Frame of Reference

A system with respect to which position or motion of a particle is described is known as frame of reference. We can classify frame of reference in two categories :

Inertial Frame of Reference

The frame for which the law of inertia is applicable is known as the inertial frame of reference.

All the frames which are at rest or moving uniformly with respect to an inertial frame, are also inertial frames.

Non-Inertial Frame of Reference

The frame for which the law of inertia is not applicable is known as the non-inertial frame of reference.

All the frames which are accelerating or rotating with respect to an inertial frame will be non inertial frames.

1.0Pseudo Force 

We use the concept of pseudo force to convert non-inertial frames into inertial frames. This force always works in the direction opposite to that of acceleration of frame and its magnitude is equal to the product of mass of body and the acceleration of the non-inertial frame.

F=-ma

Pseudo force does not follow action-reaction law.

Examples :

2.0Normal Reaction 

When a stationary body is placed on a surface then, that surface exerts a contact force on the stationary body which is perpendicular to the surface and towards the body. This force is known as 'Normal Reaction'

● Like every force, a normal force is one half of an action-reaction pair, so it is often called a ‘Normal reaction’.

● A block of mass "m" is placed on a horizontal table. Then the forces exerted on it are :

(1) ↓ Gravitational force of attraction on the body due to earth, means weight of the block (= mg). 

(2) ↑ Normal reaction exerted by the surface of the table on the block.

Since the body is in equilibrium, therefore net force on it is zero.

⇒ N+W=0 ⇒ N=-W

N = W [N= W |N|=|-W|]

∴ N=mg [W=mg]

● A body is placed on a smooth inclined plane which makes an angle θ with the horizontal. Then the forces acting on body are 

(1) Weight of the body mg acting vertically downward.

(2) Normal reaction N exerted by the surface of the plane on the block.

The weight mg of the body is resolved into two components, parallel and perpendicular to the plane as mgsinθ parallel to the plane and mgcosθ perpendicular to the plane.

Normal reaction 

N= mgcosθ  

Net force F = mgsinθ  

ma = mgsinθ 

∴ a=g sinθ    

3.0Free Body Diagram

A system diagram is a sketch of all the objects involved in a situation. A free-body diagram (FBD) is a drawing in which only the object being analysed is drawn, with arrows showing all the forces acting on the object. 

(1) Free body diagrams represent all forces acting on one object. 

(2) Forces that the object exerts on other objects do not appear in free body diagrams because they have no effect on the motion of the object itself. 

(3) In drawing a free body diagram, you can represent the object as a single dot or a simplified shape the object. 

(4) In FBD each force acting on the object is represented with an arrow. The arrow’s direction shows the direction of the force and the arrow’s relative length provides information about the magnitude of the force.

(5) Forces that have the same magnitude should be sketched with approximately the same length, forces that are larger should be longer, and smaller forces should be shorter.

(6) In case of objects in motion, the direction of acceleration should be made on the FBD in the direction of greater force (or net force).

Effective or Apparant weight of a man in lift

Case-I : If the lift is at rest or moving uniformly (a = 0), then 

N = mg

W_app =N

W_actual  =mg

So, W_app = W_actual  

Case-II : If the lift is accelerated upwards, then -

Net upward force on man = ma

N – mg = ma

N = mg + ma

∴ N = m (g +a)

W_app or N = m(g+a)

So, W_app > W_actual

Case-III : If the lift is retarding upwards, then

N – mg = m (–a)

N = mg – ma 

∴ W_app  or N = m (g-a)

So, W_app < W_actual

Case-IV : If the lift is accelerated downwards, then -      

mg – N = ma 

N = mg – ma  

W_app  or N = m (g-a)

So, W_app < W_actual

Case-V : If the lift is retarding downwards, then -

mg – N = m (–a)

mg – N = – ma  

N = ma + mg 

W_app  or N = m (g + a)

So, W_app > W_actual

Two Special Cases of downward acceleration (IV Case) 

I^st Special Case : If the lift is falling freely, means its acceleration in the vertically downward direction is equal to acceleration due to gravity. i.e. a = g, then -

W_app. = m (g–a) = m (g–g) = m × 0 = 0

∴ W_app. = 0

Means the man will feel weightless. This condition is known as the Condition of Weightlessness.

The apparent weight of any freely falling body is always zero (0).

 II^nd Special Case : If the lift is accelerating downwards with an acceleration which is greater than 'g', then the man will move up with respect to lift and he will stick to the ceiling.

Motion of bodies in contact (contact force) 

Two bodies in contact :

F – N = m₁ a .....(1)   

N = m₂ a .....(2)

On adding above equation

Free body diagrams :-

F = m₁a + m₂a   

F = a (m₁+m₂)     

⇒ $a=\frac{F}{m_1+m_2}$ ....(3)

$a=\frac{F_{net}}{m_{total}}$

putting value of 'a' from equation (3) into (2), we get;

$N=\frac{m_{2}F}{m_1+m_2}$

(here f = contact force)

Three bodies in contact :

F – N₁ = m₁ a ....(1)               

N₁ – N₂ = m₂ a ....(2)

N₂  = m₃ a ....(3)

On adding above equations

F = m₁a + m₂a + m₃a

F = a (m₁+m₂+m₃)

$a=\frac{F}{m_1+m_2+m_3}$ ....(4)

Putting value of 'a' from equation (4) into (3), we get

$N_2=\frac{F{m}_3}{m_1+m_2+m_3}$

Putting value of 'a' from (4) to (1);

F – m₁a = N₁

$N_1=F-\frac{F{m}_1}{m_1+m_2+m_3}\Rightarrow N_1=\frac{(m_2+m_3)F}{m_1+m_2+m_3}$

4.0System of Masses Tied By Strings

Tension in a String

It is an intermolecular force between the atoms of a string, which acts or reacts when the string is stretched.

Important points about the tension in a string:   

(a) Force of tension act on a body in the direction away from the point of contact or tied ends of the string.

(b) String is assumed to be inextensible so that the magnitude of accelerations of any number of masses connected through strings is always same.

(c) String is massless and frictionless so that tension throughout the string remains same.

(d) If a force is directly applied on a string as say man is pulling a tied string from the other end with some force the tension will be equal to the applied force irrespective of the motion of the pulling agent, irrespective of whether the box will move or not, man will move or not. 

(e) String is assumed to be massless unless stated, hence tension in it every where remains the same and equal to applied force. 

However, if a string has a mass, tension at different points will be different, being maximum (= applied force) at the end through which force is applied and minimum at the other end connected to a body.

(f) Every string can bear a maximum tension, i.e. if the tension in a string is continuously increased it will break if the tension is increased beyond a certain limit. The maximum tension which a string can bear without breaking is called "breaking strength". It is finite for a string and depends on its material and dimensions.

5.0Motion of Connected Bodies

Two Connected Bodies :

F - T = m₁ a ....(1)

T = m₂ a ....(2)

Free body diagram :-

On adding above equations

F = m₁a + m₂a 

F = a (m₁+m₂) 

$a=\frac{F}{m_1+m_2}$ ...(3)

$a=\frac{F_{net}}{M_{net}}$

Putting value of 'a' from equation (3) into (2), we get -

$T=\frac{m_{2}F}{m_1+m_2}$

Three Connected Bodies :

F – T₁ = m₁a .....(1)

T₁ – T₂ = m₂a .....(2)

T₂ = m₃a .....(3)

On adding above equation 

F =  (m₁ + m₂ + m₃) a

∴ $a=\frac{F}{m_1+m_2+m_3}or\frac{F_{\text{net}}}{m_{\text{total}}}$

Put the value of 'a' in equation (1) 

T₁ = F – m₁a

$T_1=F-\frac{m_{1}F}{m_1+m_2+m_3}$

$T_1=\frac{m_{1}F+m_{2}F+m_{3}F-m_{1}F}{m_1+m_2+m_3}$

$T_1=\frac{(m_2+m_3)F}{m_1+m_2+m_3}$

Now, put the value of 'a' in equation (3)

$T_2=\frac{m_{3}F}{m_1+m_2+m_3}$

Bodies Hanged Vertically 

Since the bodies are in equilibrium, therefore net force on all the bodies is zero

T₁ = m₁g ....(1)         

T₂ = T₁ + m₂g

T₂ = m₁g + m₂g            

[From equation (1)]  

T₂ = (m₁ + m₂)g ....(2)

T₃ = T₂ + m₃g

T₃ = (m₁ + m₂)g + m₃g          

[From equation (2)]

T₃ = (m₁ + m₂ + m₃)g

Bodies Accelerated Vertically Upwards 

T₁ – m₁g = m₁a          

T₁ = m₁g + m₁a

T₁ = m₁ (g + a) ....(1)

T₂ – m₂g – T₁ = m₂a

T₂ = m₂a + m₂g + T₁

T₂ = m₂ (g + a) + m₁ (g + a)   

[from equation (1)]

T₂ = (m₁ + m₂) (g + a) ....(2)

T₃ – m₃g – T₂ = m₃a

T₃ = T₂ + m₃g + m₃a

T₃ = (m₁ + m₂) (g + a) + m₃ (g + a) [from equation (2)]

T₃ = (m₁ + m₂ + m₃) (g + a) ....(3)

Bodies Accelerated Vertically Downwards

m₁g – T₁ = m₁a

T₁ = m₁g – m₁a 

T₁ = m₁ (g – a) ....(1)

m₂g + T₁ – T₂ = m₂a

T₂ = m₂g – m₂a + T₁

T₂ = m₂ (g – a) + m₁ (g – a)

[from equation (1)]

T₂ = (m₁ + m₂) (g – a) ....(2)

m₃g + T₂ – T₃ = m₃a

T₃ = m₃g – m₃a + T₂

T₃ = m₃ (g – a) + (m₁ + m₂) (g – a)   

[from equation (2)]

T₃ = (m₁ + m₂ + m₃) (g – a) ....(3)

6.0Solved Examples

1. Two blocks of mass m = 2 kg and M = 4 kg are in contact on a frictionless table. A horizontal force (F = 24 N) is applied to m. Find the force of contact between the block, will the force of contact remain same if F is applied to M ?

Solution

As the blocks are rigid under the action of a force F, both will move with same acceleration

$a=\frac{F}{m+M}=\frac{24}{2+4}=4m/s^2$

force of contact   

R_M = Ma = 4 × 4 = 16 N

If the force is applied to M then its action on m will be R_m = ma = 2 × 4 = 8 N.

From this problem it is clear that acceleration does not depends on the fact that whether the force is applied to m or M, but force of contact does.

2. Three blocks of masses m₁ = 2 kg, m₂ = 3 kg and m₃ = 4 kg are in contact with each other on a frictionless surface as shown in fig.

Find (a) horizontal force F needed to push the block as one unit with an acceleration of 3 m/s² 

(b) The resultant force on each block and 

(c) The magnitude of contact force between blocks.

Solution

(a) F = (m₁ + m₂ + m₃) a        

= (2 + 3 + 4) × 3 = 9 × 3 = 27 N      

(b) for m₁, F – R₁ = m₁a = 2 × 3  

⇒  F – R₁ = 6 ......(i)

for m₂, R₁ – R₂ = m₂ a = 3 × 3 = 9 ⇒ R₁ – R₂ = 9 N ......(ii)

for m₃, R₂ = m₃a = 4 × 3 ⇒ R₂ = 12 N ......(iii)

(c) Contact force between m₂ and m₃ = R₂ = 12 N

Contact force between m₁ and m₂ = R₁ = R₂ + 6 = 12 + 9 = 21N

3. (a) How much horizontal acceleration should be provided to the wedge so that the block of mass m on wedge remains at rest or at equilibrium?

(b) Find the force exerted by wedge on the block?

Solution

(a) For equilibrium

macosθ = mgsinθ

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

$a=gtan\theta$

(b) Normal reaction

N = masinθ + mgcosθ    

⇒  N = mg $\frac{si{n}^2\theta }{cos\theta }+mgcos\theta$

$N=mg\left[\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta }\right]$

⇒  $N=\frac{mg}{\cos \theta }$

3. A spring weighing machine inside a stationary lift reads 50 kg when a man stands on it. What would happen to the scale reading if the lift is moving upward with (i) constant velocity, and (ii) constant acceleration?

Solution

(i) In the case of constant velocity of lift, there is no pseudo force; therefore the apparent weight = actual weight. Hence the reading of machine is 50 kgwt.

(ii) In this case the acceleration is upward, the pseudo force R = ma acts downward, therefore apparent weight is more than actual weight i.e. 

W´ = W + R = m (g + a). 

Hence scale shows a reading 

$=m(g+a)=mg\left(1+\frac{a}{g}\right)$

$N=50\left(1+\frac{a}{g}\right)\text{kg wt.}$

7.0Also Read