Pulley

Ideal pulley is considered weightless and frictionless.     

Ideal string is massless and inextensible.                    

The pulley may change the direction of force in the string but not the tension.

The only function of pulley (Which has no friction on its axle to retard rotation) is to change the direction of the cord that joins the two block

1.0Cases of Pulley 

Case I

m₁ = m₂ = m

Tension in the string, 

T = mg

Acceleration 'a' = zero

Reaction at the suspension of the pulley

R = 2T = 2 mg

Case II

m₁ > m₂

now for mass m₁, 

m₁ g – T = m₁a .........(i)

for mass m₂

T – m₂ g = m₂ a .........(ii)

Adding (i) and (ii), we get

$a=\frac{(m_1-m_2)}{(m_1+m_2)}g\text{and}T=\frac{2m_1{m}_2}{(m_1+m_2)}g$

$\text{acceleration}=\frac{\text{net pulling force}}{\text{total mass to be pulled}}$

$\text{Tension}=\frac{2\times \text{Product of masses}}{\text{Sum of two masses}}g$

Reaction at the suspension of pulley  R = $2T=\frac{4m_1{m}_{2}g}{(m_1+m_2)}$

Case III

For mass m₁ : T = m₁ a        .........(i) 

For mass m₂ : m₂g – T = m₂ a . ........(ii)

Solving (i) and (ii), we get

$a=\frac{m_{2}g}{(m_1+m_2)}\text{and}T=\frac{m_1{m}_2}{(m_1+m_2)}g$

Case IV

(m₁ > m₂) 

m₁g – T₁ = m₁a .........(i)            

T₂ – m₂g = m₂a .........(ii)

T₁ – T₂ = Ma .........(iii)

by (i), (ii) and (iii)

$a=\frac{(m_1-m_2)}{(m_1+m_2+M)}g$

Case V

Mass suspended over a pulley from another mass on an inclined plane.

For mass m₁ : m₁g – T = m₁ a  

For mass m₂ : T – m₂g sinθ = m₂ a

acceleration

$a=\frac{(m_1-m_2\sin \theta )}{(m_1+m_2)}g$

and

$T=\frac{m_1{m}_2(1+\sin \theta )}{(m_1+m_2)}g$

Case VI

For mass m₁ : T₁ – m₁g = m₁a                  

For mass m₂ : m₂g + T₂ – T₁ = m₂a

For mass m₃ : m₃g – T₂ = m₃a

$a=\frac{(m_2+m_3-m_1)}{(m_1+m_2+m_3)}g$

we can calculate tensions T₁ and T₂ from above equations.

2.0Also Read