Three masses m₁ = 4 kg, m₂ = 4 kg and m₃ = 6 kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of T₁/T₂ is _____. (take g = 10 m/s ^2 )

Step-by-step solution

Let the acceleration of the system be a . The total mass on the right side is m₂ + m₃ = 4 + 6 = 10 kg, and the mass on the left side is m₁ = 4 kg. Since 10 kg > 4 kg, the right side accelerates downwards and the left side accelerates upwards. The common acceleration a of the system is given by: a = (m₂ + m₃) - m₁ m₁ + m₂ + m₃ g a = 10 - 4 4 + 4 + 6 g = 6 14 g = 3 7 g For mass m₁ , the equation of motion is: T₁ - m₁ g = m₁ a T₁ = m₁(g + a) = 4 (g + 3 7 g ) = 4 ( 10 7 g ) = 40 7 g For mass m₃ , the equation of motion