Constraint Relations

PhysicsLaws of MotionFor JEE aspirants

Pulley Mass System

(i) When unequal masses m₁ and m₂ are suspended from a pulley (m₁ > m₂)

m₁g – T = m₁a,

and T – m₂g = m₂a

On solving equations, we get

$a = \frac{( m _{1} - m _{2} )}{( m _{1} + m _{2} )} g,$

$T = \frac{2 m _{1} m _{2}}{( m _{1} + m _{2} )} g$

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(ii) When a body of mass m₂ is placed on a frictionless horizontal surface, then

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Acceleration $a = \frac{m _{1} g}{( m _{1} + m _{2} )}$

Tension in string $T = \frac{m _{1} m _{2} g}{( m _{1} + m _{2} )}$

(iii) When a body of mass m₂ is placed on a rough horizontal surface, then

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Acceleration $a = \frac{( m _{1} - μ m _{2} ) g}{( m _{1} + m _{2} )}$

Tension $T = \frac{m _{1} m _{2} ( 1 + μ ) g}{( m _{1} + m _{2} )}$

(iv) When two masses m₁ and m₂ and connected to a single mas M as shown in figure, then

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m₁g – T₁ = m₁a ….(i)

T₂ –m₂g = m₂a ….(ii)

T₁ – T₂= Ma …...(iii)

Acceleration $a = \frac{( m _{1} - m _{2} ) g}{( m _{1} + m _{2} + M )}$

Tension $T_{1} = (\frac{2 m _{2} + M}{m _{1} + m _{2} + M}) m_{1} g$

$T_{2} = (\frac{2 m _{1} + M}{m _{1} + m _{2} + M}) m_{2} g$

(v) Motion on a smooth inclined plane, then

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m₁g – T = m₁a …(i)

$T - m_{2} g sin θ = m_{2} a$ ….(ii)

Acceleration $a = (\frac{m _{1} - m _{2} sin θ}{m _{1} + m _{2}}) g$

Tension $T = \frac{m _{1} m _{2} ( 1 + sin θ ) g}{( m _{1} + m _{2} )}$

(vi) Motion of two bodies placed on two inclined planes having different angle of inclination, then

Acceleration $a = \frac{m _{1} sin θ _{1} - m _{2} sin θ _{2}}{m _{1} + m _{2}}$

Tension $T = \frac{m _{1} m _{2}}{m _{1} + m _{2}} (sin θ_{1} + sin θ_{2}) g$

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