To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that
F = ma
Where,
m = mass
a = Acceleration (Gravitational acceleration when there is action over the object of the earth)
Torque, as we know, is the force applied at a certain distance, that is,
[tex]\tau = F*d[/tex]
Where
F= Force
d = Distance
Our values are given as,
[tex]m_1 = 0.32Kg[/tex]
[tex]m_2 = 0.4Kg[/tex]
[tex]d = 8.7*10^{-2}m[/tex]
Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say
[tex]\tau = T_2-T_1[/tex]
[tex]\tau = F_2*d-F_1*d[/tex]
[tex]\tau = m_2g*d-m_1*g*d[/tex]
[tex]\tau = (m_2-m_1)g*d[/tex]
[tex]\tau = (0.4-0.32)(9.8)(8.7*10^{-2})[/tex]
[tex]\tau = 0.068N\cdot m[/tex]
Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is [tex]\tau = 0.068N\cdot m[/tex]
