For the system shown in Fig 848 m1 80 kg m2 30 kg theta 3

For the system shown in Fig 8.48, m_1 = 8.0 kg, m_2 = 3.0 kg, theta = 30 degree, and the radius and mass of the pulley are 0.10 m and 0.10 kg, respectively. (a)What is the acceleration of the masses? (neglect friction and the string\'s mass.) (b) If the pulley has a constant frictional torque of 0.050 m N when the system is in motion, What is the acceleration of the masses?

Solution

a) Apply Newton’s second law

T₂ - m₂ g = m₂ a

T₁ R - T₂ R = I alpha = 1 / 2 MR ² alpha = 1 / 2 M Ra

T₁ - T₂ = 0.5 Ma

m₁ g sin theta - T₁ = m₁ a

adding all equations

a = (m₁ sin theta - m₂) g / (m₁ + m₂ + 0.5 M)

= (8 sin 30 - 3) 9.8 / (8 + 3 + (0.5*0.1))

a = 0.89 m/s²

--------------------------------------------------

T₁ R - T₂ R - T_f = I alpha = 1 / 2 MR ² alpha = 1/2 MRa

a = (m₁ sin theta - m₂) g - T_f/R / (m₁ + m₂ + 0.5 M)

= [(8 sin 30 - 3) 9.8 - 0.05/0.1 ] / (8 + 3 + (0.5*0.1))

a = 0.84 m/s²