Answer:147 N-m
Explanation:
Given
mass of first Pulley [tex]m_1=300 kg[/tex]
mass of second Pulley [tex]m_2=450 kg[/tex]
radius of Pulley [tex]r=10 cm[/tex]
If the system is in equilibrium then tension in the [tex]m_1[/tex] side mass is
[tex]T_1=m_1g[/tex]
Tension in [tex]m_2[/tex] side block
[tex]T_2=m_2g[/tex]
Net torque is
[tex]\tau =(T_2-T_1)\times r[/tex]
[tex]\tau =(m_2-m_1)g\times r[/tex]
[tex]\tau =(450-300)\times 9.8\times 0.1[/tex]
[tex]\tau =147 N-m[/tex]
therefore Frictional Force must apply a force of 147 N-m in order to system remains in equilibrium