown in figure, acceleration of cart is a = g/7; such that m2 is in a critical equilibrium of just to go upwards. For ideal pulley and string find on the contact surface of m2 (Also m1 rests on a frictionless surface). Find the co-efficient of friction between m2 and cart.

Solution:

Drawing FBD

Since m2 is about to go up therefore R is acting downwards

For stability of Block m2

m2g + μR = T

Here R = m2a  is the reaction force (refer to illustration of Newton's III law of motion)

For Block m1                       T = m1a.

μR = m1a - m2g

=> μm2(g/7) = m1(g/7) - m2g

m2 = m1 - m2 (7)

μ = m1/m2 - 7 = 0.5(·.· m1/m2 = 7.5)

Example 18:

In a configuration as shown in figure masses of 10, 20, 40 kg are hanging by pulley's all strings and pulleys are ideal. Find the acceleration of the masses.

Solution:

The FBD is shown in figure

a1 = acceleration of 40kg Block on pulley (1)

a2 = acceleration of 10, 20 kg blocks w.r.t.

pulley (2)

Hence the actual acceleration of 10kg and 20kg blocks will be (a2+a1) and (a2-a1) respectively.

Writing down equations with the help of FBD's.

for 40kg. block

40g - T1 = 40a1                           ...... (1)

for 20 kg. block

20g - T2 = 20 (a2 - a1)                 ...... (2)

for 10 kg. block

T2 - 10g = 10(a2 + a1)                  ...... (3)

also for pulley (massless)

T1 = 2T2                                     ...... (4)

from equation (1) & (4)

T1 = 40 (g - a1)

T2 = 20 (g - a1)                           ...... (5)

substituting T2 in (2) & (3)

20a1 = 20(a2 - a1)

a1 = a2/2

substituting T2 in (3)

10g - 20 a1 = 10(a2 + a1)

·.·     a2 = 2a1

=>    10g = 20a1 + 30a1

a1 = g/5 = 2 m/sec2

a2 = 4m/sec2

Example 19:

In the arrangement shown (figure) mass m and M are of small Block and wedge respectively. If wedge angle shown is α, and masses of pulley and thread (inextensible) are negligible; Find the acceleration of the wedge. [plane is frictionless]

Solution:

In the given problem we have two moving blocks M & m. M moves in horizontal directions w.r.t. ground. To observe its motion we can choose a reference frame fixed on the ground, whose X-axis is parallel to the horizontal surface and Y axis is perpendicular i.e. which is convenient in this case and we shall designate it with subscript '1'. Smaller block 'm' is sliding down the inclined plane of 'M', which is itself moving towards right. With the experience of drawing FBD, one intuitively picks up a titled coordinate axis whose x-axis is parallel to the inclined plane and Y-axis is perpendicular to it. Now, confusion arises where can we fix this coordinate axis? Our choices are.

(i) 'm' (ii) 'M' and (iii) ground.

We can logically eliminate first two choices. Suppose 'm' is chosen then obviously we will not be able to see the motion of smaller block as the frame is fixed on it. So we cannot fix axis on it. Suppose we try the bigger block 'M' on which 'm' is sliding down. The choice is somewhat wiser but in the given situation we cannot ascertain that the block 'M' is not accelerating and therefore it may become a non inertial frame of references which is not the desired frame for analysis.

Now, at last we have no other convenient choice but the ground. On it we can fix the titled axis and proceed to solve the problem. We shall designate this coordinate system (reference frame) by subscript '2'.

1.     This is natural and we deignate it with subscript 1

2.     This also we fix on ground but keep x-axis parallel to the inclined plane of wedge and y axis | to wedge's inclined plane, we designate it with subscript 2.

Suppose a1 = acceleration of wedge w.r.t. ground in 1st coordinate system

a2 = acceleration of block m w.r.t. wedge in 2nd coordinate system

Since thread is inextensible the distance traveled downwards by block m on inclined surface is equal to distance traveled by wedge towards right. Therefore magnitudes of acceleration are equal.

a1 = a2 = a

acceleration 'm' w.r.t. ground is as per relative concept given by vector aN, figure.

= a√2(1-cosα)

aN is the net (resultant) acceleration. We can put it in 2nd coordinate system and resolve it as shown in figure.

Here notice that x-axis is parallel to the inclined plane and therefore the angle of aNwith it is same as what it makes with a2.

aN2x = a√2(1-cosα) cos(90 - α/2)

= a√2(1-cosα) sin(α/2) = a√2(1-cosα) √((1-cosα)/2)

= a(1 - cos α)

Similarly aN2Y = a√2(1-cosα) sin(90 - α/2)

aN2Y = a sin α

Let's draw FBS's as shown in figure and notice that we have marked aN2X and aN2Yacceleration of mass 'm' w.r.t. ground in the coordinate system 2.

Now, let's write down equation of motion for wedge M,

T - T cos α + N sin α = Ma1 = Ma                                    ...... (1)

For block m

mg sin α - T = maN2X = ma (1 - cosα)                              ...... (2)

mg cos - N = maN2Y = ma sin α                                      ...... (3)

Solving them simultaneously, we get

a = mgsinα/M+2m(1-cosα)

which is the acceleration of the wedge.