IIT JEE Main / Advanced Physics 2018 Tips & Trends – Mechanics > ‘Pulley systems’

We are starting a new post series highlighting specific sections selected from the last 30 years of iit jee physics exam papers. Here , we will be looking at typical questions frequently asked by the exam board , and will be elucidating and explaining the broad theoretical principles behind them , some key questions asked , sample problems and easy techniques to solve them.

Starting the series is the ‘ pulley ‘ system related problems from mechanics..

The Atwoods machine – It represents one of the basic pulley systems studied in theoretical physics. It was invented in 1784 by George Atwood as a laboratory experiment to verify the mechanical laws of motion with constant acceleration. It ideally consists of two objects of mass m1 and m2, connected by an inextensible ,  massless string over an ideal massless pulley.   Naturally , when  m1 = m2 , the system is in neutral equilibruim , irrespective of the position of the masses , and when  m1 ≠   m2 ,   both masses undergo uniform acceleration.

Atwoods machine

Now , let us have a look at the equation derivation part….

Points to note

  1. string is inextensible – the length of the connecting chord is constant.
  2. string is massless – it doesn’t interfere with the acceleration of the connected masses.
  3. pulley is frictionless – pulley doesn’t add any factor to the equations.
  4. pulley is massless – it doesn’t interfere with the acceleration of the connected masses
  5. what does the pulley do – it demarcates the forces acting via the connected string – mass combination.
  6. after demarcating the concerned forces – give any one group a + ve sign .
  7. after determining the forces , conceptualize the corresponding tensions involved.
  8. also note that the cartesian co-ordinates ( x,y,z ) are positioned correctly.

forces affecting m1:

\; m_1g-T=m_1a

Forces affecting m2:

\; T-m_2g=m_2a

and adding the two previous equations we obtain

\; m_1g-m_2g=m_1a+m_2a,

and our concluding formula for acceleration

a = g{m_1-m_2 \over m_1+m_2}

Conversely, the acceleration due to gravity, g, can be found by timing the movement of the weights, and calculating a value for the uniform acceleration a:

 d = {1 \over 2} at^2 .         Finding tension – To evaluate tension  the equation for acceleration is substituted in either of the 2 force equations.

a = g{m_1-m_2 \over m_1 + m_2}

For example substituting into m1a = m1gT, we get

T={2 g m_1 m_2 \over m_1 + m_2}

The tension can be found in using this method.

If the pulley has Inertia and Friction

For very small mass differences between m1 and m2, the rotational inertia I of the pulley of radius r cannot be neglected. The angular acceleration of the pulley is given by the no-slip condition:

 \alpha = {a\over r}

In that case, the net torque for the pulley becomes:

\tau_{net}=\left(T_1 - T_2 \right)r - \tau_{friction} = I \alpha

Combining with Newton’s second law for the hanging masses, and solving for T1, T2, and a, we get:


 a = {{g (m_1 - m_2) - {\tau_{friction} \over r}} \over {m_1 + m_2 + {{I} \over {r^2}}}}

Tension in string segment nearest m1:

 T_1 = {{m_1 g (2 m_2 + {{I} \over {r^2}} + {{\tau_{friction}} \over {r g}})} \over {m_1 + m_2 + {{I} \over {r^2}}}}

Tension in string segment nearest m2:

 T_2 = {{m_2 g (2 m_1 + {{I} \over {r^2}} + {{\tau_{friction}} \over {r g}})} \over {m_1 + m_2 + {{I} \over {r^2}}}}

Should bearing friction be negligible (but not the inertia of the pulley and not the traction of the string on the pulley rim), these equations simplify as the following results:


 a = {{g (m_1 - m_2)} \over {m_1 + m_2 + {{I} \over {r^2}}}}

Tension in string segment nearest m1:

 T_1 = {{m_1 g (2 m_2 + {{I} \over {r^2}})} \over {m_1 + m_2 + {{I} \over {r^2}}}}

Tension in string segment nearest m2:

 T_2 = {{m_2 g (2 m_1 + {{I} \over {r^2}})} \over {m_1 + m_2 + {{I} \over {r^2}}}}

Pulley Systems  – Deviations

1.Multiple pulley systems –  Here we deal with multiple pulleys connected to multiple masses. The pulleys / masses may be horizontally / vertically connected, it doesn’t necessarily be hanging. Below are typical cases  –

Finding the acceleration of blocks A and B
Solution : Demarcating the forces,  I1 + I2  = I3+I4+I5+I6 ,
when at balance, both sides have a value of zero.
Suppose block A is moving to the left with a velocity of 8 m/s²,
then ,  -ap – ap + 8 = 0 , hence ap = 4 m/s² , also ap-ab-ab-ab = 0 , hence ab = 4/3 m/s²

Finding the acceleration of block A

Solution : The figure indicates that the block B is moving towards right with acceleration b and we have to find the acceleration of block A . Here, the acceleration [ b ] of B to the right is balanced by the acceleration of A in the horizontal direction . Similarly , the acceleration of A in the vertical direction is balanced by the acceleration of A with respect to b in upwards direction  and hence  a = 4b ( note the tension chords opposing the acceleration of B ) . Therefore , the net acceleration of A  = bˆi  + 4bˆj

2 . Pulley on a wedge

Here , the difference from above case is that forces involved get divided into angular components with respect to the wedge.

Finding the acceleration of the above system

Solution – We infer that the block m2 is heavier and that the block m1 is on a frictionless incline . ( Of course , the string is massless and inextensible ) . The weight of m1 has been split into the components m1gsinβ  parallel to the incline, and m1gcosΒ perpendicular to it.The block m1 accelerates upward .Sum of forces on m1 in the direction of the incline: T-m1gsinβ=m1a
Sum of vertical forces on m2 = m2g-T = m2a
And adding both equations we get m2g-m1gsinβ = a(m1+m2)
Or a = g(m2-m1sinβ)/(m1+m2)

What if there is friction involved and the coefficient of kinetic friction between m1 and the plane is 0.2.  ( the angle β = 30° and  m1 = 2 kg, m2 = 2.5 kg ). Finding the accelerations of m1 and m2 –

The magnitude of the frictional  force is Ff = FN. FN is obtained using:
Sum of forces perpendicular to the incline = FN-m1gcosβ = 0
or FN = m1gcosβ. The friction force is then Ff = m1gcosβ
The sum of forces on m1 in the incline is now
(1) T-m1gsinβ-m1gcosβ = m1a
The sum of vertical forces on m2 is
(2) m2g-T = m2a

Adding (1) and (2) we get m2g-m1g(sinβ+cosβ) = a(m1+m2)
Replacing values, with = 0.2, m1 = 2 kg, m2 = 2.5 kg and β = 30° results in a = 2.51 m/s2 .

Given below is an interesting case , where eventhough two wedges are involved, angular calculations isn’t necessary . (The crucial point is how the forces are distributed )

Find the acceleration of block B

Let the strings be of m1 , m2 and m3 metres in length. Note that for block B , this would be m2 + m2 or 2 m2.

Here, the total length of ths string is m1 + 2 m2 + m3 . Differentiating with respect to time, we get –

O = d²m1/ dt² + 2 d²m2/ dt² + d²m3/dt²  = -aA + 2aB + aC => acceleration of block B = aA – aC/ 2

Find the acceleration of the masses ( coefficient of friction, μ = 0.4 ) –

Solution :Block A causes a normal reaction on the horizontal surface –

Ra = mass of A x g = 4x 9.8N = 39.2 , hence, frictional force –

Fa = μ x Ra = 0.4 x 39.2 = 15.68 N.

Now, as block B is at an angle of 30° , Rb = mass of B x g cos α = 42.43

and Fb = μ x Rb =  16.97 N . Now , the tensions involved –

From the diagram itself it’s understood that Ta is acting towards left , which is opposed by the frictional force of 15.68( acting towards right ). Hence , the resultant force is – Ta – 15.68 N  – eq.1

Let the acceleration of block A be x , then , force on block A = 4x ,

Therefore ,  Ta = 4x + 15.68 . Let us shift our attention to block B – here,the frictional force opposing the movement of block B is acted against / opposed by Tb and mB g sin α . Therefore –

resultant force = Tb + mB sin 30° – 16.97   =  Tb + 7.53

Now , this resultant force is balanced by that due to acceleration of block B , hence,

Tb + 7.53 = 5x  , hence  Tb = 5x – 7.53.

Considering block C , it’s acceleration downwards due to gravity is opposed both by Ta and Tb , therefore ,

Rc = mass of c xg – ( Ta + Tb ) = 147 – ( Ta + Tb ) . Now , the acceleration of block C due to the system = 15x , therefore,

15x = 147 – ( Ta + Tb ) , from this x or acceleration of the system can be found out and it’s  x = 5.8 m/s².

3. Pulley + Spring combinations –

In the third group, our review deals with systems where spring(s) are added to pulleys.

In the above figured system , the pulley and string are massless,and no friction is present. The system is released from rest. Find the speed of 10 kg block as the 2kg leaves the ground.( spring constant,k= 40 N/m and take g = 10 m/s² ) . Solution :

As the 2 kg block leaves the ground , the spring gets extended , say , by x  metres and the spring experiences a tension which is equal to the weight of the 2 kg block. Therefore –

kx = 2g , and x = 2×10/40 = 1/2 m . Based on the conservation of ( mechanical) energy ,

mgx = 1/2 kx² + 1/2 mv² , hence v = √ 2gx – kx²/m  = 3 m/s

In the above system, the pulley and the string are massless ( of course!) and a block A of mass m rests on an incline with coefficient of friction μ .The angle of inclination is θ. As the block is released from rest to slide downwards, the spring gets extended.Calculate the speed of the block as it reaches the end of the wedge.

Solution :  It can be easily deduced that the net result depends on the law of conservation of energy. Accordingly,it can be seen that the potential energy of the block due to it’s height h is balanced by the other forces acting as the block slides down. Hence,

mgh = 1/2 mv² + 1/2 k [ h/sinθ]² + μmgcosθ[h/sinθ]. Therefore –

v = √ 2/m [mgh – 1/2k(h/sinθ)²- μmghcosθ

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